Category: Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.1

10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Answer:
Given that
△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.
Among the given lengths CA is longest.
Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.
i.e., ∠B = 90°.
With reference to ∠A, we have opposite side = BC = 15 cm
adjacent side = AB = 8 cm
and hypotenuse = AC = 17

sin A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{BC}{AC}\) = \(\frac{15}{17}\)
cos A = \(\frac{\text { Adjacent side of } \angle \mathrm{A}}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{8}{17}\)
tan A = \(\frac{\text { Opposite side of } \angle \mathrm{A}}{\text { Adjacent side of } \angle \mathrm{A}}\) = \(\frac{BC}{AB}\) = \(\frac{15}{8}\)
∴ sin A = \(\frac{15}{17}\);
cos A = \(\frac{8}{17}\)
tan A = \(\frac{15}{8}\)

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.
Answer:

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.
By Pythagoras theorem QR² = PQ² + PR²
(25)² = (7)² + PR²
PR² = (25)² – (7)² = 625 – 49 = 576
PR = √576 = 24 cm
tan Q = \(\frac{PR}{PQ}\) = \(\frac{24}{7}\);
tan R = \(\frac{PQ}{PR}\) = \(\frac{7}{24}\)
∴ tan Q – tan R = \(\frac{24}{7}\) – \(\frac{7}{24}\)
= \(\frac{(24)^{2}-(7)^{2}}{168}\)
= \(\frac{576-49}{168}\)
= \(\frac{527}{168}\)

Question 3.
In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Answer:
Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem
AC² = AB² + BC²
(25)² = AB² + (24)²
AB² = 25² – 24² = 625 – 576
AB² = 49
AB = √49 = 1
With reference to ∠BAC = θ, we have
Opposite side to θ = BC = 24 units.
Adjacent side to θ = AB = 7 units.
Hypotenuse = AC = 25 units.
Now
cos θ = \(\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}\) = \(\frac{AB}{AC}\) = \(\frac{7}{25}\)
tan θ = \(\frac{\text { Opposite side of } \theta}{\text { Adjacent side of } \theta}\) = \(\frac{BC}{AB}\) = \(\frac{24}{7}\)
Hence cos θ = \(\frac{7}{25}\) and tan θ = \(\frac{24}{7}\)

Question 4.
If cos A = \(\frac{12}{13}\), then find sin A and tan A.
Answer:
From the identity
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A
= 1 – \(\left(\frac{12}{13}\right)^{2}\)
= 1 – \(\frac{144}{169}\)
= \(\frac{169-144}{169}\)
= \(\frac{25}{169}\)
∴ sin A = \(\sqrt{\frac{25}{169}}\) = \(\frac{5}{13}\)

∴ sin A = \(\frac{5}{13}\); tan A = \(\frac{5}{12}\)

Question 5.
If 3 tan A = 4, then find sin A and cos A.
Answer:
Given 3 tan A = 4
⇒ tan A = \(\frac{4}{3}\)
From the identify sec² A – tan² A = 1
⇒ 1 + tan² A = sec² A

If cos A = \(\frac{3}{5}\) then from
sin² A + cos² A = 1
We can write sin²A = 1 – cos²A
= 1 – \(\left(\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}\)
⇒ sin² A = \(\frac{16}{25}\)
⇒ sin A = \(\frac{4}{5}\)
∴ sin A = \(\frac{4}{5}\); cos A = \(\frac{3}{5}\)

Question 6.
In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.
Answer:

In the given triangle, cos A = cos X
⇒ \(\frac{AC}{AX}\) = \(\frac{XC}{AX}\)
⇒ AC = XC
⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = \(\frac{7}{8}\), then evaluate

Answer:

cot² θ = (cot θ)²
= \(\left(\frac{7}{8}\right)^{2}\) = \(\frac{49}{64}\) …… (1)

= sec θ + tan θ
So cot θ = \(\frac{7}{8}\)
⇒ tan θ = \(\frac{8}{7}\)
⇒ tan² θ = \(\left(\frac{8}{7}\right)^{2}\) = \(\frac{64}{49}\)
From sec² θ – tan² θ = 1
⇒ 1 + tan² θ = sec² θ

Question 8.
In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Answer:
Given, tan A = \(\frac{\sqrt{3}}{1}\)
Hence \(\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{\sqrt{3}}{1}\)
Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,
AC² = AB² + BC²
(By Pythagoras theorem)
⇒ AC² = (1k)² + (√3k)²
⇒ AC² = 1k² + 3k²
⇒ AC² = 4k²
∴ AC = \(\sqrt{4 k^{2}}\) = 2k

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.2

10th Class Maths 13th Lesson Probability Ex 13.2 Textbook Questions and Answers

Question 1.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ? (ii) not red?
Answer:
i) Total number of balls in the bag = 3 red + 5 black = 8 balls.
Number of total outcomes when a ball is drawn at random = 3 + 5 = 8
Now, number of favourable outcomes of red ball = 3.
∴ Probability of getting a red ball = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{3}{8}\)
ii) If P( E) is the probability of drawing no red ball, then P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – P(E)= 1 – \(\frac{3}{8}\) = \(\frac{5}{8}\)

Question 2.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Answer:
Total number of marbles in the box = 5 red + 8 white + 4 green = 5 + 8 + 4= 17
Number of total outcomes in drawing a marble at random from the box =17.
i) Number of red marbles = 5
Number of favourable outcomes in drawing a red ball = 5
∴ Probability of getting a red ball P(R) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
P(R) = \(\frac{5}{17}\)

ii) Number of white marbles = 8
Number of favourable outcomes in drawing a white marble = 8
∴ Probability of getting a white marble
P(W) = \(\frac{8}{17}\)

iii) Number of ‘non-green’ marbles = 5 red + 8 white = 5 + 8 = 13
Number of outcomes favourable to drawing a non-green marble =13.
∴ Probability of getting a non- green marble
P(non – green) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
P(non – green) = \(\frac{13}{17}\)
Probability of getting a green ball = \(\frac{\text { No. of green balls }}{\text { Total no. of balls }}\) = \(\frac{4}{17}\)
Now P(G) = \(\frac{4}{17}\)
and P(G) + P(\(\overline{\mathrm{ G}}\)) = 1
∴ P(\(\overline{\mathrm{G}}\)) = 1 – P(G)
= 1 – \(\frac{4}{17}\)
= \(\frac{13}{17}\)

Question 3.
A Kiddy bank contains hundred 50p coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
Answer:
i) Number of 50 p coins = 100
Number of Rs. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
Total number of coins = 180
Number of total outcomes for a coin to fall down = 180.
Number of outcomes favourable to 50 p coins to fall down = 100.
∴ Probability of a 50 p coin to fall down = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{100}{180}\) = \(\frac{5}{9}\)

ii) Let P(E) be the probability for a Rs. 5 coin to fall down.
Number of outcomes favourable to Rs. 5 coin = 10.
∴ Probability for a Rs. 5 coin to fall down = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{10}{180}\) = \(\frac{1}{18}\)
Then P(\(\overline{\mathrm{E}}\)) is the probability of a coin which fall down is not a Rs. 5 coin.
Again P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\))= l-P(E)
= 1 – \(\frac{1}{18}\)
= \(\frac{17}{18}\).

Question 4.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (See figure). What is the probability that the fish taken out is a male fish?
Answer:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 m + 8 f
= 13 fishes.
∴ Number of total outcomes in taking a fish at random from the aquarium =13.
Number of male fish = 5
∴ Number of outcomes favourable to male fish = 5.
∴ The probability of taking a male fish = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{5}{13}\)
= 0.38

Question 5.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Answer:
Number of total outcomes are (1,2,……….., 8) = 8

i) Number of outcomes favourable to 8 = 1.
∴ P(8) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{8}\)

ii) Number of ‘odd numbers’ on the spinning wheel = (1, 3, 5, 7) = 4
∴ Number of outcomes favourable to an odd number.
∴ Probability of getting an odd number = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4}{8}\) = \(\frac{1}{2}\)

iii) Number greater than 2 are (3, 4, 5, 6, 7, 8)
Number of outcomes favourable to ‘greater than 2’ are = 6.
Probability of pointing a number greater than 2
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

iv) Number less than 9 are: (1,2, 3, 4, 5, 6, 7, 8 …… 8)
∴ Number of outcomes favourable to pointing a number less than 9 = 8.
∴ Probability of a number less than 9
P(E) = \(\frac{\text { No. of outcomes favourable to less than } 9}{\text { No.of total outcomes }}\)
= \(\frac{8}{8}\) = 1
Note : This is a sure event and hence probability is 1.

Question 6.
One card is drawn from, a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Answer:
Total number of cards = 52.
∴ Number of all possible outcomes in drawing a card at random = 52.
i) Number of outcomes favourable to the king of red colour = 2(♥ K, ♦ K)
∴ Probability of getting the king of red colour
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

ii) Number of face cards in a deck of cards = 4 × 3 = 12 (K, Q, J)
Number of outcomes favourable to select a face card = 12.
∴ Probability of getting a face card
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)

iii) Number of red face cards = 2 × 3 = 6.
∴ Number of outcomes favourable to select a red face card = 6.
∴ Probability of getting a red face
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{6}{52}\) = \(\frac{3}{26}\)

iv) Number of outcomes favourable to the jack of hearts = 1.
∴ Probability of getting jack of hearts
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{52}\)

v) Number of spade cards = 13
∴ Number of outcomes favourable to ‘a spade card’ = 13.
∴ Probability of drawing a spade
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)

vi) Number of outcomes favourable to the queen of diamonds = 1.
∴ Probability of drawing the queen of diamonds
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{52}\)

Question 7.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked is (a) an ace? (b) a queen?
Answer:
Total number of cards = 5.
∴ Number of total outcomes in picking up a card at random = 5.
i) Number of outcomes favourable to queen = 1.
∴ Probability of getting the queen
= \(\frac{\text { No.of outcomes favourable to the ‘Q’ }}{\text { No.of total outcomes }}\)
= \(\frac{1}{5}\)

ii) When queen is drawn and put aside, remaining cards are four.
∴ Number of total outcomes in drawing a card at random = 4.
a) Number of favourable outcomes to ace 1
Probability of getting an ace
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{1}{4}\)

b) Number of favourable outcomes to ‘Q’ = 0 (as it was already drawn and put aside)
∴ Probability that the card is Q = \(\frac{0}{4}\) = 0
After putting queen aside, selecting the queen from the rest is an impossible event and hence the probability is zero.

Question 8.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
Number of good pens = 132
Number of defective pens = 12
Total number of pens = 132 + 12 = 144
∴ Total number of outcomes in taking a pen at random = 144.
No. of favourable outcomes in taking a good pen = 132.
∴ Probability of taking a good pen
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 9.
A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer:
Given :
4 out of 20 bulbs are defective
(i.e.) Number of defective bulbs = 4
Number of non-defective bulbs = 20 – 4 = 16
If a bulb is drawn at random, the total outcomes are = 20
Number of outcomes favourable to ‘defective bulb’ = 4
∴ Probability of getting a defective bulb
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4}{20}\) = \(\frac{1}{5}\)
Suppose a non-defective bulb is drawn and not replaced, then the bulbs remaining are = 19
∴ Total outcomes in drawing a bulb from the rest = 19
Number of favourable outcomes in drawing non-defective bulb from the rest = 16 – 1 = 15
∴ Probability of getting a non-defective bulb in the second draw
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{15}{19}\)

Question 10.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Answer:
Total number of discs in the box = 90
∴ Number of total outcomes in drawing a disc at random from the box = 90.

i) Number of 2-digit numbers in the box (10, 11,….., 90) = 81
i.e., Number of favourable outcomes in drawing a 2 – digit numbers = 81
∴ Probability of selecting a disc bearing a 2 – digit number
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{81}{90}\) = \(\frac{9}{10}\) = 0.9

ii) Number of perfect squares in the box (1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64 and 9² = 81) = 9
i.e., Number of favourable out-comes in drawning a disc bearing a perfect square = 9
∴ Probability of drawning a disc with a perfect square
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{9}{90}\) = \(\frac{1}{10}\)

iii) Number of multiples of 5 from 1 to 90 are (5, 10, 15, ……….., 90) = 18
i.e., Number of favourable outcomes in drawing a disc with a multiple of 5 = 18
∴ Probability of drawing a disc bearing a number divisible by 5
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{18}{90}\) = \(\frac{1}{5}\)

Question 11.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Answer:
3 m.
Length of the given rectangle = 3 m.
and its breadth = 2 m.
Area of the rectangle
= length × breadth = 3 × 2 = 6 m²
∴ Total area of the region for landing = 6 m².
Diameter of the given circle = 1 m.
Area of the circle = \(\frac{\pi \mathrm{d}^{2}}{4}\)
= \(\frac{22}{7} \times \frac{1 \times 1}{4}\left[\text { or } \pi r^{2}=\frac{22}{7} \times \frac{1}{2} \times \frac{1}{2}\right]\)
= \(\frac{22}{28}\)
∴ Probability of the coin to land on the circle
= \(\frac{\frac{22}{28}}{6}\)
= \(\frac{22}{28×6}\)
= \(\frac{11}{28×3}\)
= \(\frac{11}{84}\)

Question 12.
A lot consists of 144 ball pens of which 20 are defective and the others are good. The shopkeeper draws one pen at random and gives it to Sudha. What is the probability that (i) She will buy it? (ii) She will not buy it?
Answer:
Given : 20 out of 144 are defective i.e., no. of defective ball pens = 20
no. of good ball pens = 144 – 20 = 124
∴ Total outcomes in drawing a ball pen at random = 144.

i) Sudha buys it if it is not defective / a good one.
No. of outcomes favourable to a good pen = 124.
∴ Probability of buying it
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{124}{144}\) = \(\frac{31}{36}\)

ii) Sudha will not buy it-if it is a defective pen
No. of outcomes favourable to a defective pen = 20
∴ Probability of not buying it
= \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{20}{144}\) = \(\frac{5}{36}\)

!! (not buying) = 1 – P (buying)
= 1 – \(\frac{31}{36}\) = \(\frac{5}{36}\)

Question 13.
Two dice are rolled simultaneously and counts are added
(i) Complete the table given below:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.
Answer:
When two dice are rolled, total number of outcomes = 36 (see the given table).
(i)
(ii) The above (given) argument is wrong [from the above table].
The sum 2, 3, 4, ………… and 12 have different no. of favourable outcomes, moreover total number of outcomes are 36.

Question 14.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer:
When a coin is tossed for n – times, the total number of outcomes = 2ⁿ.
∴ If a coin is tossed for 3 – times, then the total number of outcomes = 2³ = 8
Note the following :
TTT
TTH
THT
HTT
HHT
HTH
THH
HHH
Of the above, no. of outcomes with different results = 6.
Probability of losing the game
= \(\frac{\text { No. of favourable outcomes to lose }}{\text { No. of total outcomes }}\)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 15.
A dice is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up atleast once? [Hint : Throwing a dice twice and throwing two dice simultaneously are treated as the same experiment].
Answer:
If a dice is thrown n-times or n-dice are thrown simultaneously then the total
number of outcomes = 6×6×6….×6
(n – times) = 6ⁿ.
No. of total outcomes in throwing a dice for two times = 6² = 36.
i) Let E be the event that 5 will not come up either time, then the favourable outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3/2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6) = 25.
∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{25}{36}\)

ii) Let E be the event that 5 will come up atleast once.
Then the favourable outcomes are (1,5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) = 11 No. of favourable outcomes = 11
∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{11}{36}\)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.2

10th Class Maths 14th Lesson Statistics Ex 14.2 Textbook Questions and Answers

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
Maximum number of patients joined in the age group 35 – 45.
∴ Modal class is 35 – 45.
Lower limit of the modal class ‘l’ = 35
Class size h = 10
Frequency of modal class, f₁ = 23
Frequency of the class preceding the modal class f₀ = 21
Frequency of the class succeeding the modal class f₂ = 14

∴ Mode = \(l+\frac{\left(f_{1}-f_{0}\right)}{2 f_{1}-f_{0}-f_{2}} \times h\)
\(\begin{array}{l}
=35+\left(\frac{23-21}{2 \times 23-21-14}\right) \times 10 \\
=35+\left(\frac{2}{46-35}\right) \times 10
\end{array}\)
= 35 + \(\frac{2}{11}\) × 10
= 35 + 1.81818……
= 36.8 years.
Mean x = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2830}{80}\)
= 35.37 years.
Interpretation: Mode age is 36.8 years, Mean age = 35.37 years.
Maximum number of patients admitted in the hospital are of the age 36.8 years, while on an average the age of patients admitted to the hospital is 35.37 years. Mode is less than the Mean.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.
Answer:

Since the maximum frequency 61 is in the class 60 – 80, this is the required modal class.
Modal class frequency, f₁ = 61.
Frequency of the class preceding the modal class f₀ = 52
Frequency of the class succeeding the modal class f₂ = 38
Lower boundary of the modal class, l = 60
Height of the class, h = 20
∴ Mode (Z) = \(l+\frac{\left(f_{1}-f_{0}\right)}{2 f_{1}-f_{0}-f_{2}} \times h\)
\(=60+\left[\frac{61-52}{2 \times 61-(52+38)}\right] \times 20\)
= 60 + \(\frac{9}{122-90}\) × 20
= 60 + \(\frac{9}{32}\) × 20
= 60 + 5.625
= 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Answer:

Since the maximum families 40 lies in the class 1500 – 2000, this is the required modal class.
Lower boundary of the modal class (l) = 1500
Frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class f₀ = 24
Frequency of the class succeeding the modal class f₂ = 33
Height of the class, h = 500
Hence, modal monthly income = Rs. 1847.83.
Assumed mean (a) = 3250
∑f_i = 200, ∑u_if_i = -235
Mean monthly income = \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
= 3250 – \(\frac{235}{200}\) × 500
= 3250 – 587.5
= Rs. 2662.50

Question 4.
The following distribution gives the state-wise, teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Answer:

Since the maximum number of states ‘10’ lies in the class interval 30-35, this is the modal class.
Lower boundary of the modal class, l = 30
Frequency of the modal class, f₁ = 10
Frequency of the class preceding the modal class = f₀ = 9
Frequency of the class succeeding the modal class = f₂ = 3
Height of the class, h = 5
∴ Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
\(=30+\frac{10-9}{(10-9)+(10-3)} \times 5\)
= 30 + \(\frac{1×5}{1+7}\)
= 30 + \(\frac{5}{8}\)
= 30 + 0.625
= 30.625
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
a = assumed mean = 32.5; h = height of the class = 5
∴ x = 32.5 – \(\frac{23}{35}\) × 5
= 32.5 – 3.28
= 29.22
Mean = 30.625
Mode = 29.22
Mode states have a students – teacher ratio 29.22 and on an average this ratio is 30.625.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.
Answer:

Maximum number of batsmen are in the class 4000 – 5000.
∴ Modal class is 4000 – 5000.
Lower boundary of the modal class ‘l’ = 4000
Frequency of the modal class, f₁ = 18
Frequency of the class preceding the modal class, f₀ = 4
Frequency of the class succeeding the modal class, f₂ = 9
Height of the class, h = 1000
Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
Mode (Z) = \(4000+\frac{18-4}{(18-4)+(18-9)} \times 1000\)
= 4000 + \(\frac{14}{14+9}\) × 1000
= 4000 + \(\frac{14000}{23}\)
= 4000 + 608.695
= 4608.69
≃ 4608.7 runs

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.

Find the mode of the data.
Answer:

Since the maximum frequency is 20, the modal class is 40 – 50.
Lower boundary of the modal class ‘l’ = 40
Frequency of the modal class, f₁ = 20
Frequency of the class preceding the modal class, f₀ = 12
Frequency of the class succeeding the modal class, f₂ = 11
Height of the class, h = 10;
Mode (Z) = \(l+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{\left(\mathrm{f}_{1}-\mathrm{f}_{0}\right)+\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)}\right) \times \mathrm{h}\)
Mode (Z) = \(40+\frac{(20-12)}{(20-12)+(20-11)} \times 10\)
= 40 + \(\frac{8}{8+9}\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.70588
= 44.705
≃ 44.7 cars

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.2

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.2 Textbook Questions and Answers

Question 1.
A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road.
Answer:

Let the height of the tower = h mts say
Width of the road be = x m.
Distance between two points of observation = 10 cm.
Angles of elevation from the two points = 60° and 30°.
From the figure
tan 60° = \(\frac{h}{x}\)
√3 = \(\frac{h}{x}\)
⇒ h = √3x …….(1)
Also tan 30° = \(\frac{h}{10+x}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{10+x}\)
⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2)
From equations (1) and (2) h
h = √3x = \(\frac{10+x}{\sqrt{3}}\)
∴ √3x = \(\frac{10+x}{\sqrt{3}}\)
√3 × √3x = 10 + x
⇒ 3x – x = 10
⇒ 2x = 10
⇒ x = \(\frac{10}{2}\) = 5m
∴ Width of the road = 5 m
Now Height of the tower = √3x = 5√3 m.

Question 2.
A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.
Answer:

Height of the temple = 30 m
Height of the man = 1.5 m
Initial distance between the man and temple = d m. say
Let the distance walked = x m.
From the figure
tan 30° = \(\frac{30-1.5}{d}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{28.5}{d}\)
∴ d = 28.5 × √3m ………(1)
Also tan 60° = \(\frac{28.5}{d-x}\)
⇒ √3 = \(\frac{28.5}{d-x}\)
⇒ √3(d-x) = 28.5
⇒ √3(28.5 × √3-x) = 28.5
⇒ 28.5 × 3 – √3x = 28.5
⇒ √3x = 3 × 28.5-28.5
⇒ √3x = 2 × 28.5 = 57
∴ x = \(\frac{57}{\sqrt{3}}=\frac{19 \times 3}{\sqrt{3}}\) = 19√3
= 19 × 1.732
= 32.908 m.
∴ Distance walked = 32.908 m.

Question 3.
A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.
Answer:

Height of the pedestal = 2 m.
Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.
Angle of elevation of top of the pedestal = 45°.
Let the distance between the point of observation and foot of the pedestal = x m.
From the figure
tan 45° = \(\frac{2}{x}\)
1 = \(\frac{2}{x}\)
∴ x = 2 m.
Also tan 60° = \(\frac{2+h}{x}\)
⇒ √3 = \(\frac{2+h}{x}\)
⇒ 2√3 = 2 + h
⇒ h = 2√3 – 2
= 2(√3-1)
= 2(1.732 – 1)
= 2 × 0.732
= 1.464 m.

Question 4.
From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.
Answer:

Angle of elevation of the top of the tower = 60°.
Angle of depression to the foot of the tower = 45°.
Distance between tower and building = 7 m.
Let the height of the building = x m and tower = y m.
From the figure
tan 45° = \(\frac{x}{7}\)
1 = \(\frac{x}{7}\)
∴ x = 7 m.
Also tan 60° = \(\frac{y-x}{7}\)
⇒ √3 = \(\frac{y-x}{7}\)
⇒ 7√3 = y – 7
∴ y = 7 + 7√3
= 7 (√3 + 1)
= 7(1.732 + 1)
= 2.732 × 7
= 19.124 m.

Question 5.
A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ?
Answer:

Length of the wire = 18 m
Let the length of the wire removed = x
Height of the pole be = h
From the figure
sin 30° = \(\frac{h}{18}\)
⇒ \(\frac{1}{2}\) = \(\frac{h}{18}\)
⇒ h = \(\frac{18}{2}\) = 9 m
Also sin 60° = \(\frac{h}{18-x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{9}{18-x}\)
√3(18-x) = 9 × 2
18√3 – √3x = 18
√3x = 18√3 – 18
√3x = 18(√3-1)
x = \(\frac{18(\sqrt{3}-1)}{\sqrt{3}}\)
= \(\frac{6 \times 3(\sqrt{3}-1)}{\sqrt{3}}\)
= 6√3(√3-1)
= 6(3-√3)
= 18 – 6√3
= 18 – 10.392
= 7.608 m.

Question 6.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.
Answer:

Height of the tower = 30 m
Angle of elevation of the top of the tower = 60°.
Angle of elevation of the top of the building = 30°.
Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.
From the figure
tan 60° = \(\frac{30}{d}\)
√3 = \(\frac{30}{d}\)
⇒ d = \(\frac{30}{\sqrt{3}}=\frac{10 \times 3}{\sqrt{3}}\) = 10√3m
Also tan 30° = \(\frac{x}{d}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{x}{10 \sqrt{3}}\)
⇒ x = \(\frac{10 \sqrt{3}}{\sqrt{3}}\) = 10 m
∴ Height of the building = 10 m

Question 7.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:

Width of the road = 120 f.
Angle of elevation of the top of the 1st tower = 60°.
Angle of elevation of the top of the 2 tower = 30°.
Let the distance of the point from the 1st pole = x.
Then the distance of the point from
the 2nd pole = 120 – x.
and height of each pole = h say.
From the figure
tan 60° = \(\frac{h}{x}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ h = √3x ……..(1)
Also tan 30° = \(\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{120-\mathrm{x}}\)
⇒ h = \(\frac{120-x}{\sqrt{3}}\)
From (1) and (2)
√3x = \(\frac{120-x}{\sqrt{3}}\)
⇒ √3.√3x = 120-x
⇒ 3x = 120 – x
⇒ 3x + x = 120
⇒ 4x = 120
⇒ x = \(\frac{120}{4}\) = 30 ft
Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet
∴ Distances of the poles = 30 ft. and 120 – 30 fts = 90 ft.
Height of each pole = 51.96 ft.

Question 8.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.
Answer:

Let the height of the tower = h m.
Angles of elevation of the top of the tower from two points = x° and (90° – x)
From the figure
tan x = \(\frac{h}{4}\) ……. (1)
Also tan (90° – x) = \(\frac{h}{9}\)
⇒ cot x = \(\frac{h}{9}\)
⇒ \(\frac{1}{\tan x}\) = \(\frac{h}{9}\)
∴ tan x = \(\frac{9}{h}\) …….. (2)
From (1) and (2)
tan x = \(\frac{h}{4}\) = \(\frac{9}{h}\)
∴ \(\frac{h}{4}\) = \(\frac{9}{h}\)
h × h = 9 × 4
⇒ h² = 36
⇒ h = 6 m

Question 9.
The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)
Answer:

Height of the plane from the ground PM = RN = 1500√3 m.
Angle of elevation are 30° and 60°.
From the figure
tan 60° = \(\frac{PM}{QM}\)
√3 = \(\frac{1500 \sqrt{3}}{\mathrm{QM}}\)
QM = \(\frac{1500 \sqrt{3}}{\sqrt{3}}\) = 1500 m
Also tan 30° = \(\frac{RN}{QN}\)
\(\frac{1}{\sqrt{3}}=\frac{1500 \sqrt{3}}{\mathrm{QM}+\mathrm{MN}}\)
QM + MN = 1500√3 × √3
1500 + MN = 1500 × 3
MN = 4500 – 1500
MN = 3000 mts.
∴ Distance travelled in 15 seconds = 3000 mts.
∴ Speed of the jet plane = \(\frac{\text { distance }}{\text { time }}=\frac{3000}{15}\) = 200 m/s
= 200 × \(\frac{18}{5}\) kmph
= 720 kmph
Speed = 200 m/sec. or 720 kmph.

AP SSC 10th Class Maths Textbook Solutions

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 14 Carbon and its Compounds Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 14th Lesson Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds Textbook Questions and Answers

Improve Your Learning

Question 1.
Name the simplest hydrocarbon. (AS1)
Answer:
The simplest hydrocarbon is alkane called Methane (CH₄). It’s an aliphatic, saturated compound of Hydrogen and Carbon.

Question 2.
What are the general molecular formulae of alkanes, alkenes and alkynes? (AS1)
Answer:
General molecular formula of alkane is C_nH_2n+2.
General molecular formula of alkene is C_nH_2n.
General molecular formula of alkyne is C_nH_2n-2.

Question 3.
Name the carboxylic acid used as a preservative. (AS1)
Answer:
Vinegar with chemical formula CH₃COOH is used as preservative. 5 – 8% of solution of acetic acid or ethanoic acid in water is called vinegar and it is used widely as preservative in pickles.

Question 4.
Name the product other than water formed on burning of ethanol in air. (AS1)
Answer:
C₂H₃OH + 3O₂ → 2CO₂ + 3H₂O + Energy
So, the product other than water formed on burning of ethanol in air is carbon dioxide (CO₂).

Question 5.
Give the IUPAC name of the following compounds. If more than one compound is possible, name all of them. (AS1)
i) An aldehyde derived from ethane.
ii) A ketone derived from butane.
iii) A chloride derived from propane.
iv) An alcohol derived from pentane.
Answer:
i) An aldehyde derived from ethane is ethanal. Its formula is CH₃CHO.
ii) A ketone derived from butane. Its IUPAC name is Butanone.
Its chemical formula is CH₃COCH₂CH₃
It is also known as methyl ethyl ketone. (Its general name)

iii) A chloride derived from propane.
A) 1-Chloro propane. Its formula is CH₃CH₂CH₂Cl.

iv) An alcohol derived from pentane :
A) 1-Pentanol. Its formula is CH₃CH₂CH₂CH₂CH₂OH.
B) 2-Pentanol. Its formula is CH₃CHOH CH₂CH₂CH₃
C) 3-Pentanol. Its formula is CH₃CH₂ CHOH CH₂CH₃

Question 6.
A mixture of oxygen and ethyne is burnt for welding ; can you tell why a mixture of ethyne and air is not used? (AS1)
Answer:

• Ethyne when burnt in the presence of oxygen gives enough heat that can be used for welding.
• Whereas if it is burnt in air which contains nitrogen, CO₂ and other inactive gaseous contents, sufficient oxygen is not available for burning ethyne to give the required heat.

Question 7.
Explain with the help of a chemical equation, how an addition reaction is used in vegetable ghee industry. (AS1)
Answer:

• The addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon is called hydrogenation. The process of hydrogenation takes place in the presence of nickel or palladium metals as catalyst.
• The process of hydrogenation has an important industrial application. It is used to prepare vegetable ghee (or vanaspati ghee) from vegetable oils.
• Vegetable oils are unsaturated fats having double bonds between some of their carbon atoms.
• When a vegetable oil (like groundnut oil) is heated with hydrogen in the presence of finely divided nickel as catalyst, a saturated oil called vegetable ghee (or vanaspati ghee) is formed. This a reaction is called hydrogenation of oils and it can be represented as follows.

Here vegetable oil is a liquid whereas vegetable ghee is a solid (or a semi solid).

Question 8.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆O? (AS1)
b) Give the IUPAC names of the above possible compounds and represent them in structures. (AS1)
c) What is the similarity in these compounds? (AS1)
Answer:
a) They are CH₃COCH₃and CH₃ CH₂ CHO

b) i) The IUPAC name of CH₃COCH₃ is propanone.
ii) The IUPAC name of CH₃ CH₂ CHO is propanal.

Question 9.
Name the simplest ketone apfl write its molecular formula. (AS1)
Answer:
Acetone is the simplest ketone. Its molecular formula is CH₃COCH₃ Its IUPAC name is propanone.

Question 10.
What do we call the Self linking property of carbon? (AS1)
Answer:
The property of self combination (or linking) of carbon atoms to form long chains is useful to us because it gives rise to an extremely large number of carbon compounds (or organic compounds). This is known as catenation.

Question 11.
Name the compound formed by heating ethanol at 443 K with excess of cone. H₂SO₄. (AS1)
(OR)
What is the compound formed when ethyhalcohol (Ethanol) is dehydrated ? Write the chemical equation of the reaction.
Answer:
1. When ethanol is heated with excess of cone. H₂SO₄ at 443 K (170° C), it gets dehydrated to form ethene (which is an unsaturated hydrocarbon).

2. During dehydration of ethanol molecules (CH₃ – CH₂OH), H from the CH₃ group and OH from CH₂OH group are removed in the form of a water molecule (H₂O) regulating in the formation of this molecule (CH₂ = CH₂).
3. In this reaction concentrated sulphuric acid acts as a dehydrating agent.

Question 12.
Give an example for esterification reaction. (AS1)
Answer:
The reaction between carboxylic acid and an alcohol in the presence of cone. H₂SO₄ to form a sweet odoured substance, ester with the functional group

is called esterification.

Ex: Ethanoic acid (carboxylic acid) reacts with Ethanol (alcohol) and forms ethyl acetate.

Question 13.
Name the product obtained when ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate. (AS1)
(OR)
If the ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate, what is the product obtained from them?
Answer:
Ethanol (Ethyl alcohol) undergoes oxidation to form the product of Acetaldehyde and finally Acetic acid.

Question 14.
Write the chemical equation representing the reaction of preparation of ethanol from ethane. (AS1)
Answer:
1. Ethane in the absence of air on heating forms ethene

2. Then Ethanol is prepared on large scale from ethene by the addition of water vapour to it in the presence of catalyst like P₂O₅, Tungsten oxide at high pressure and temperature.

Question 15.
Write the IUPAC name of the next homologous of CH₃OHCH₂CH₃. (AS1)
Answer:
The IUPAC name of the next homologous of CH₃OHCH₂CH₃ is HO-CH₃CH₂CH₂CH₃ 1 – butanol.

Question 16.
Define homologous series of carbon compounds. Mention any two characteristics of homologous series. (AS1)
Answer:
1. The series of carbon compounds in which two successive compounds differ by – CH₂ unit is called homologous series.
Ex : 1) CH₄, C₂H₆, C₃H₈, ………………..
2) CH₃OH, C₂H₅OH, C₃H₇OH, ………………..

2. If we observe above series of compounds, we will notice that each compound in the series differs by – CH₂ unit by its successive compound.

3. Characteristics of homologous series :
i) They have one general formula.
Ex : alkanes (C_nH_2n+2), alkynes (C_nH_2n-2), alcohols (C_nH_2n+1) OH, etc.
ii) Successive compounds in the series possess a difference of (-CH₂) unit.
iii) They have similar chemical properties.

Question 17.
Give the names of functional groups
(i) – CHO
(ii) – C = O. (AS1)
(OR)
Write the names of the given functional groups
(i) – CHO
(ii) – C = O
Answer:
i) – CHO → aldehyde
ii) – C = O → ketone

Question 18.
Why does carbon form compounds mainly by covalent bonding? (AS1)
Answer:
Since carbon atoms can achieve the inert gas electron arrangements only by the sharings of electrons, therefore, carbon always forms covalent bonds.

Question 19.
Allotropy is a property shown by which class substance: elements, compounds or mixtures? Explain allotropy with suitable examples. (AS1)
Answer:

1. Allotropy is a property shown by the elements.

2. The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy.

3. The different forms of the element are called allotropes and are formed due to the difference in the arrangement of atoms.

4. Example for allotropes : Allotropes of carbon.

Allotropes of carbon are classified into two types. They are
1) Amorphous forms,
2) Crystalline forms.

5) Amorphous forms of carbon:
Coal, coke, wood, charcoal, animal charcoal, lampblack, gas carbon, petroleum coke, sugar charcoal.

6) Crystalline forms of carbon :
Diamond, graphite and buckminsterfullerene.

Question 20.
Explain how sodium ethoxide is obtained from ethanol. Give chemical equations. (AS1)
Answer:
As ethanol is similar to water molecule (H₂O) with C₂H₅ group in place of hydrogen, it reacts with metallic sodium to liberate hydrogen and form sodium ethoxide.

Question 21.
Describe with chemical equation how ethanoic acid may be obtained from ethanol. (AS1)
Answer:
Ethyl alcohol (Ethanol) undergoes oxidation to form the product Acetaldehyde and finally acetic acid (Ethanoic acid).

Question 22.
Explain the cleansing action of soap. (AS1)
Answer:
When a dirty cloth is put in water containing dissolved soap, the hydrocarbon ends of the soap molecules in the micelle attach to the oil or grease particles present on the surface of dirty clothes.

Question 24.
Explain the structure of graphite in terms of bonding and give one property based on this structure. (AS1)
(OR)
Why does graphite act as lubricant?
Answer:

• Graphite forms a two dimensional layer structure with C – C bonds within the layers.
• There are relatively weak interactions between the layers.
• In the layer structure, the carbon atoms are in a trigonal planar environment.
• This is consistent with each carbon atom in sp² hybridisation.
• Interactions between the sp² orbitals (overlaps) lead to the formation of C – C bonds.
• Each carbon atom is with one unhybridised ‘p’ orbital.
• The unhybridised ‘p’ orbitals interact to form a π system that is delocalised over the whole layer.
• The interactions known as London dispersion forces between the layers which are separated by a distance of 3.35 A° are weakened by the presence of water molecules so that it is easy to cleave graphite.
• For this reason graphite is used as lubricant and as the lead in pencils.

Question 25.
Name the acid present in vinegar. (AS1)
Answer:
1) The acid present in vinegar is Ethenoic acid or acetic acid (CH₃COOH).
2) 5 – 8% solution of acetic acid in water is called vinegar.

Question 26.
What happens when a small piece of sodium is dropped into ethanol? (AS2)
Answer:
Ethanol reacts with sodium to liberate hydrogen and form sodium ethoxide.

Question 27.
Two carbon compounds A and B have molecular formula C₃H₈ and C₃H₆ respectively. Which one of the two is most likely to show addition? Justify your answer. (AS2)
Answer:

• It is a saturated hydrocarbon. It shows substitution reaction.

• This is an unsaturated hydrocarbon. Hence it shows addition to become saturated. During the reactions, addition of reagent takes place at the double bonded carbon atoms.

Justification :
In the following, C₃H₆ undergoes addition reaction.

Question 28.
Suggest a test to find the hardness of water and explain the procedure. (AS3)
(OR)
How do you test whether a given water sample is soft or hard?
Answer:

• Take about 10 ml hard water (well water or hand pump water) in a test tube.
• Add five drops of soap solution to it.
• Shake the test tube vigorously.
• We see that no lather is formed at first.
• Only a dirty white curd like scum is formed on the surface of water.
• From this, we conclude that soap does not form lather easily with hard water.
• We have to add much more soap to obtain lather with hard water.

Question 29.
Suggest a chemical test to distinguish between ethanol and ethanoic acid and explain the procedure. (AS3)
Answer:

1. Take ethanol and ethanoic acid in two different test tubes.
2. Add nearly 18 g of sodium bicarbonate (NaHCO₃) to each test tube.
3. Lots and lots of bubbles and foam will be observed from the test tube containing ethanoic acid. This is due to release of CO₂.
   NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
4. Ethanol will not react with sodium bicarbonate and thus we won’t observe any change in the test tube containing ethanol.
   Thus we can separate ethanol from ethanoic acid.

Question 30.
An organic compound ‘X’ with a molecular formula C₂H₆O undergoes oxidation with alkaline KMnO₄ and forms the compound ‘Y’, that has molecular formula C₂H₄O₂. (AS3)
i) Identify ‘X’ and ‘Y’.
Answer:
X is Ethanol is CH₃CH₂OH and T is Ethanoic acid, i.e., CH₃COOH.

ii) Write your observation regarding the product when the compound X is made to react with compound IT which is used as a preservative for pickles.
Answer:
Ethyl alcohol undergoes oxidation to form the product Acetaldehyde and finally Acetic acid.

Here CH₃COOH is used as preservative for pickles.

When X reacts with Y it forms ethyl acetate and water which is called esterification reaction.
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 31.
Prepare models of methane, ethane, ethene and ethyne molecules using clay balls and matchsticks. (AS4)
Answer:
Stick and ball model :
1) Methane (CH₄) :

2) Ethane (C2H₆):

3) Ethene (C₂H₄):

4) Ethyne (C₂H₂)

Question 32.
Collect information about artificial ripening of fruits by ethylene. (AS4)
Answer:

• Seasonal fruits like mango, banana, papaya, sapota and custard apple are often harvested in nature. But due to unripe condition they are subsequently allowed to ripen by natural release of ripening harmone (ethylene) from the fruit.
• However, natural ripening in some fruits is a slow process, which leads to high weight loss, desiccation of fruits and under ripening. With the rapid development of fruit trade, artificial ripening has become essential and the methods practised earlier by small traders are smoking and calcium carbide treatment.
• Fruits ripened with calcium carbide though seem attractive and colourful are inferior in taste, flavour and spoil faster.
• Government of India has banned the use of calcium carbide for artificial ripening of fruits under PFA Act 8-44AA, 1954.
• Artificial ripening of fruits by using the above steps spoils the health of consumers, so we should not use such type of fruits.
• Government has to take serious action on the fruit sellers who are practising the above said methods.

Question 33.
Draw the electronic dot structure of ethane molecule (C₂H₆). (AS6)
Answer:
C₂H₆:

Question 34.
How do you appreciate the role of esters in everyday life? (AS6)
Answer:

• Esters are usually volatile liquids having sweet or pleasant smell.
• They are also said to have fruity smell.
• Esters are used in making artificial perfumes.
• This is because of the fact that most of the esters have a pleasant smell.
• Esters are also used as flavouring agents.
• This means that esters are used in making artificial flavours and essences used in ice-cream, sweets and cool drinks.
• The alkaline hydrolysis of esters is known as saponification (Soap making).
• That’s why we can appreciate the role of esters in everyday life.

Question 35.
How do you condemn the use of alcohol as a social practice? (AS7)
Answer:

• Consumption of alcohol in the form of beverages is harmful to health.
• It causes severe damage to blood circulation system.
• Addiction to alcohol drinking leads to heart diseases and damages the liver.
• It also causes ulcers in small intestines due to increased acidity and damages the digestive system.
• Alcohol which is consumed in raw form under the names liquor, gudumba which is more harmful to health due to adulteration.
• Alcohol mixed with pyridine is called denatured spirit. Consumption of denatured spirit causes blindness and death.
• Hence use of alcohol is a social evil which harms the society.

Question 36.
An organic compound with molecular formula C₂H₄O₂ produces brisk effervescence on addition of sodium carbonate/bicarbonate.
Answer the following :
a) Identify the organic compound. (AS1)
Answer:
The organic compound is Ethanoic acid (CH₃COOH).

b) Write the chemical equation for the above reaction. (AS1)
Answer:

c) Name the gas evolved. (AS2)
Answer:
CO₂

d) How will you test the gas evolved? (AS3)
Answer:
1) Pass the evolved gas through lime water in a test tube.
2) We will find that lime water turns milky.
3) Only CO₂ gas can turn lime water milky.

e) List two important uses of the above compound. (AS1)
Answer:
1) Dilute ethanoic acid (CH₃COOH) is used as a food preservative in the preparation of pickles and sauces.
2) Ethanoic acid is used for making cellulose acetate which is an important artificial fibre.

Question 37.
1 ml glacial acetic acid and 1 m/of ethanol are mixed together in, a test tube. Few drops of concentrate sulphuric acid is added in the mixture are warmed in a water bath for 5 min.
Answer the following:
a) Name the resultant compound formed.
b) Represent the above change by a chemical equation.
c) What term is given to such a reaction?
d) What are the special characteristics of the compound formed?
Answer:
a) Ethyl acetate.

c) Esterification
d) It has fruity smell or pleasant smell.

Fill In The Blanks

1. Carbon compounds containing double and triple bonds are called ………………….
2. A compound which is basic constituent of many cough syrups ………………………
3. Very dilute solution of ethanoic acid is ………………..
4. A sweet odour substance formed by the reaction of an alcohol and a carboxylic acid is ………………
5. When sodium metal is dropped in ethanol …………………. gas will be released.
6. The functional group present in methanol is …………………….
7. IUPAC name of alkene containing 3 carbon atoms is ………………….
8. The first member of homologous series among alkynes is ……………………
9. The product that is formed by dehydration of ethanol in cone, sulphuric acid is ………………….
10. Number of single covalent bonds in ammonia are ………………..
11. Type of reactions shown by alkanes is ……………….
Answer:

1. unsaturated compounds
2. ethanol
3. vinegar
4. ester
5. H₂
6. – OH (Alcohol)
7. propene
8. ethyne (C₂H₂)
9. ethene (C₂H₄)
10. 3
11. substitutional

Multiple Choice Questions

1. Which of the four test tubes containing the following chemicals shows the brisk effervescence when dilute acetic acid was added to them?
i) KOH
ii) NaHCO₃
iii) K₂CO₃
iv) NaCl
A) i & ii
B) ii & iii
C) i & iv
D) ii & iv
Answer:
B) ii & iii

2. Which of the following solution of acetic acid in water can be used as preservative?
A) 5-10%
B) 10-15%
C) 15-20%
D) 100%
Answer:
A) 5-10%

3. The suffix used for naming an aldehyde is
A) – ol
B) – al
C) – one
D) – ene
Answer:
B) – al

4. Acetic acid, when dissolved in water, it dissociates into ions reversibly because it is a
A) weak acid
B) strong acid
C) weak base
D) strong base
Answer:
A) weak acid

5. Which one of the following hydrocarbons can show isomerism?
A) C₂H₄
B) C₂H₆
C) C₃H₈
D) C₄H₁₀
Answer:
D) C₄H₁₀

6. Combustion of hydrocarbon is generally accompanied by the evolution of
A) Heat
B) Light
C) Both heat and light
D) Electric current
Answer:
C) Both heat and light

7. 2 ml of ethanoic acid was taken in each of the three test tubes A, B and C and 2 ml, 4 ml and 8 ml water was added to them respectively. A clear solution is obtained in:
A) Test tube A only
B) Test tubes A & B only
C) Test tubes B and C only
D) All the test tubes
Answer:
D) All the test tubes

8. If 2 ml of acetic acid was added slowly in drops to 5 ml of water then we will notice
A) The acid forms a separate layer on the top of water
B) Water forms a separate layer on the top of the acid
C) Formation of a clear and homogenous solution
D) Formation of a pink and clear solution
Answer:
C) Formation of a clear and homogenous solution

9. A few drops of ethanoic acid were added to solid sodium carbonate. The possible results of the reactions are
A) A hissing sound was evolved
B) Brown fumes evolved
C) Brisk effervescence occurred
D) A pungent smelling gas evolved
Answer:
C) Brisk effervescence occurred

10. When acetic acid reacts with ethyl alcohol, we add cone. H₂SO₄, it acts as and the process is called
A) Oxidizing agent, saponification
B) Dehydrating agent, esterification
C) Reducing agent, esterification
D) Acid and esterification
Answer:
B) Dehydrating agent, esterification

10th Class Chemistry 14th Lesson Carbon and its Compounds InText Questions and Answers

10th Class Chemistry Textbook Page No. 254

Question 1.
Can carbon get helium configuration by losing four electrons from the outer shell?
Answer:

• If carbon loses four electrons from the outer shell, it has to form C⁴⁺ ions.
• This requires huge amount of energy which is not available normally.
• Therefore C⁴⁺ formation is also a remote possibility.
• Carbon has to satisfy its tetravalency by sharing electrons with other atoms.
• It has to form four covalent bonds either with its own atoms or atoms of other elements.

10th Class Chemistry Textbook Page No. 255

Question 2.
How do carbon atoms form bonds in so many different ways?
Answer:
As per valence bond theory, the four unpaired electrons in a carbon atom is main cause to form many bonds.

Question 3.
Explain the four unpaired electrons in carbon atom through excited state.
Answer:
Electronic configuration of carbon (ground state):

Electronic configuration of carbon (excited state):

10th Class Chemistry Textbook Page No. 256

Question 4.
Where does this energy to excite electron come from?
Answer:

• We have to understand that free carbon atom would not be in excited state under normal conditions.
• When the carbon atom is ready to form bonds with other atoms, the energy required for excitation is taken up from bond energies, which are the liberated energies when bonds are formed between carbon atom and other atoms.

Question 5.
In methane (CH₄) molecule all four carbon – hydrogen bonds are identical and bond angle HCH is 109°28′. How can we explain this?
Answer:
In excited state, carbon atom has three unpaired electrons in p-orbitals and one electron in s-orbital. These four valence electrons are with different energies. These orbitals combine to form four identical orbitals. Four hydrogen atoms form four identical C -H bonds with bond angle 109° 28′. This is called hybridisation.

Question 6.
How do these energetically unequal valence electrons form four equivalent covalent bonds in methane molecule?
Answer:
1) When bonds are formed, energy is released and the system becomes more stable. If carbon forms four bonds rather than two, still more energy is released and so the resulting molecule becomes even more stable.

2) The energy difference between the 2s and 2p orbitals is very small. When carbon atom is ready to form bonds it gets a small amount of energy from bond energies and gets excited to promote an electron from the 2s to the empty 2p to give four unpaired electrons.

3) We have got four unpaired electrons ready for bonding, but these electrons are in two different kinds of orbitals and their energies are different.

4) We are not going to get four identical bonds unless these unpaired electrons are in four identical orbitals.

10th Class Chemistry Textbook Page No. 257

Question 7.
How to explain the four orbitals of carbon containing unpaired electrons as energetically equal?
Answer:
With hybridisation we explai n the four orbitals of carbon containing unpaired electrons are energetically equal.
Ex : Methane (CH₄).

10th Class Chemistry Textbook Page No. 258

Question 8.
How do you explain the ability of C – atom to form two single covalent bonds and one double bond?
Answer:
Ethylene (CH₂ = CH₂) explains the ability of carbon atom to form two single covalent bonds and one double bond.
Ex:

10th Class Chemistry Textbook Page No. 259

Question 9.
How do you explain the ability of carbon atom to form one single bond and one triple bond?
Answer:
Ethyne (HC \(\equiv\) CH) explains the ability of carbon atom to form one single bond between one hydrogen and carbon, and one triple bond between carbon and carbon.
Ex : H – C \(\equiv\) C – H.

10th Class Chemistry Textbook Page No. 260

Question 10.
What are bond angles H\(\widehat{\mathbf{C}}\)H in CH₄, C₂H₄ and C₂H₂ molecules?
Answer:

10th Class Chemistry Textbook Page No. 262

Question 11.
How do you understand the markings (writings) of a pencil on a paper?
Answer:

1. When we write with a pencil, the inter layer attractions breakdown and leave graphite layers on the paper.
2. It is easy to remove pencil marks from paper with an eraser because, the layers do not bind strongly to the paper.

10th Class Chemistry Textbook Page No. 265

Question 12.
Allotting completely one special branch in chemistry to compounds of only one element. Is it justified when there are so many elements and their compounds but not with any special branches?
Answer:

1. We understand that all molecules that make life possible carbohydrates, proteins, nucleic acids, lipids, hormones, and vitamins contain carbon.
2. The chemical reactions that take place in living systems are of carbon compounds.
3. Food that we get From nature, various medicines, cotton, silk and fuels like natural gas and petroleum almost all of them are carbon compounds.
4. Synthetic fabrics, plastics, synthetic rubber are also compounds of carbon.
5. Hence, carbon is a special element with the largest number of compounds:

10th Class Chemistry Textbook Page No. 266

Question 13.
What are hydrocarbons?
Answer:
The compounds containing only carbon and hydrogen in their molecules are called hydrocarbons.

Question 14.
Do all the compounds have equal number of C and H atoms?
Answer:
No. All the compounds do not have equal number of C and H atoms.

10th Class Chemistry Textbook Page No. 269

Question 15.
Observe the following two structures.
a) CH₃ – CH₂ – CH₂ – CH₃
b)

i) How about their structures? Are they same?
Answer:
No, they are not same compounds.

ii) How many carbon and hydrogen atoms are there in (a) and (b) structures?
Answer:
Carbon – 4 ; Hydrogen – 10.

iii) Write the condensed molecular formulae for (a) and (b), do they have same molecular formulae?
Answer:
C₄H₁₀; Yes.

Question 16.
Can carbon form bonds with the atoms of other elements?
Answer:
Carbon forms compounds not only with atoms of hydrogen but also with atoms of other elements like oxygen, nitrogen, sulphur, phosphorus, halogens, etc.

10th Class Chemistry Textbook Page No. 272

Question 17.
What do you mean by nomenclature of Organic componds?
Answer:
Nomenclature of organic chemistry is systematic method of naming organic compound.

Question 18.
What is the basis for nomenclature?
Answer:
The basic of the nomenclature is number of carbons in the parent chain in a compound.

10th Class Chemistry Textbook Page No. 273

Question 19.
What are the word – root and suffix?
Answer:
Word root:
Word root indicates the number of carbon atoms in the longest possible continuous carbon chain also known as parent chain.

Suffix :
Suffix is added immediately after the word root. It is two types

1) Primary Suffix :
It is used to indicate the degree of saturation or unsaturation of the main chain.

2) Secondary Suffix :
It is used to indicate the main functional group in the organic compound.

10th Class Chemistry Textbook Page No. 274

Question 20.
What do you mean by the term ‘alkyl’?
Answer:
Alkyl:
Alkyl is a substituent, that is attached to the molecular fragment.
General formula of alkyl is C_nH_{2n + 1}

10th Class Chemistry Textbook Page No. 278

Question 21.
Can we write the structure of a compound if the name of the compound is given?
Answer:
Yes, we can write the structure of a compound if the name of the compound is given.

10th Class Chemistry Textbook Page No. 279

Question 22.
Why do sometimes cooking vessels get blackened on a gas or kerosene stove?
Answer:
Because of the inlets of air getting closed, the fuel gases do not completely undergo combustion. Hence, it forms a sooty carbon form which gets coated over the vessels.

10th Class Chemistry Textbook Page No. 280

Question 23.
Do you know what is a catalyst?
Answer:
A catalyst is a substance which regulates the rate of a given reaction without itself finally undergoing any chemical change.

10th Class Chemistry Textbook Page No. 281

Question 24.
Do you know how the police detect whether suspected drivers have consumed alcohol or not?
Answer:

1. The police officer asks the suspect to blow air into a plastic bag through a mouth piece of the detecting instrument which contains crystals of potassium-di-chromate (K₂Cr₂O₇).
2. As K₂Cr₂O₇ is a good oxidizing agent, it oxidizes any ethanol in the driver’s breath to ethanal and ethanoic acid.
3. Orange Cr₂O₇^2- changes to bluish green Cr³⁺ during the process of the oxidation of alcohol.
4. The length of the tube that turned into green is the measure of the quantity of alcohol that had been drunk.
5. The police even use the IR Spectra to detect the bonds C – OH and C – H of CH₃ – CH₂OH.

10th Class Chemistry Textbook Page No. 283

Question 25.
What are esters?
Answer:

The compounds which contain the functional group and the general formula R – COO – R’, where R and R’ are alkyl groups or phenyl groups, are known as “Esters”.

10th Class Chemistry Textbook Page No. 284

Question 26.
What is a true solution?
Answer:
A true solution is that in which the solute particles dispersed in the solvent are less than 1 nm in diameter.

10th Class Chemistry Textbook Page No. 286

Question 27.
What is the action of soap particles on the greasy cloth?
Answer:

• Soaps and detergents make oil and dirt present on the cloth come out into water, thereby making the cloth clean.
• Soap has one polar end and one non-polar end.
• The polar end is hydrophilic in nature and this end is attracted towards water.
• The non-polar end is hydrophobic in nature and it is attracted towards grease or * . ; oil on the cloth, but not attracted towards water.
• When soap is dissolved in water, its hydrophobic ends attach themselves to dirt and remove it from the cloth.
• The hydrophobic end of the soap molecules move towards the dirt or grease particles. ’
• The hydrophobic ends attach to the dirt particle and try to pull out.
• The molecules of soap surround the dirt particle at the centre of the cluster and form a spherical structure called micelle.
• These micelles remain suspended in water like particles in a colloidal solution.
• The various micelles present in water do not come together to form a precipitate as each micelle repels the other because of the ion-ion repulsion.
• Thus, the dust particles remain trapped in micelles and are easily rinsed away with water.
• Hence, soap micelles remove dirt by dissolving it in water.

10th Class Chemistry Textbook Page No. 280

Question 28.
Why we are advised not to use animal fats for cooking?
Answer:

• Animal fats have recently been implicated as the cause of heart disease and obesity. So, we are advised not to use animal fats for cooking.
• Excess animal fat is stored in lipocytes, which expand in size until the fat is used for fuel.

Question 29.
Which oil is recommended for cooking? Why?
Answer:
Canola oil :

• A recent entrant into the Indian market Canola is flying off the shelves.
• Canola oil which is made from the crushed seeds of the Canola plant, is said to be amongst the healthiest of cooking oils.
• It has the lowest saturated fat content of any oil.
• It’s seen as a healthy alternative as it’s rich in monosaturated fats and is high in omega-3 and omega a fats.
• It has a medium smoking point and is an oil that works well for fruits, baking, sauteing, etc.

10th Class Chemistry 14th Lesson Carbon and its Compounds Activities

Activity – 1

Question 1.
Observe the structural formula of the following hydro carbons and write their names in your notebook.
Answer:
1) CH₃ – CH₂ – CH = CH₂
Sol. But-l-ene

Sol. 2-Methyl butane

3) CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – CH₃
Sol. Hexane

Sol. 3-Methyl, but-l-ene

5)

Sol. Prop-l-yne

Activity-2

Question 2.
Read the names of the following hydro carbons and draw their structures in your notebook.
1. 2,2-Dimethyl hexane
Sol.

2. But-l-yne
Sol. CH₃ – CH₂ – C = CH

3. 3-Methyl Pent-2-ene
Sol.

4. But-1.2-diene
Sol. CH₃ – CH₃ = c = CH₂

5. Hept-2 en, 4-yne
Sol.

Activity – 3

Question 3.
Write an activity to show esterification reactions.
Answer:
The compound formed is ester. The process is called esterification.

1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
2. Warm it in a water bath or a beaker containing water for at least five minutes.
3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
4. We will notice that the resulting mixture is sweet odoured subatance.
5. This substance is nothing but ethyl acetate, an ester.
6. This reaction is called esterification reaction.

Activity – 4

Question 4.
Write an activity to show soap solution separates oil from water.
Answer:

1. Take about 10 ml of water each in two test tubes.
2. Add a drop of oil to both the test tubes.
3. Label them as A and B.
4. Add a few drops of soap solution to test tube B.
5. Now shake both the test tubes vigorously for the same period of time.
6. We can see the oil and water layers separately in both the test tubes immediately after we stop shaking them.
7. Leave the test tubes undisturbed for sometime and observe.
8. The oil layer separates out first in which test tube we added drops of soap solution.

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 13 Principles of Metallurgy Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 13th Lesson Principles of Metallurgy

10th Class Chemistry 13th Lesson Principles of Metallurgy Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Can you mention some articles that are made up of metals?
Answer:
Jewellery, conducting wires, utensils, etc.

Question 2.
Do metals exist in nature in the form same as that we use in our daily life?
Answer:
No, they exist as ores and minerals.

Question 3.
Have you ever heard the words like ore, mineral and metallurgy?
Answer:
Yes, these words are related to extraction of metals.

Question 4.
Do you know how these metals are obtained?
Answer:
These metals are generally extracted from their ores.

Improve Your Learning

Question 1.
List three metals that are found in nature as oxide ores.
Answer:
The three metals that are found in nature as oxide ores are

1. Bauxite (Al₂O₃ 2H₂O)
2. Haematite (Fe₂O₃)
3. Zincite (ZnO).

Question 2.
List three metals that are found in nature in uncombined form.
Answer:
The three metals that are found in nature in uncombined form are

1. Gold
2. Silver
3. Platinum.

Question 3.
Write a note on dressing of ore in metallurgy.
(OR)
What is concentration of Ore? List various physical methods that are used to enrich the ore.
Answer:

• Ores that are mined from the earth are usually contaminated with large amount of impurities such as soil and sand, etc.
• Concentration or dressing means, simply getting rid of as much of the unwanted rocky material as possible from the ore. The impurities like sand and clay are called gangue.

The physical methods adopted in dressing of the ore or enriching the ore depends upon the difference between physical properties of ore and gangue.

Methods of dressing or concentration of the ore:
1. Hand picking :
If the ore particles and the impurities are different in one of the properties like colour, size, etc. using that property, the ore particles are handpicked separating them from other impurities.

2. Washing:

• We use washing method with water to separate dust from rice, dal and vegetable fruits, etc.
• Ore particles are crushed and kept on a slopy surface. They are washed with controlled flow of water. Less densive impurities are carried away by water flow, leaving the more densive ore particles behind.

3. Froth flotation:
This method is mainly useful for sulphide ores. The ore with impurities is finely powdered and kept in water taken in a flotation cell. Air under pressure is blown to produce froth in water. Froth so produced, takes the ore particles to the surface whereas impurities settle at the bottom. Froth is separated and washed to get ore particles.

4. Magnetic separation:
If the ore or impurity, one of them is magnetic substance and the other is non-magnetic substance, they are separated using electromagnets.

Question 4.
What is an ore? On what basis a mineral is chosen as an ore?
Answer:
Ore :
A mineral from which a metal can be extracted economically and conveniently is called ore.
A mineral is chosen as an ore if the mineral is economical and profitable to extract.

Example:
Aluminium is the common metal in the Earth’s crust in all sorts of minerals. It is economically feasible and profitable to extract from bauxite which contains 50-70% of aluminium oxide.

Question 5.
Write the names of any two ores of iron.
Answer:
The names of two ores of iron :

1. Haematite (Fe₂O3)
2. Magnetite (Fe₃O₄).

Question 6.
How do metals occur in nature ? Give examples to any two types of minerals.
Answer:

• The earth’s crust is the major source of metals.
• Sea water also contains some soluble salts such as sodium chloride and magnesium chloride etc.
• Some metals like gold (Au), silver (Ag) and copper (Cu) are available in nature in free state as they are least reactive.
• Other metals are found in nature in the combined form due to their more reactivity.
• The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals.

Examples :
Minerals in oxide form :
Bauxite, Zincite, Magnetite, etc.

Minerals in sulphide form :
Copper iron pyretes, Galena, etc.

Question 7.
Write short notes on froth flotation process.
(OR)
Which method is useful for concentration of sulphide ore? Explain the method.
Answer:
Froth Floatation process :

• This method is mainly useful for sulphide ores which have no wetting property whereas impurities get wetted.
• The ore with impurities is finely powdered and kept in water taken in a floatation cell.

Froth floatation process for the concentration of sulphide ores

• Air under pressure is blown to produce froth in water.
• Froth so produced takes the ore particles to the surface whereas impurities settle at the bottom.
• Froth is separated and washed to get ore particles.

Question 8.
When do we use magnetic separation method for concentration of an ore? Explain with an example.
(OR)
Write the name of the method we use to separate the ore or impurity in which one of them is magnetic substance. Draw a neat diagram indicating the method.
Answer:
If the ore or impurity, one of them is magnetic substance and the other non-magnetic substance they are separated using electromagnets.
Ex :
Iron from iron ore (Fe₃O₄) is separated from its impurity by passing through a magnetic field. The field attracts magnetic ore (Iron) and repels the non-magnetic impurities.

Question 9.
Write short notes on each of the following :
i) Roasting
ii) Calcination
iii) Smelting
Answer:
i) Roasting :
Roasting is a pyrochemical process in which the ore is heated in the presence of oxygen or air, below its melting point. Generally, reverberatory furnace is used for roasting.
Ex:
Zinc blende on heating with oxygen in reverberatory furnace forms zinc oxide as solid and liberating sulphur dioxide as gas.
2Zns_(s) + 3O_2(g) → 2ZnO_(s) + 2SO_2(g)

ii) Calcination :
Calcination is a pyrochemical process in which the ore is heated in the absence of air. The ore gets generally decomposed in this process.
Ex: MgCO_3(s) → MgO_(s) + CO_2(g)

iii)Smelting:
Smelting is a pyrochemical process, in which the ore is mixed with flux and fuel, then is strongly heated.

Question 10.
What Is the difference between roasting and calcination? Give one example for each.
(OR)
Roasting and Calcination are the methods to extract crude metals from ores. What is the difference between Roasting and Calcination?
Answer:

Question 11.
Define the terms:
i) gangue
ii) slag.
Answer:
i) Gangue:
The impurity present in the ore is called gangue.
(or)
Unwanted impurity associated with ore.

ii) Slag:
The impurities found from molten metal during poling process of refining are called slag.

Question 12.
Magnesium is an active metal if it occurs as a chloride in nature, which method of reduction is suitable for its extraction?
Answer:

• The method of reduction which is useful for chloride of magnesium which is active is electrolytic reduction.
• Fused MgCl₂ is electrolysed with steel cathode (-) and graphite anode (+). The metal (Mg) will be deposited at cathode and chlorine gas liberates at the anode.
  At cathode : Mg²⁺ + 2 e^– → Mg
  At anode : 2 Cl^– → Cl₂ + 2e^–

Question 13.
Mention two methods which produce very pure metals.
Answer:
Methods which produce very pure metals are :

1. Electrolytic Reduction
2. Smelting.

(OR)

1. Distillation
2. Poling.

Question 14.
Which method do you suggest for extraction of high reactivity metals? Why?
Answer:
1.The only method which is suitable for extraction of high reactivity metals is electrolysis of their fused compounds.

2. Other methods are not suitable due to following reasons :
a) Simple reduction methods like heating with C, CO, etc. to reduce the ores of these metals are not suitable because the temperature required for the reduction is too high and more expensive.

b) Electrolysis of their aqueous solutions are also not preferable because water in the solution would be discharged at the cathode in preference to the metal ions.

Question 15.
Suggest an experiment to prove that the presence of air and water is essential for corrosion. Explain the procedure.
(OR)
Write the experimental procedure to prove that water and air are essential for rusting of iron articles.
(OR)
How can you prove that the presence of air and humid are essential for corrosion?
(OR)
Explain in brief, an experiment to prove that the presence of air and water are essential for corrosion.
Write the precautions to be taken in the experiment to show air and water are essential for rusting iron articles and also write the experimental procedure.
Answer:
Aim :
To prove that the presence of air and water is essential for corrosion or for rusting of iron articles.

Apparatus :
3 boiled test tubes, 3 corks, boiled distilled water, anhydrous calcium chloride, clean iron nails.

Procedure :

• Take three test tubes and place clean iron nails in each of them.
• Label these test tubes A, B and C. Pour some water in test tube A and cork it.
• Pour boiled distilled water in test tube B, add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
• Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture.
• Leave these test tubes for a few days and then observe.
• We will observe that iron nails rust in test tube A, but they do not rust in test tubes B and C.

Observation :

• In test tube A, the nails are exposed to air and water. Hence, the nails rusted.
• In test tube B, the nails are exposed only to water, but not to air, because the oil float on water and prevent the air from dissolving in the water. Hence, the nails are not rusted.
• In test tube C, the nails are exposed to dry air, because anhydrous calcium chloride will absorb the moisture, if any, from the air. Hence, the nails are not rusted.

Conclusion :
From the above experiment we can prove that air and water are essential for corrosion.

Question 16.
Collect information about extraction of metals of low reactivity silver, platinum and gold and prepare a report.
(OR)
Prepare a report with the collected information about extraction of metals of low reactivity silver, platinum and gold.
Answer:
Extraction of Silver:

1. Silver can be extracted from Ag2S by using displacement from aqueous solution.
2. Ag₂S is dissolved in KCN solution to get dicyanoargentate (I) ions.
3. From these ions Ag is precipitated by treating with Zinc dust powder.
   Ag₂S + 4 CN^– → 2 [Ag(CN)₂]^– + S^2-
   2[Ag(CN)₂]^–_(aq) + Zn_(s) → [Zn(CN)₄]^2-_(aq) + 2Ag_(s)

Extraction of Gold :

1. Gold is extracted from gold ore like electrum. Impurities are separated from the gold by treating gold ore with a weak cyanide solution.
2. Zinc is added and a chemical reaction takes place which separates the gold from ore.
3. Pure gold is removed from the solution with a filter press.

Extraction of Platinum:

1. The extraction of platinum from ore is a complex process and includes milling the ore, a froth flotation process, and smelting at high temperatures.
2. This removes the base metals, notably iron and sulphur and concentrate platinum.

Question 17.
Draw the diagram showing
i) Froth flotation
ii) Magnetic separation.
(OR)
Draw a neat diagram and label the parts that shows froth floatation process for the concentration of sulphide ore.
Answer:
i)

ii)

Question 18.
Draw a neat diagram of reverberatory furnace and label it neatly.
(OR)
Draw a neat labelled diagram of a reverberatory furnace.
Answer:

Question 19.
What is activity series? How does it help in extraction of metals?
Answer:
Activity Series :
We can arrange metals in descending order of their reactivity. This series of writing metals is called activity series.

Uses of activity series the extraction of metals :

• Activity series is extremely useful in extraction of metals because we can judge the nature of metal and how it exists.
• High reactive metals like K, Na, Ca, Mg and Al are so reactive that they are never found in nature in free state.
• The moderate reactive metals like Zn, Fe, Pb, etc. are found in the earth’s crust mainly as oxides, sulphides and carbonates.
• The least reactive metals like Au, Ag, Pt are found even in free state in nature.

Question 20.
What is thermite process? Mention its applications in daily life.
Answer:
Thermite Process :

• Thermite process is the chemical reaction which takes place between metal oxides and aluminium.
• When highly reactive metals such as sodium, calcium, aluminium, etc. are used as reducing agents, they displace metals of lower reactivity from the compound.
• This reaction is highly exothermic. The amount of heat evolved is so high that the metals can he directly converted into molten state.

Applications in daily life :

• The reaction if Iron (III) oxide (Fe203) with aluminium is used to join railing of railway tracks or cracked machine parts.
  2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe + Heat.
• And also used for joining of cracked metal utensils in the house.

Question 21.
Where do we use hand picking and washing methods in our daily life ? Give examples.
How do you correlate these examples with enrichment of ore?
Answer:
Daily life examples for hand picking:

• Separating mud particles from rice is an example for hand picking because the colour and size of these two are different.
• Similarly, the ore particles and the impurities are different in one of the properties like colour, size, etc. are separated by hand picking.

Daily life examples for washing :

• We can clean some vegetables like potatoes by controlled flow of water. Less densive impurities are carried away by the flow leaving the more densive potatoes.
• Similarly, ores are washed with controlled flow of water. Less densive impurities i are carried away by water flow, leaving the more densive ore particles behind.

Fill In The Blanks

1. The method is suitable to enrich the sulphide ores.
2. Arranging metals in the decreasing order of their reactivity is called
3. The method suitable for purification of low boiling metals.
4. Corrosion of iron occurs in the presence of and
5. The chemical process in which the ore is heated in the absence of air is called
Answer:

1. Froth flotation
2. activity series
3. distillation
4. air, water
5. calcination

Multiple Choice Questions

1. The impurity present in the ore is called ………………….
A) Gangue
B) Flux
C) Slag
D) Mineral
Answer:
A) Gangue

2. Which of the following is a carbonate ore?
A) Magnesite
B) Bauxite
C) Gypsum
D) Galena
Answer:
A) Magnesite

3. Which of the following is the correct formula of Gypsum?
A) CuSO₄ • 2 H₂O
B) CaSO₄ • ½ H₂O
C) CuSO₄ • 5 H₂O
D) CaSO₄ • 2 H₂O
Answer:
D) CaSO₄ • 2 H₂O

4. The oil used in the froth flotation process is
A) kerosene oil
B) pine oil
C) coconut oil
D) olive oil
Answer:
B) pine oil

5. Froth flotation is method used for the purification of ………………. ore.
A) sulphide
B) oxide
C) carbonate
D) nitrate
Answer:
A) sulphide

6. Galena is an ore of ………………..
A) Zn
B) Pb
C) Hg
D) Al
Answer:
B) Pb

7. The metal that occurs in the native form is ………………
A) Pb
B) Au
C) Fe
D) Hg
Answer:
B) Au

8. The most abundant metal in the earth’s crust is …………………
A) silver
B) aluminium
C) zinc
D) iron
Answer:
B) aluminium

9. The reducing agent in thermite process is ………………….
A) Al
B) Mg
C) Fe
D) Si
Answer:
A) Oxidise

10. The purpose of smelting an ore is to ……………….. it.
A) Oxidise
B) Reduce
C) Neutralise
D) None of these
Answer:
B) Reduce

10th Class Chemistry 13th Lesson Principles of Metallurgy InText Questions and Answers

10th Class Chemistry Textbook Page No. 238

Question 1.
How are the metals present in nature?
Answer:
Some metals like gold (Au), silver (Ag) and copper (Cu) are available in nature in free sjate. Other metals mostly are found in nature in the combined form.

10th Class Chemistry Textbook Page No. 240

Question 2.
What metals can we get from the ores mentioned in the Table – 1?
Answer:
The metals are Aluminium (Al), Copper (Cu), Magnesium (Mg), Silver (Ag), Manganese (Mn), Iron (Fe), Zinc (Zn), Sodium (Na), Mercury (Hg), Lead (Pb), Calcium (Ca).

Question 3.
Can you arrange metals in the order of their reactivity?
Answer:
The order of reactivity is like this : Ag < Cu < Pb < Mn < Fe < Zn < Al < Mg < Ca < Na.

Question 4.
What do you notice in Table – 2?
Answer:
We notice that ores of many metals are oxides and sulphides. ______

Question 5.
Can you think how we get these metals from their ores?
Answer:
We can get metals from their ores by using various extracting techniques.

Question 6.
Does the reactivity of a metal and form of its ore (oxides, sulphides, chlorides, carbonates, sulphates) has any relation with process of extraction?
Answer:
Yes, they have relation. Metals like K, Na, Ca, Mg and Al are so reactive. They exist in all forms whereas moderate reactive metals like Zn, Fe, Pb, etc. exist as oxides, sulphides and carbonates. The least reactive metals are found even in free state.

Question 7.
How are metals extracted from mineral ores?
Answer:
The extraction of a metal from its ores involves mainly in three states. They are :

1. Concentration or Dressing
2. Extraction of crude metal
3. Refining or purification of the metal.

10th Class Chemistry Textbook Page No. 248

Question 8.
Do you know why corrosion occurs?
Answer:
Corrosion occurs due to reaction of metal with both air and water.

Question 9.
What are the conditions under which iron articles rust?
Answer:
Iron articles get rust due to both air and water.

10th Class Chemistry Textbook Page No. 251

Question 10.
What is the role of furnace in metallurgy?
Answer:
Furnace is used to carry out pyrochemical process in metallurgy.

Question 11.
How do furnaces bear large amounts of heat?
Answer:
Furnaces have metallic lining. So they bear large heats.

Question 12.
Do all furnaces have same structure?
Answer:
No, they have different structures.

10th Class Chemistry Textbook Page No. 239

Question 13.
Do you agree with the statement “All ores are minerals but all minerals need not be ores.? Why?
Answer:

• Yes, I agree with the statement. The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals whereas ore is a mineral from which the metal is profitably extracted.
• For example, aluminium exists in two mineral forms that is clay and bauxite. But aluminium is mainly extracted from bauxite which contains 70% aluminium oxide.
• So Bauxite is an ore of aluminium whereas clay is not ore.
• So all ores are minerals but all minerals need not be ores.

10th Class Chemistry 10th Lesson Principles of Metallurgy Activities

Activity – 1
1. How do you classify ores based on their formula?
Answer:
1) Look at the following ores.

2) Identify the metal present in each ore.
3) Classification of ores as oxides, sulphides, chlorides, carbonates, and sulphates as follows :

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 11 Electric Current Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 11th Lesson Electric Current

10th Class Physics 11th Lesson Electric Current Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What do you mean by electric current?
(OR)
Define electric current.
Answer:
Electric current is defined as the amount of charge crossing any cross-section of the conductor in one second.

Question 2.
Which type of charge (positive or negative) flows through an electric wire when it is connected in an electric circuit?
Answer:
Negative type of charge flows through an electric wire when it is connected in an electric circuit.

Question 3.
Is there any evidence for the motion of charge in daily life situations?
Answer:
Yes, lightning is a live example.

Improve Your Learning

Question 1.
Explain how electron flow causes electric current with Lorentz – Drude theory of electrons. (AS1)
(OR)
How does electron flow cattle elfectric current with Lorentz – Drude theory of electrons? Explain.
Answer:
Lorentz – Drude theory :

1. Lorentz – Drude proposed that conductors like metals contain a large number of free electrons.
2. The positive ions are fixed in their locations. The arrangement of the positive ions is called lattice.
3. The negative ions (electrons) move randomly in lattice in an open circuit.
4. When the lattice is closed the electrons are arranged in ordered motion.
5. When the electrons are in order motion, there will be a net charge (crossing through any cross section.
6. This order motion of electrons is called electric current.

Question 2.
How does a battery work? Explain. (AS1)
(OR)
How does a battery maintain a constant potential difference between its terminals?
Answer:
Working of a battery :

• A battery consists of two metal plates (positive electrode = anode and negative electrode = cathode) and a chemical (electrolyte).
• The electrolyte between the two metal plates consists of positive and negative ions which move in opposite directions.
• The electrolyte exerts a chemical force on these ions and makes them move in a specified direction.
• Depending upon the nature of the chemical, positive ions move towards one of the plates and accumulate on that plate.
• As a result of this accumulation of charges on this plate it becomes anode.
• Negative ions move in a direction opposite to the motion of positive ions and accumulate on the other plate.
• As a result of this the plate becomes negatively charged called cathode.
• This accumulation of different charges on respective plates continues till both plates are sufficiently charged.
• But the ions in motion experience electric force when sufficient number of charges are accumulated on the plates.
• The motion of ions continues towards their respective plates till the chemical force is equal to electric force.
• Thus the battery works.

Question 3.
Write the difference between potential difference and emf. (AS1)
Answer:
Potential Difference:
Work done by the electric force on unit charge is called potential difference.
\(\mathbf{V}=\frac{\mathbf{W}}{q}=\frac{\mathbf{F} l}{\mathbf{q}}\)

Electromotive force (emf):
The work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.
\(\varepsilon=\frac{W}{q}=\frac{F d}{q}\)

Question 4.
How can you verify that the resistance of a conductor is temperature dependent? (AS1)
(OR)
How do you prove increase in temperature affects the resistance with an activity?
Answer:
Resistance :
The resistance of a conductor is the obstruction offered to the flow of electrons in a conductor.

Resistance is temperature dependent:
Aim:
To show that the value of resistance of a conductor depends on temperature for constant voltage between the ends of the conductor.

Materials required :

1. A bulb
2. A battery
3. Key
4. Insulated wire
5. Multimeter

Procedure :

1. Take a bulb and measure the resistance when it is in open circuit using a multimeter.
2. Note the value of resistance in your notebook.
3. Connect a circuit with components as shown in figure.
4. Switch on the circuit. After few minutes, measure the resistance of the bulb again.
5. Note this value in your notebook.

Observation :

1. The value of resistance of the bulb in second instance is more than the resistance of the bulb in open circuit.
2. The bulb gets heated.

Result:
The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.

Question 5.
What do you mean by electric shock? Explain how it takes place. (AS1)
Answer:
Electric shock:
The electric shock is combined effect of potential difference, electric current, and resistance of the human body.

• An electric shock can be experienced when there exists a potential difference between one part of the body and another part.
• When current flows through human body, it chooses the path which offers low resistance.
• The resistance of a body is not uniform throughout it.
• As long as current flow continues inside the body, the current and resistance of human body go on changing inversely.
• This is called the electric shock.

Question 6.
Derive \(\mathbf{R}=\frac{\rho l}{\mathbf{A}}\). (AS1)
(OR)
What are laws of resistance and derive a formula for resistance.
Answer:
Resistance of a conductor is directly proportional to the length of the conductor,
i.e., R ∝ l ………………….. (1)
Resistance of a conductor is inversely proportional to the cross-section area of the conductor.
i.e., R ∝ \(\frac{1}{\mathrm{~A}}\) ………………….. (2)
From (1) and (2) R ∝ \(R \propto \frac{l}{A} \Rightarrow R=\frac{\rho l}{A}\)
where ρ is a constant,
ρ is called specific resistance or resistivity.

Question 7.
How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross-section area and temperature? (AS1)
Answer:

• Collect manganin wires of different lengths with the same cross-sectional areas.
• Make a circuit as shown in figure.
• Connect one of the manganin wires between the ends P and Q.
•  Measure the value of the current using the ammeter.
• Repeat the same for other lengths of the wires.
• Note the values of currents.
• We notice that the current decreases with increase in the length of the wire.
  ∴ R ∝ l (at constant temperature and cross-section area) …………… (1)
• Do the same with manganin wires with equal lengths but different cross-section area.
• We notice that the resistance was more when the cross-section area was less.
  ∴ R ∝ \(\frac{1}{\mathrm{~A}}\) ………………. (2)
  ∴ R ∝ \([latex]\frac{1}{\mathrm{~A}}\)[/latex]
  Thus we verify l and A.

Question 8.
Explain Kirchhoff’s laws with examples. (AS1)
(OR)
Write two examples of Kirchhoffs laws and explain it.
Answer:
Kirchhoff’s laws :
Two simple rules called Kirchhoff’s rules are applicable to any DC circuit containing batteries and resistors connected in any way.
The two laws are (i) Junction law and (ii) Loop law.

i) Junction law :

Here P is called junction point where conducting wires meet. The junction law states that, at any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents leaving the junction.
i.e., I₁ + I₄ + I₆ = I₂ + I₃ + I₅
This law is based on the conservation of charge.

ii) Loop law:

Loop law states that, the algebraic sum of the increases and decreases in potential difference (voltage) across various components of the circuit in a closed circuit loop must be zero.

This law is based on the conservation of energy.

Question 9.
What is the value of 1 KWH in Joules? (AS1)
Answer:
1 KWH = 1 KW x 1h
= 1000 W × 60 min = 1000 W × 60 × 60 s = 3.6 × 10₆ Ws = 3.6 × 10₆ J.
∴ 1 KWH = 3.6 × 10₆ J.

Question 10.
Explain overloading of household circuit. (AS1)
Answer:

• Electricity enters our homes through two wires called lines. These lines have low resistance and the potential difference between the wires is usually about 240 V.
• All electrical devices are connected in parallel in our home. Hence, the potential drop across each device is 240 V.
• Based on the resistance of each electric device, it draws some current from the supply. Total current drawn from the mains is equal to the sum of the currents passing through each device.
• If we add more devices to the household circuit the current drawn from the mains also increases.
• This leads to overheating and may cause a fire. This is called “overloading”.

Question 11.
Why do we use fuses in household circuits? (AS1)
(OR)
What is the use of fuses?
Answer:

• The fuse consists of a thin wire of low melting point.
• When the current in the fuse exceeds 20 A, the wire will heat up and melt.
• The circuit then becomes open and prevents the flow of current into the household circuit.
• Hence all the electric devices are saved from damage that could be caused by overload.
• Thus we can save the household wiring and devices by using fuses.

Question 12.
Deduce the expression for the equivalent resistance of three resistors connected in series. (AS1)
(OR)
Derive R = R₁ + R₂ + R₃
(OR)
The second end of a first resistor is connected to first end of second resistor. Then how are the resistors connected? Derive the expression for the resultant resistance of this connection.
Answer:
Series connection:
In series connection of resistors, there is only one path for the flow of current in the circuit. Hence, the current in the circuit is equal to I.
According to Ohm’s law,
∴ V₁ = IR₁ ; V₂ = IR₂ and V₃ = IR₃.
⇒ Let R be the equivalent resistance of the combination of resistors in series.

Also V = I R_eq
V = V₁ + V₂ + V₃
I R_eq = IR₁ + IR₂ + IR₃
⇒ I R_eq = I (R₁ + R₂ + R₃)
⇒ R_eq = R₁ + R₂ + R₃
∴ The sum of individual resistances is equal to their equivalent resistance when the resistors are connected in series.

Question 13.
Deduce the expression for the equivalent resistance of three resistors connected in parallel. (AS1)
(OR)
Derive : \(\frac{1}{\mathbf{R}}=\frac{1}{\mathbf{R}_{1}}+\frac{1}{\mathbf{R}_{2}}+\frac{1}{\mathbf{R}_{3}}\)
(OR)
Explain the expression for the equivalent resistance of three resistors which are connected in parallel.
(OR)
If all the first ends of resistors are connected to and second ends are connected to another point, then what type of connection is this? Derive the resultant resistance for this connection.
Answer:
Parallel Connection :
In parallel connection of resistors, there is same potential difference at the ends of the resistors. Hence the voltage in the circuit is equal to V.
Let Ip I2 and I3 be the currents flowing through R₁, R_2, and R₃ resistors respectively.
Hence, we can write I = I₁ + I₂ + I₃.
According to the Ohm’s law,

∴ The equivalent resistance of a parallel combination is less than the resistance of each of the resistors.

Question 14.
Silver is a better conductor of electricity than copper. Why do we use copper wire for conduction of electricity? (AS1)
Answer:
Silver is costlier than copper. So, we use copper wire for conduction of electricity even though silver is a better conductor of electricity.

Question 15.
Two bulbs have ratings 100 W, 220 V and 60 W, 220 V. Which one has the greater resistance? (AS1)
Answer:

∴ The second bulb possessing 60 W, 220 V has the greater resistance.

Question 16.
Why don’t we use series arrangement of electrical appliances like bulb, television, fan, and others in domestic circuits? (AS1)
Answer:

• If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases. To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series. They are connected in parallel.
• In series combination same current passes through all resistors. This is not suggestable for household appliances. Hence, they are connected in parallel.

Question 17.
A wire of length 1 m and radius 0.1 mm has a resistance of 100 Ω. Find the resistivity of the material. (AS1)
Answer:
1) Given l = 1 m, r = 0.1 mm = 10^-4 m, R = 100 Ω

Question 18.
Why do we consider tungsten as a suitable material for making the filament of a bulb? (AS2)
(OR)
What is the reason for using Tungsten as a filament in electric bulb?
Answer:
Tungsten has higher resistivity values and melting point. So, we consider tungsten as a suitable material for making the filament of a bulb.

Question 19.
Are the head lights of a car connected in series or parallel? Why? (AS2)
Answer:
The headlights of a car are connected in parallel.
Reason :

• When they are connected in parallel, same voltage (RD) will be maintained in the two lights.
• If one of the light damaged, the other will work without any disturbance.

Question 20.
Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?
Answer:

• The electric appliances are connected in parallel in a household circuit. Because in parallel wiring if any electric appliance is switched off, other appliances don’t get off.
• If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases.
• To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series.

Question 21.
Suppose that you have three resistors each of value 30Ω. How many resistors can you obtain by various combinations of these three resistors? Draw diagrams in support of your predictions. (AS2)
Answer:
Let R₁ = 30Ω, R₂ = 30Ω, R₃ = 30Ω
We get different resistors by different combinations as shown below.

Question 22.
State Ohm’s law. Suggest an experiment to verify it and explain the procedure. (AS3)
How do you prove experimentally the ratio V/l is a constant for a given conductor?
Answer:
Ohm’s law :
The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.

Verification :
Aim :
To verify Ohm’s law or to show that \(\frac{\mathrm{V}}{\mathrm{I}}\) = constant for a conductor.

Materials required :
6V Battery eliminator, 0 to 1A Ammeter, 0 – 6V volt meter, copper wires, 50 cm manganin coil, Rheostat, switch and 3V LED, etc.

Procedure :

• Complete the circuit as shown in figure. Knob should be adjusted to 4.5V at battery eliminator.
• Using Rheostat change the potential difference between two ends of manganin wire from 0V to 4.5V (maximum).
• By using Rheostat adjust the potential difference 1V between two ends of manganin wire.
• Now observe the electric current through Ammeter in the circuit and note down in the following table.

• Using Rheostat change the potential difference with different values upto 4.5V and note down the current value (I) in the table.
• Take atleast five values of V and I and note down in the table.
• Find \(\frac{\mathrm{V}}{\mathrm{I}}\) for each set of values.
• We notice that \(\frac{\mathrm{V}}{\mathrm{I}}\) is a constant.
  V ∝ I ⇒ \(\frac{\mathrm{V}}{\mathrm{I}}\) = constant
  This constant is known as resistance of the conductor, denoted by R.
  ⇒ \(\frac{\mathrm{V}}{\mathrm{I}}\) = R
  ∴ Ohm’s law is verified.

How to Make Rheostat:

Make two holes at the two ends of 30cm Wooden scale. Through these holes fix two bolts with the help of nuts.Then take iron box filament i. e., nichrome wire and tie one end of thewire to the first bolt and wound wire with equal distance on the wooden scale to other end of the second bolt. Place this scale on the other scale perpendicularly as shown in the figure and stick them with glue. Now Rheostat is ready. Take support of your teacher to know the connection and functioning of Rheostat.

Question 23.
a) Take a battery and measure the potential difference. Make a circuit and measure the potential difference when the battery is connected in the circuit. Is there any difference in potential difference of battery? (AS4)
b) Measure the resistance of a bulb (filament) in open circuit with a multi-meter. Make a circuit with elements such as bulb, battery of 12 V and key in series. Close the key. Then again measure the resistance of the same bulb (filament:) for every 30 seconds. Record the observations in a proper table. What can you conclude from the above results? (AS4)
Answer:

a) When the battery is connected in a circuit, the voltage slowly decreases due to consumption of it. So, there is difference in voltage before using and after connecting.

b) After connecting battery (12 V), key in ammeter and bulb as shown in figure, we measure current (I) using the ammeter and voltage using multi-meter or voltmeter.

Note these values in the following table. Measure the resistance of the same bulb for every 30 seconds.

We conclude that the resistance is constant.

Question 24.
Draw a circuit diagram for a circuit in which two resistors A and B are connected in series with a battery and a voltmeter is connected to measure the potential difference across the resistor A. (AS5)
Answer:

V : Volt meter
A and B : Resistors
B : Battery
K: Key

Question 25.
How can you appreciate the role of a small fuse in house wiring circuit in preventing damage to various electrical appliances connected in the circuit? (AS7)
(OR)
We can save the household wiring and devices by using fuses. Write any four points by appreciating the role of fuse.
Answer:

• The fuse consists of a thin wire of low melting point. When the current in the fuse exceeds 20 A, the wire will heat up and melt.
• The circuit then becomes open and prevents the flow of current into the household circuit. So all the electric devices are saved from damage that could be caused by overload.
• Thus we can save the household wiring and devices by using fuses.
• In this way a small fuse prevents a great damage to costly electrical appliances in the circuit.

Question 26.
In the figure, the potential at A is………….. when the potential at B is zero. (AS7)

Answer:
Potential difference at A = V
Potential difference atB = V + 5 × 1 + 2 = 0 ⇒ V + 7V = 0
∴ V = – 7V

Question 27.
Observe the circuit and answer the questions given below. (AS7)

i) Are resistors C and D in series?
ii) Are resistors A and B in series?
iii) Is the battery in series with.any resistor?
iv) What is the potential drop across the resistor C?
v) What is the total emf in the circuit if the potential drop across resistor A is 6 V?
Answer:
The given circuit is written / drawn as
i) Yes, resistors ‘C’ and ‘D’ are connected in series. (Because, passing of the current is same to those resistors)
ii) No, resistors A’ and ‘B’ are not in series. (Because, different currents are passing through A and B. i.e., I₁ and I₂)
iii) The battery is in series with the resistor ‘A’. (Because, same current is passing through battery and resistor ‘A’, i.e., I)

iv) Potential drop across the resistor ‘C’
V₂ = V₃ + V₄
14V = V₃ + 8V
V₃ = 6V
Potential drop = 6V

v) Total emf
emf of combination of V₃ and V₄ = 14V ……………….. (1)
emf of combination of (1) and V₂ = 14 V ………………. (2)
emf of combination of (2) and V₁ = 6V + 14V = 20V
(Given, emf of ‘A’ = 6V)
Total emf = 20V

Question 28.
If the resistance of your body is 100000 Cl, what would be the current that flows in your body when you touch the terminals of a 12 V battery? (AS7)
Answer:
We know that, \(I=\frac{V}{R}\); here V = 12 V, R = 1,00,000Ω.
∴ The current passing through our body \(I=\frac{12 \mathrm{~V}}{100000 \Omega}\) = 0.00012 Ampere.

Question 29.
A uniform wire of resistance 100 Ω is melted and recast into wire of length double that of the original. What would be the resistance of the new wire formed? (AS7)
Answer:
Given R = 100 Ω
When ‘l = l’, R = 100 Ω.
When’l = 2l’, A’ = A / 2.

∴ Resistance is increased by four times.
∴ R = 4 × 1ooΩ = 400Ω.

Question 30.
A house has 3 tube lights, two fans and a Television. Each tube light draws 40 W. The fan draws 80 W and the Television draws 60 W. On the average, all the tube lights are kept on for five hours, two fans for 12 hours and the television for five hours every day. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWh. (AS7)
Answer:
Given 3 tube lights, two fans and a television.
Power consumed by 1 tube light = 40 W
∴ Power consumed by 3 tube lights = 3 × 40W = 120W
3 tube lights are kept on for five hours. So, consumption of power by 3 tube lights
= 5 × 120 W = 600 W ……………. (1)
Power consumed by 1 fan = 80 W
∴ Power consumed by 2 fans = 2x80W=160W
2 fans are kept on for 12 hours. So, consumption of power by 2 fans
= 12 × 160 W = 1920 W ……………. (2)
Power drawn by TV = 60 W
TV is kept on for 5 hours = 5 x 60 W = 300 W ………………. (3)
∴ Consumption of power in one day = (1) + (2) + (3)
= 600W+ 1920 W + 300 W = 2820 W = 2.820 KW
∴ Total consumption of power in 30 days at Rs. 3 per KW
= 2.820 × 30 × 3 = Rs. 253.80/-

Fill in The Blanks

1. The kilowatt hour is the unit of …………………..
2. A thick wire has ………………….. resistance than a thin wire.
3. An unknown circuit draws a current of 2 A from a 12 V battery. Its equivalent resistance is …………………..
4. The SI unit of potential difference is …………………..
5. The SI unit of current is …………………..
6. Three resistors of values 2Ω, 4Ω, 6Ω are connected in series. The equivalent resistance of combination of resistors is ……………………
7. Three resistors of values 2Ω, 4Ω, 6Ω are connected in parallel. The equivalent resistance of combination of resistors is ……………………
8. The power delivered by a battery of emf, 10 V is 10 W. Then the current delivered by the battery is ……………………
Answer:

1. electrical energy
2. less
3. 6 Ω
4. volt
5. Ampere
6. 12 Ω
7. \(\frac{11}{12} \Omega\)
8. 1 ampere

Multiple Choice Questions

1. A uniform wire of resistance 50 Ω. is cut into five equal parts. These parts are now connected in parallel. Then the equivalent resistance of the combination is
A) 2 Ω
B) 12 Ω
C) 250 Ω
D) 6250 Ω
Answer:
A) 2 Ω

2. A charge is moved from a point A to a point B. The work done to move unit charge during this process is called
A) potential at A
B) potential at B
C) potential difference between A and B
D) current from A to B
Answer:
C) potential difference between A and B

3. Joule/ coulomb is the same as
A) 1 – watt
B) 1 – volt
C) 1- ampere
D) 1 – ohm
Answer:
B) 1 – volt

4. The current in the wire depends
A) only on the potential difference applied
B) only on the resistance of the wire
C) on potential difference and resistance
D) none of them
Answer:
C) on potential difference and resistance

5. Consider the following statements.
a) In series connection, the same current flows through each element.
b) In parallel connection, the same potential difference gets applied across each element
A) both a and b are correct
B) a is correct but b is wrong
C) a is wrong but b is correct
D) both a and b are wrong
Answer:
A) both a and b are correct

10th Class Physics 11th Lesson Electric Current InText Questions and Answers

10th Class Physics Textbook Page No. 179

Question 1.
Does motion of charge always lead to electric current?
Answer:
Yes, it does.

Question 2.
Take a bulb, a battery, a switch and few insulated copper wires to the terminals of the battery through the bulb and switch. Now switch on the circuit and observe the bulb. What do you notice?
Answer:
The bulb glows.

10th Class Physics Textbook Page No. 180

Question 3.
Can you predict the reason for the bulb not glowing in situations 2 and 3?
Answer:
Yes, in situation 2 – there is no charge to travel in the circuit as the battery is disconnected. So, the bulb isn’t glowing.

In situation 3, we replaced the copper wires with nylon wires. Nylon is not a conductor. So, the bulb isn’t glowing.

Question 4.
Why do all materials not act as conductors?
Answer:
In conductors the gap between the atoms is very less. So, the transfer of energy is easy. But in other materials the gap is more. So, the transfer of energy is not possible.

Question 5.
How does a conductor transfer energy from source to bulb
Answer:

• A source has chemical energy which transfers electrons to the conductor.
• The conductor carries the electrons to the bulb when connected.
• Thus, the conductor transfers energy from source to bulb.

Question 6.
What happens to the motion of electrons when the ends of the conductor are connected to the battery?
Answer:
The energy transfer takes place from battery to the bulb through conductor.

10th Class Physics Textbook Page No. 181

Question 7.
Why do electrons move in specified direction?
Answer:
The electrons move in specified direction when the ends of the conductpr are connected to the terminals of a battery.
A uniform electric field is set up throughout the conductor. This field makes the electrons move in a specified direction.

Question 8.
In which direction do the electrons move?
Answer:
In a direction opposite to the direction of the field.

Question 9.
Do the electrons accelerate continuously?
Answer:
No, they lose energy and are again accelerated by the electric field.

Question 10.
Do they move with a constant speed?
Answer:
Yes, they move with a constant average Speed.

Question 11.
Why does a bulb glow immediately when we switch on?
Answer:
When we switch on any electric circuit, irrespective of length of the conductor, an electric field is set up throughout the conductor instantaneously due to the voltage of the source connected to the circuit.

Question 12.
How can we decide the direction of electric current?
Answer:
By the signs of the charge and drift speed.

10th Class Physics Textbook Page No. 183

Question 13.
How can we measure electric current?
Answer:
An ammeter is used to measure electric current.

Question 14.
Where do the electrons get energy for their motion from?
Answer:
From an electric field set up throughout the conductor.

Question 15.
Can you find the work done by the electric force?
Answer:
Yes. With the help of the formula W = F_el, we can find the work done by the electric force.

Question 16.
What is the work done by the electric force on unit charge?
Answer:
Work done by the electric force on unit charge \(\mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}=\frac{\mathrm{F}_{\mathrm{e}} l}{\mathrm{q}}\). It is called Potential difference.

10th Class Physics Textbook Page No. 184

Question 17.
What is the direction of electric current in terms of potential difference?
Answer:
Electrons move from low potential to high potential.

Question 18.
Do positive charges move in a conductor? Can you give an example of this?
Answer:
No, they don’t move. They are fixed in the lattice.
Eg : battery.

Question 19.
How does a battery maintain a constant potential difference between its terminals?
Answer:
We know that a battery consists electric force (Fe) and chemical force (Fc). These two forces are balanced in a battery. Due to this reason a battery maintains a constant , potential difference between its terminals.

Question 20.
Why does the battery discharge when its positive and negative terminals are connected through a conductor?
Answer:
A conductor permits the charges to pass through it. Due to this the exhaustion of charges is created after completion of all charges. So, when a battery is connected with a conductor it discharges.

10th Class Physics Textbook Page No. 185

Question 21.
What happens when the battery is connected in a circuit?
Answer:
A potential difference is created between the ends of the conductor, when the battery is connected in a circuit.

10th Class Physics Textbook Page No. 186

Question 22.
How can we measure potential difference or emf?
Answer:
With the help of a voltmeter, we measure potential difference or emf.

10th Class Physics Textbook Page No. 187

Question 23.
Is there any relation between emf of battery and drift speed of electrons in the conductor connected to a battery?
Answer:
Yes, when emf of a battery is more the drift speed of electrons will be more.

10th Class Physics Textbook Page No. 189

Question 24.
Can you guess the reason why the ratio of V and I in case of LED is not constant?
Answer:
This is due to forward voltage and maximum continuous current rating characters of LEDs.

Question 25.
Do all materials obey Ohm’s law?
Answer:
No, some materials don’t obey Ohm’s law.

Question 26.
Can we classify the materials based on Ohm’s law?
Answer:
Yes, the materials which obey Ohm’s law are conductors and others are same conductors or non-conductors.

Question 27.
What is resistance?
Answer:
The obstruction offered to the flow of electrons in a conductor is called the resistance.

Question 28.
Is the value of resistance the same for all materials?
Answer:
Yes, it varies.

Question 29.
Is there any application of Ohm’s law in daily life?
Answer:
Yes, this law is used in wiring.

Question 30.
What causes electric shock in the human body – current or voltage?
Answer:
Current with sufficient voltage.

10th Class Physics Textbook Page No. 190

Question 31.
Do you know the voltage of mains that we use in our household circuits?
Answer:
Yes, I know the voltage of mains that we use in our household circuits is 120 V.

Question 32.
What happens to our body if we touch live wire of 240 V?
Answer:
240 V current disturbs the functioning of organs inside the body. It is called electric shock. If the current flow continues further, it damages the tissues of the body which leads to decrease in resistance of the body. When this current flows for a longer time, damage to the tissues increases and thereby the resistance of human body decreases further. Hence, the current through the human body will increase. If this current reaches 0.07 A, it effects the functioning of the heart and if this much current passes through the heart for more than one second it could be fatal.

If this current flows for a longer time, the person in electric shock will be killed.

10th Class Physics Textbook Page No. 191

Question 33.
Why doesn’t a bird get a shock when it stands on a high voltage wire?
Answer:
There are two parallel lines carrying 240 V current. The voltage current will pass through the body if both the wires are touched at the same time. But, when the bird stands on only one wire, there is no potential difference between the legs. So, no current passes through the bird. Hence, it doesn’t feel any electric shock.

10th Class Physics Textbook Page No. 192

Question 34.
What could be the reason for increase in the resistance of the bulb when current flows through it?
Answer:
The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.

Question 35.
What happens to the resistance of a conductor if we increase its length?
Answer:
The resistance of a conductor increases with the increase of its length.

10th Class Physics Textbook Page No. 193

Question 36.
Does the thickness of a conductor influence its resistance?
Answer:
Yes, as the thickness of the conductor increases the resistance decreases.

10th Class Physics Textbook Page No. 195

Question 37.
How are electric devices connected in circuits?
Answer:
Electric devices are connected either in series or parallel in circuits.

Question 38.
When bulbs are connected (resistors) in series, what do you notice
Answer:
We notice that, the sum of the voltages of the bulbs (resistors) is equal to voltage across the combination of the resistors.

10th Class Physics Textbook Page No. 196

Question 39.
What do you notice when bulbs (resistors) are connected in series to the current?
Answer:
The current is not changing

Question 40.
What do you mean by equivalent resistance?
Answer:
If the current drawn by a resistor is equal to the current drawn by the combination of resistors, then the resistor is called equivalent resistor.

Question 41.
What happens when one of the resistors in series breaks down?
Answer:
The circuit becomes open and flow of current will be broken down.

Question 42.
Can you guess in what way household wiring has been done?
Answer:
Parallel connection.

10th Class Physics Textbook Page No. 197

Question 43.
How much current is drawn from the battery if the resistors are connected in parallel?
Is it equal to individual currents drawn by the resistors?
Answer:
It is the sum of currents flowing through each resistor. No, it is the sum of individual currents drawn by the resistors.

10th Class Physics Textbook Page No. 199

Question 44.
How could the sign convention be taken in a circuit?
Answer:
The potential difference across the resistor is taken as negative when we move along the direction of electric current through the resistor, and it is taken as positive when we move against the direction of electric current through the resistor.

10th Class Physics Textbook Page No. 201

Question 45.
You might have heard the sentences like “this month we have consumed 100 units of current”. What does ‘unit’ mean?
Answer:
Unit (or) kilo watt hour is the consumption of electric power in one hour by our electric appliances.

Question 46.
A bulb is marked 60 W and 120 V. What do these values indicate?
Answer:
It means, the resistance of the bulb is

Question 47.
What is the energy lost by the charge in 1 sec.?
Answer:
It is equal to \(\frac{\mathrm{W}}{\mathrm{t}}\).

10th Class Physics Textbook Page No. 202

Question 48.
What do you mean by overload?
Answer:
When a high current flows through the wire which is beyond the rating of wire then heating of wire takes place. This phenomenon is called overloading.

Question 49.
Why does it (overloading) cause damage to electric appliances?
Answer:
Due to overload the heat increases in the circuit and this melts the parts of the appliances. Thus overload causdt damage to the electric appliances.

10th Class Physics Textbook Page No. 203

Question 50.
What happens when this current (overloading) increases greatly to the household circuit?
Answer:
It causes fire.

Question 51.
How can we prevent damage due to overloading?
Answer:
To prevent damages due to overloading we connect an electric fuse to the household circuit.

10th Class Physics Textbook Page No. 203

Question 52.
What do you mean by short circuit?
Answer:

• The line wires that are entering the meter have a voltage of 240 V.
•  The minimum and maximum limit of current that can be drawn from the mains is 5 to 20 A.
• Thus, the maximum current that we can draw from the mains is 20 A.
• When the current drawn from the mains is more than 20 A, overheating occurs and may cause a fire. This is called overloading.
• A short circuit is an electrical circuit that allows a current to travel along an unintended path often where essentially no electrical impedance is encountered.

Question 52.
Why does a short circuit damage electric wiring and devices connected to it?
Answer:
In a short circuit the current drawn from the main exceeds the maximum limit 20 A. This will lead to overloading which can damage the electrical appliances.

10th Class Physics 11th Lesson Electric Current Activities

Activity – 1

Question 1.
Write an activity to check when a bulb glows in a circuit.
(OR)
How do you prove a source of energy is required to glow a bulb in a circuit?
Answer:
Aim :
To check when a bulb glows in a circuit.

Materials required:

1. A bulb
2. a battery
3. a switch
4. few insulated copper wire

Procedure (1) :

1. Take a bulb, a battery, a switch and few insulated copper wires.
2. Connect the ends of the copper wires to the terminals of the battery through the bulb and switch.
3. Now switch on the circuit.
   Observation (1) : The bulb glows.

Procedure (2) :

1. Remove the battery from the circuit and connect the remaining components to make a complete circuit.
2. Again switch on the circuit and observe the bulb.

Observation (2): The bulb does not glow.

Procedure (3) :
Replace the copper wires with nylon wires and connect the nylon wires to the terminals of the battery through a bulb and switch. Now switch on the circuit. We observe that the bulb does not glow. Because the wires are not conductors.

Observation (3) : The bulb does not glow.

Result:
The battery contains charges which glow the bulb.

Activity – 3

Question 2.
Write an activity to show that the values of current are different for different wires for a constant voltage.
(OR)
The resistance of a conductor depends on the material of the conductor. Prove this through an activity.
(OR)
List out the material required in the experiment to show that the electric resistance depends upon the nature of the material and write experimental procedure.
Answer:
Aim:
To show that the values of current are different for different wires for a constant voltage. Materials required : (wires of the same length and some cross-sectional area).

• Copper rod
• Nichrome rod
• Battery
• Ammeter
• Key
• Manganin Wire

Procedure :

1. Make a circuit as shown in figure.
2. Connect one of the wires between the ends P and Q.
3. Switch on the circuit. Measure the electric current for a fixed voltage, using the ammeter connected to the circuit. Note it in your notebook.
4. Repeat this experiment with other wires and note the current in your notebook.

Observation :
The values of current are different for different wires for a constant voltage.

Conclusion:
The resistance of a conductor depends on the material of the conductor.

Activity – 5

Question 3.
Write an activity to show that resistance is inversely proportional to the c section area of the conductor.
(OR)
What happens to resistance if the area of a cross-section of conductor is increased? Explain with an activity.
Aim :
To prove that resistance is inversely proportional to the cross-section area of the conductor.

Materials required :

1. A Battery
2. Mangnin Wires
3. Ammeter
4. Key
5. Manganin wires with different cross-section areas (lengths are same).

Procedure:

1. Make the circuit as given figure.
2. Connect one of the wires between points P and Q.
3. Switch on me circuit. Note the ammeter reading in your notebook.
4. Continue the experiment with different wires of same length but different cross-section areas. Note the ammeter readings in your notebook.

Observation :
As the cross-section area of the rods increases, the current increases.

Result (Conclusion) :
Resistance is inversely proportional to cross-section area of the conductor.

Activity – 6

Question 4.
Write an activity to prove that the sum of the potential differences of the bulb is equal to voltage across the combination of the resistors. (OR)
Prove that during series connection potential difference is distributed among the resistors experimentally.
Answer:
Aim:
To prove that the sum of the potential differences of the bulbs is equal to potential difference across the combination of the resistors.

Materials required :

1. Bulbs
2. Voltmeters
3. Insulated wires
4. Ammeter
5. Key

Procedure :

1. Take different bulbs. Using a multimeter measure their resistances. Note them as R,, R2 and Rv
2. Connect them as shown in figure.
3. Measure the voltage between terminals of the battery connected to the circuit.
4. Measure the voltages between the ends of each bulb and note them as Vj, V2 and V3 from voltmeters in your notebook.
5. Compare them.

Observation :
We notice that the’sum of the voltages of the bulbs is equal to voltages across the combination of the resistors.

Activity – 7

Question 5.
Write an activity to prove that the current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.
(OR)
Prove that during parallel connection the current is distributed among the resistances by using an experimental activity.
Answer:
Aim:
To prove that the current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.
Materials required :

1. Bulbs
2. Ammeters
3. Buttery
4. Key
5. Wires

Procedure :

1. Connect the bulbs in parallel connection as shown in the given circuit.
2. Measure the voltage across each bulb using a voltmeter or multimeter.
3. Note these values in your notebook.

Observation :

1. The voltage at the ends of each bulb is the same.
2. Measure electric currents flowing through each bulb using ammeters. Note these values.
3. Measure the current (I) drawn from the battery using the ammeter 1.

Result (Conclusion) :
The current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 9 Our Environment.

AP State Syllabus SSC 10th Class Biology Important Questions 9th Lesson Our Environment

10th Class Biology 9th Lesson Our Environment 1 Mark Important Questions and Answers

Question 1.
Suggest one alternative method in place of pesticides to protect crops?
Answer:
Alternative methods for using pesticides to save the crops from pests:

1. Rotation of crops
2. Studying the life histories of pests
3. Biological Control
4. Sterility
5. Genetic Strains

Question 2.
By taking two plants of your surroundings as examples, explain how they protect themselves against the animals which eat them.
Answer:

1. Neem Tree: Neem leaves contain an alkaloid Nimbin to protect themselves from the animals which eat them.
2. Cactus: They have thorns to protect themselves.
3. Datura: Datura leaves gives bad odour.

Question 3.
Identify one food chain from your surroundings. Name the producers and different levels of consumers in that food Chain.
Answer:
Grass → Insects → Frog → Snake.
Producers – Grass.
Primary Consumers – Insects.
Secondary Consumers – Frog
Tertiary consumers – Snake.

Aquatic Plants → Insects → Fish → Crane.
Producers – Aquatic Plant.
Primary Consumers – Insects
Secondary Consumers – Fish
Tertiary consumers – Crane

Question 4.
Write the names of producers and consumers in the food chain, you have observed.
Answer:
Producers – Plants, Green Algae
Consumers – All Animals.

Question 5.
Write any two slogans to promote awareness among the people about Ecofriendly programs.
Answer:
a) Lets go green to get global clean.
b) If you disturb the nature, the nature will disturb you.
c) The best solution to arrest pollution is plantation.
d) Reduce the usage of plastic and reduce the pollution.

Question 6.
What happens if decomposers are removed from the food web?
Answer:

1. If decomposers are removed from the food web then the biological cycles are not completed.
2. If the decomposers are not present in an ecosystem the remains of the other organisms accumulate.

Question 7.
Observe the following given below. Draw the pyramid of numbers.
Grass → Goat → Man
Answer:

Question 8.
We can’t expect the world without sparrows. So how should be our concern towards their conservation?
Answer:

1. Sparrows are useful to control harmful insects like locust which damage food grains.
2. Chemical pesticides are the cause for destruction of sparrows and useful insects.
3. By using biological methods we can conserve the sparrow population.

Question 9.
Human being is modifying agriculture lands and lakes into residential areas. What is its effect on Bio-diversity?
Answer:

1. The shelter may not be provided for migratory birds.
2. Food chain get disturbed.
3. Decrease in the ground water level.

Question 10.
How do you protect the plants, which were planted in “Haritha Haaram” programme in your school?
Answer:
We protect the plants:

1. Watering of plants at regular intervals.
2. Fencing or gaurding of plants.
3. Adoption of plants.
4. Providing organic manure.

Question 11.
The figure given below represents a food pyramid. Study it and answer the following questions.

i) Which trophic level has maximum energy?
Answer:
T₁ (or) Primary producers (green plants)
ii) Give one example for T₄ trophic level.
Answer:
Lion, tiger, hawk, etc.

Question 12.
“We can’t imagine the world without insects and birds, conserve them.
Answer:
Methods to conserve insects and birds:

1. Avoid indiscriminate usage of pesticides
2. Protect the natural habitats of insects and birds.
3. Development of bird sanctuaries
4. Everyone should follow environmental ethics.

Question 13.
What is environment?
Answer:
The sum of physical and biological factors along with their chemical interactions that affect an organism is called environment.

Question 14.
What is biosphere?
Answer:
The world of living things is called biosphere.
(OR)
The life supporting zone on the earth is called biosphere.

Question 15.
What are the physical or abiotic factors in a biosphere?
Answer:
Land, air, water, sunlight, humidity etc. are the physical or abiotic factors in a biosphere.

Question 16.
What does a food chain show?
Answer:
Food chain shows that how the energy is passed from one organism to another.

Question 17.
How the terrestrial ecosystems are determined?
Answer:
The terrestrial ecosystems are determined largely by the variations in climatic conditions between the poles and equator.

Question 18.
Where did the Kilimanjaro mountain located?
Answer:
The Kilimanjaro mountain is located in equatorial Africa (present in Tanzania, Africa).

Question 19.
What is the main source of energy for all the organisms in an ecosystem?
Answer:
Sunlight is the main source of energy for all the organisms in an ecosystem.

Question 20.
What is food web?
Answer:
The network of a large number of food chains existing in an ecosystem is called a food web.

Question 21.
What does a food web indicate?
Answer:
A food web indicates that the number of possible links for food in an ecosystem and reflects the fact that the whole community is a complex inter-connected unit.

Question 22.
What is ‘niche’?
Answer:
The position of organisms in a food web.

Question 23.
What does the word ‘niche’ denotes?
Answer:
‘Niche’ is the term used to describe not only the animals position in the food web and what it eats but also its mode of life.

Question 24.
What is an ecological pyramid?
Answer:
The graphic representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called “Ecological pyramid”.

Question 25.
Who was the first one to introduce “Ecological pyramid”?
Answer:
Ecological pyramid was first introduced by a British Ecologist Charles Elton in 1927.

Question 26.
In ecological pyramids the producers are represented at?
Answer:
The producers are represented at the base of the ecological pyramids.

Question 27.
What is pyramid of number?
Answer:
A graphical representation designed to show the number of organisms at each tropic level in a given ecosystem is called “pyramid of number”.

Question 28.
What does the pyramid of number show?
Answer:
Pyramid of number shows the population of organisms at each tropic level in a food chain.

Question 29.
What does each bar represent in a pyramid of number?
Answer:
In a pyramid of number, each bar represents the number of individuals at each tropic level in a food chain.

Question 30.
When does the pyramid of number not look like a pyramid at all?
Answer:
If the producer is a large plant such as tree or if one of the organisms at any tropic level is very small, then the pyramid of number does not look like a pyramid.

Question 31.
What is biomass?
Answer:
Any type of plant or animal material that can be converted into energy is called biomass.

Question 32.
What is biofuels?
Answer:
The materials which are used for energy production are known as biofuels.

Question 33.
What is Pyramid of biomass?
Answer:
A graphical representation designed to show the quantity of living matter at each tropic level in a given ecosystem is called “Pyramid of biomass”.

Question 34.
Why Pyramid of biomass inverted in case of aquatic ecosystem?
Answer:

1. In an aquatic ecosystem, the biomass of phytoplankton is quite negligible as compared to that of the crustaceans and small herbivorous fish that feed on these producers.
2. The biomass of large carnivorous fish living on small fishes is still greater. This makes the pyramid of biomass inverted.

Question 35.
How much percentage of the biomass is transferred from one tropic level to the next in a food chain?
Answer:
The percentage of the biomass transferred from one tropic level to the next level in food chain is nearly 10 – 20%.

Question 36.
When does the species at the top of the pyramid get more energy?
Answer:
The species at the top of the pyramid get, more energy when the steps in a food chain are fewer.

Question 37.
What are bio-geochemical cycles?
Answer:
Flow of materials between organisms and their environment is called cycling of materials or bio-geochemical cycles.

Question 38.
What is ecological efficiency?
Answer:
The ratio between energy flows at different tropic levels among the food chain expressed as percentage is called ecological efficiency.

Question 39.
What is ten per cent law?
Answer:
During the transfer of energy from one tropic level to the next, only about ten per cent of the energy from organic matter is stored as flesh. This is called “Ten per cent law”.

Question 40.
Where do Kolleru one of the largest fresh water lakes in India exists?
Answer:
Kolleru is one of the largest fresh water lakes in India exists between West Godavari and Krishna districts of Andhra Pradesh.

Question 41.
What is the catchment area of the lake Kolleru?
Answer:
A catchment area of the Kolleru lake extends up to 6121 Km2.

Question 42.
Through which The lake Kolleru discharges its excess water into Bay of Bengal?
Answer:
The lake Kolleru discharges its excess water into Bay of Bengal through the twisty channel called Upputeru which is about 65 km long.

Question 43.
When did Government of Andhra Pradesh had declared the lake as Bird Sanctuary?
Answer:
In November 1999, Government of Andhra Pradesh had declared the lake as Bird Sanctuary.

Question 44.
What is the number of species of birds being hosted by the Kolleru?
Answer:
The Kolleru lake is hosting 193 species of birds.

Question 45.
What are the major sources of pollution in Kolleru lake?
Answer:
The major sources of pollution are agricultural runoff containing residues of several agrochemicals, fertilizers, fish tank discharges, industrial effluents containing chemical residues and different types of organic substances, municipal and domestic sewage.

Question 46.
What is the objective of “Operation Kolleru” by the ministry of environment and forest, Government of India?
Answer:
The objective of operation Kolleru by the ministry of environment and forest, Government of India is to bring back the ecological balance of Kolleru lake which is a gift of nature.

Question 47.
What is Bioaccumulation?
Answer:
The process of entering of pollutants in a food chain is known as Bioaccumulation.

Question 48.
What is Biomagnification?
Answer:
The tendency of pollutants to concentrate as they move from one tropic level to the next is known as Biomagnification.

Question 49.
What are pesticides?
Answer:
The chemical materials used to control pests that attack crop plants or live as parasites on the body of farm animals are called pesticides.

Question 50.
What is a perfect pesticide?
Answer:
The perfect pesticide is one which destroys a particular pest and is completely harmless to every other form of life.

Question 51.
Why did the egg breaking among the peregrines increase?
Answer:
Egg breaking among the peregrines increased due to their disturbed behaviour caused by the nerve poisons that entered into their tissues through food chain.

Question 52.
Why did the aquatic biota is being contaminated?
Answer:
The aquatic biota is being contaminated with heavy metals due to industrialization and anthropogenic activities.

Question 53.
Why fish are considered to be the bioindicators of metal contamination?
Answer:
Fish are considered to be the bioindicators of metal contamination in environmental monitoring because fish species are strongly respond to stress conditions.

Question 54.
Where did Edulabad water reservoir located?
Answer:
Edulabad water reservoir is located in urban areas of Ranga Reddy district of Telangana.

Question 55.
Which fish species is grown in Edulabad water reservoir?
Answer:
Cyprinus carpio (common scale carp) is the fish species grown in Edulabad water reservoir.

Question 56.
What are the effects of bioaccumulation of metals in human beings that eat cyprinus carpio?
Answer:
The bioaccumulation of various metals in cyprinus carpio cause disorders. Such as hypertensions, sporadic fever, renal damage, nausea, etc.

Question 57.
In which country sparrows were hunted extensively in 1958?
Answer:
In China sparrows were hunted extensively in 1958.

Question 58.
In your opinion what are the effective methods to control pests?
Answer:
Rotation of crops, biological control, developing genetically modified plants are the effective methods to control pests in my opinon.

Question 59.
Why the temperatures are very high during the day and cold during the nights in deserts?
Answer:
In deserts, the rainfall and humidity are very low, so the sun’s rays easily penetrate the atmosphere making ground temperatures very high during the day. But the nights are often cold as the earth loses heat rapidly.

Question 60.
How can we draw a food chain?
Answer:
We can draw a food chain by connecting the pictures or names of organisms by putting arrows between them. These arrows should always point from food to the feeder.

Question 61.
How many types of ecological pyramids are there in practice? Name them.
Answer:
There are three types of ecological pyramids. They are:

1. Pyramid of number
2. Pyramid of biomass and
3. Pyramid of energy.

Question 62.
Which process helps to convert the solar energy into suitable form of energy for animals to consume?
Answer:
Photosynthesis helps to convert the solar energy into suitable form of energy (food) for animals to consume.

10th Class Biology 9th Lesson Our Environment 2 Marks Important Questions and Answers

Question 1.
Grass → Grasshopper → Frog → Snake → Hawk
What will happen if we remove Frog from the above food chain? Explain.
Answer:

1. Frog is secondary consumer in this food chain.
2. If we remove frog from the food chain, the number of grasshopper will increase on other hand the number of snakes which depend on frogs will decrease.
3. Hence, ecological balance may be damaged.

Question 2.
Observe the diagram and answer the following.
i) Write any two food chains from the diagram.
ii) What are the secondary consumers in the food chain that are written by you?
Answer:
i) a) Plants → Goat → Tiger
b) Plants → Rabbit → Wolf / Fox
ii) Tiger, Wolf / Fox

Question 3.
Study the given paragraph and answer the questions.

Solar energy from sun enters into the producers of an ecosystem. No organisms except green plants and photosynthetic bacteria can absorb solar energy and convert it into chemical energy.

A) What are the producers mentioned in the given paragraph?
B) What form is energy converted into photosynthesis? In Photosynthesis, which form is energy converted into?
Answer:
A) Green plants and photosynthetic bacteria.
B) In photosynthesis, the light (or) solar energy is converted into chemical energy.

Question 4.
Explain the flow chart given below.
Answer:
It is the pyramid of biomass.

1. In this pyramid 10% of the food will reach to the next trophical level and so on at each level.
2. It would take 1000 kg of phytoplankton to provide 100 kg of zooplankton and to form 1 kg of human tissue, 10 kg of frog is needed.
3. The fewer the steps in the food chain, the more energy will be for the species at the top.

Question 5.
Observe the pyramid of number which is given below and answer the questions.
i) As per the number of organisms in the tropic level, which group of organisms
are more in number and which are less in number?
ii) What happens if Secondary consumers disappear?
Answer:
i) If producers are more in number, then tertiary consumers are less in number,
ii) If secondary consumers disappear the primary consumers increase in number and the tertiary consumers found no food to live. It leads to death.

Question 6.
Explain in brief about the alternate methods to be followed to prevent the harmful effects of over usage of pesticides.
(OR)
Mention any four effective methods of controlling pests, which are less harmful on environment based on biological principles.
Answer:
Some of alternative pest control methods are

1. Rotation of Crop: Growing different crops on a particular piece of land in successive years.
2. Studying the life histories of the pests: When this is done it is sometimes possible to sow the crops at a time when least damage will be caused.
3. Biological Control: Introducing Natural predator or parasite of the pest.
4. Sterility: Rendering the males of a pest species sterile.
5. Genetic Strains: The development of genetic strains (genetically modified plants) which are resistant to certain pest.
6. Environmental ethics: People need to know besides laws regarding environment there are some basic ethics what is right and what is wrong in view of environment.

Question 7.
Write any 4 slogans on the necessity of forests and on their conservation.
Answer:

1. Save the trees, save the earth. We are the guardians of nature’s birth.
2. Don’t destroy the greenary and don’t spoil the scenery.
3. Don’t make trees rare, we should keep them with care.
4. To live for future in rest, saving forest is the best.

Question 8.
How does the given below concepts differs?
(a) Bioaccumulation b) Biomagnification
Answer:
a) Bioaccumulation: The process of entry of pollutants into a food chain is known as bioaccumulation.
b) Biomagnification: It is the tendency of pollutants to concentrate as they move from one tropic level to the next is known as biomagnification.

Question 9.
The biomass of a producer in an ecosystem is calculated as 3500 kgs. Calculate the biomass of primary, secondary, tertiary consumers.
Answer:
In a food chain roughly 90% of the food is lost at each step. So if the biomass of a producer in an ecosystem is calculated as 3500 kgs. the biomass of primary consumer as will be 350 kgs. and of secondary consumer is 35 kgs and biomass of tertiary consumer is 3.5 kgs.

Question 10.
Write a short note on food chain and food web.
Answer:

1. Food chain is a pathway along which food is transferred from one tropic level to another tropic level beginning with producers.
2. It shows who eats what in a particular habitat.
3. The arrows between each item in the chain always point from the food to the feeder.
4. For example
   Grass → Rabbit → Snake → Hawk
5. The elaborate interconnected feeding relationships in an ecosystem is said to be food web.
6. Many of the food chains in an ecosystem are crosslinked to form food web.
7. For example,
   
8. Food chain and food web help us to understand the food relations among living things.

Question 11.
Write a short notes on ecological pyramids.
Answer:

1. The graphical representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called “Ecological pyramid”.
2. It was first introduced by a British Ecologist Charles Elton in 1927.
3. In the ecological pyramid, the producers (First tropic level) are represented at the base, and the successive tropic levels (primary, secondary and tertiary consumers) are represented one above the other with top carnivores at the tip.
4. There are three types of pyramids.
   i) Pyramid of number ii) Pyramid of biomass and iii) Pyramid of energy.
5. Pyramid of number shows the population of organisms at each tropic level in a food chain.
6. Pyramid of biomass represents the available food as a source of energy at each tropic level in the food chain.
7. Pyramid of energy represents the available energy at each tropic level in food chain.

Question 12.
Write a short notes on pyramid of number.
Answer:

1. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.
2. The shape of this pyramid varies from eco-system to ecosystem.
   
3. In aquatic and grassland ecosystems, numerous small autotrophs support lesser herbivores which support further small number of carnivores and hence the pyramid structure is upright.
   
4. In forest ecosystem, less number of producers support greater number of herbivores who in turn support a fewer number of carnivores. Hence the pyramid structure is partly upright.
5. In parasitic food chain, one primary producer supports numerous parasites which support still more hyperparasites. Hence the pyramid structure is inverted.
   

Question 13.
Write a short notes on pyramid of biomass.
Answer:

1. Pyramid of biomass is a graphical representation designed to show the quantity of living matter (biomass) at each tropic level in a given ecosystem.
2. In terrestrial ecosystems, the biomass progressively decreases from producers to top carnivores. Hence the pyramid structure is upright.
   
3. In an aquatic ecosystem, the biomass of phytoplankton (producers) is quite negligible as compared to that of crustaceans and small herbivorous fish that feed on these producers. The biomass of large carnivorous fish living on small fishes is still greater. This makes the pyramid of biomass inverted.
   

Question 14.
How do pesticides cause Bioaccumulation and Biomagnification?
(OR)
What are the effects of pesticides on environment?
Answer:

1. Pesticides are the toxic chemicals used to destroy pest and insects which damage our crops and stored foods.
2. These pesticides vary in their length of life as toxic materials.
3. Some of the pesticides are degradable that can be broken down into harmless substances in a comparatively short time and others are non-degradable.
4. Non-degradable pesticides accumulate in the bodies of animal and pass right through food web.
5. Thus the pesticides cause bioaccumulation.
6. These accumulated pesticides concentrate as they move from one tropic level to the next, thus leads to biomagnification.

Question 15.
List out some human activities which altered the communities of plants and animals in their natural ecosystem.
Answer:

1. Industrialization
2. Damming rivers
3. Draining marshes
4. Re-claiming land from the sea
5. Cutting down forests
6. Using chemical fertilisers and pesticides
7. Building towns, cities, canals and motor ways.

Question 16.
What kind of changes may come in 2m ecosystem due to development of a large town?
Answer:
The following changes are expected due to development of a large town.

1. Some plants and animal species will die out.
2. Some will adapt to the new conditions sufficiently to survive in reduced numbers.
3. Some will benefit by the new conditions and will increase in numbers.

Question 17.
Write a comparative note on pyramid of number and pyramid of biomass.
(OR)
Write the differences between pyramid of number and pyramid of biomass.
Answer:

Question 18.
Write a comparative note on pyramid of biomass and pyramid of energy.
(OR)
What are the differences between pyramid of biomass and pyramid of energy?
Answer:

Question 19.
Write a comparative note on pyramid of number and pyramid of energy?
(OR)
What are the differences between pyramid of number and pyramid of energy?
Answer:

Question 20.
What is ecological efficiency? Write a short notes on Ten per cent law?
Answer:
Ecological efficiency: The ratio between energy flows at different tropic levels along the food chain expressed as percentage is called “ecological efficiency”.
Ten per cent law:

1. The amount of energy transferred decreases with successive tropic levels.
2. Slobodkin (1959) suggested that the transfer of energy from one tropic level to the next is of the order of 10% and this is called “Gross ecological efficiency”.

Question 21.
Geetha said “a given species may occupy more than one tropic level in the same ecosystem at the same time”. Do you support her or not? Explain your answer with example.
Answer:
I support her for the following reason.

1. A snake eating a mouse in a field or lawn occupies the third tropic level.
   Plant → Mouse → Snake
2. When the snake eats a frog in the same field, it occupies the fourth tropic level in a food chain. It is because the frog feed on some of the insects that depend on the plants.
   Plant → Insect → Frog → Snake
3. Thus, a given species may occupy more than one tropic level in the same ecosystem at the same time.
4. This is to satisfy its food needs, as it cannot do so by occupying one tropic level.

Question 22.
If we introduce a man into a forest ecosystem, at which level of food chain we will place him? Explain your answer.
Answer:

1. If we introduce a man into a forest ecosystem, he can fit for any level of consumers of food chain.
2. He may feed on plant parts such as fruits. Then we can place him at primary consumer level.
   Plant → Man
3. He may feed on some of the herbivorous organisms such as rabbit, then we can place him at secondary consumers level.
   Plant → Rabbit → Man
4. He may also feed on some of the carnivorous organisms such as insectivorous birds then we can place him at tertiary consumers level.
   Plant → Insect → Bird → Man
5. This is possible to place him at any level of consumers, as he is an omnivore, who feed on both plant originated and animal originated foods.

Question 23.
Draw the ecological pyramids for the given food chain.
Banyan Tree → Herbivorous birds → Carnivorous birds.
Answer:
1) Pyramid of number

2) Pyramid of biomass

3) Pyramid of energy

Question 24.
Write briefly about Minamata disease.
Answer:

1. Minamata disease was first discovered in Minamata city in Kumamoto prefecture, Japan, in 1956.
2. It was caused by the release of methyl mercury in the industrial waste water from the Chisso corporation’s chemical factory, which continued from 1932 to 1968.
3. This highly toxic chemical bioaccumulated in shellfish and fish in Minimata Bay and the Shiranui Sea, which, when eaten by the local populace, resulted in mercury poisoning.
4. While cat, dog, pig and humans death continued for 36 years.

Question 25.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:

1. Trophic levels is the feeding position in a food chain.
2. It is the functional level occupied by an organism in a food chain.
3. Examples of trophic levels include ‘herbivores’ and ‘decomposers’
4. An example of food chain depicting various trophic levels is as follows:
   

Question 26.
What is the role of decomposers in the ecosystem? (OR)
How decomposers help in cleaning the environment?
Answer:

1. If the decomposers are not present in an ecosystem the remains of the other organisms accumulate.
2. Eventually the world would run out of carbon dioxide or nitrate or phosphate or other inorganisms material essential for life.
3. The decomposers breakdown the organic waste products and dead remains of organisms into the inorganic substances needed by the producers.
4. Most decomposition is carried out by saprophytic fungi, by bacteria and by invertebrates.

10th Class Biology 9th Lesson Our Environment 4 Marks Important Questions and Answers

Question 1.
Prepare some slogans about ‘Vanam – Manam’ programme to display in your school rally?
Answer:

1. Save paper – Save trees
2. Plant a tree – Plant a life
3. Saving trees is our duty
4. Think green – Go green
5. If we protect plants – they protects us
6. Conserve plants – Conserve life
7. Plant a tree – get the air free
8. Plant a tree – Reduce the pollution.
9. Tree on – Global warming gone.
10. If cut a tree – It kills a life.

Question 2.
Read the information about Kolleru lake in the given table and answer the following questions.

a) In which year, lake water spread area is more?
b) Why do you think weeds are more in the lake?
c) Guess the reasons for decrease in the lake area.
d) What measures are to be taken to control pollution in the lake?
Answer:
a) In the year 1967
b) Addition of excessive nutrients from aquaculture ponds and rice fields.
c) Aquaculture ponds, Ricefields and Encroachment are the reasons for decrease in lake area.
d) Anthropogenic activities are to be controlled in the lake catchment area. (Or)
Fish ponds are to be removed in the lake catchment area. (Or)
Agricultural practices in the lake area should be minimised as per the norms of government.

Question 3.
Observe the following pyramid of biomass and answer the following questions.
a) This pyramid shows a decrease in the biomass as we move up, why the biomass is decreasing?
Answer:
The pyramid of biomass for the given food chain, at each step 90% of the food is lost. That means 1000 kg of phytoplankton to produce 100 kg of Zooplankton to form 10 kg of fish to produce 1kg of human tissues. The fewer the steps in the food chain the more energy will be for the species at the top.

b) Give some examples of producers and primary consumers.
Answer:
Examples for producers: Plants, Grass, Diatoms.
Examples for primary consumers: Grasshopper, Rabbit, Deer,

c) Where do producers get the energy from?
Answer:
From the sun.

d) How much biomass is lost at each step?
Answer:
90%

Question 4.
Observe the following diagram and answer the following questions.
i) Name the primary producers in the given food web.
ii) Prepare any one food – chain from the diagram.
iii) What are the tertiary consumers?
iv) Write names of any two herbivores.
Answer:
i) Plants, Grass, Trees Phytoplanktons etc.
ii) Grass → Rabbit → Fox → Tiger
iii) Tiger, Vulture, Crane, Owl, Peacock etc.
(OR)
The animals which are at 4th trophic level in a food chain are called as Tertiary Consumers.
iv) Rabbit, Deer, Goat, Cow

Question 5.
What is number pyramid? What does it indicate?
Answer:

1. The number of organisms in a food chain can be represented graphically in a pyramid of number.
2. Each bar represents the number of individuals at each tropic level in a food chain.
3. At each link in a food chain, from the first order consumers to the large carnivores, there is normally an increase in size but decrease in number.
4. For example in a wood, the aphids are very small and occur in astronomical numbers.
5. The lady birds which feed on them are distinctly larger and not so numerous.
6. The insectivorous birds which feed on the lady birds are larger still and are only present in a small number and there may only be a single pair of hawks of much larger size than the insectivorous birds on which they prey.

Question 6.
Draw the diagram of number pyramid keeping foxes as third consumers. What are the consequences if their number increases?
Answer:

1. If the number of foxes increases, then the competition for food will be very severe and less amount of food will be available for them.
   
2. As a result some of the foxes may not get enough food and die due to starvation.
3. This reduces the population of foxes and very few foxes will be left in the forest.
4. This increases the chances of survival of secondary consumers birds, hence their number increases.
5. This increases the availability of food for foxes. Very soon a balance will be established between the number of secondary consumers and foxes.

Question 7.
What reasons are responsible for decrease in number of top carnivores and biomass starting production in a food chain?
(OR)
Why the number of organisms get decreased as we move from producers to consumer levels?
Answer:

1. In a food chain as we move from producers to different levels of consumers the energy available will decrease gradually.
2. Only ten percent of the energy present in one tropic level transfer to another tropic level.
3. Biomass also decreases gradually as only 10 – 20% of the biomass is transferred from one tropic level to the next in a food chain.
4. As there is less energy of less biomass available at top levels, number of organisms also less generally.
5. So the number of organisms get decreased as we move from producers to different level of consumer.

Question 8.
Show food chain of different organisms, number of pyramid of your school.
Answer:
Food chain of different organisms in our school:
Plant → aphids → spiders thirds.

1. The pyramid of organisms in a food chain can be represented graphically in a pyramid of number.
2. Each bar represents the number of individuals at each tropic level in a food chain.
3. At each link in a food chain, from the first order consumers to the large carnivores, there is normally an increase in size, but decrease in number.

Question 9.
What determines the terrestrial ecosystems on the earth?
Answer:

1. The terrestrial ecosystems on the earth are being determined largely by the variations in climatic conditions between the poles and equator.
2. The main climatic influences which determine these ecosystems are rainfall, temperature and availability of light from the sun.
3. For instance, forests are usually associated with high rainfall, but the type Is influenced by temperature and light.
4. The same applies to deserts which occur in regions where rainfall is extremely low.
5. Thus, the climatic conditions along the horizontal climatic regions determined the terrestrial ecosystems on the earth.
6. If we move from equatorial region to the polar region, we can come across tropical rain forests, savannah, deciduous forest, coniferous forests and then tundras respectively.
7. Similarly altitude of the place is also a determining factor.
8. If we climb a mountain such as Kilimanjaro in equatorial Africa, we can go through a comparable system of ecosystems, starting with tropical rain forest at the base and ending with perpetual snow and ice at the summit.

Question 10.
“All the energy in the ecosystem is ultimately derived from sunlight.” Justify.
Answer:

1. All the organisms in an ecosystem derive energy from food.
2. The food by its nature is the chemical energy and by in its stored form, it is the potential energy.
3. In an ecosystem, all the consumers at any level depend upon producers for their food either directly or indirectly.
4. The producers in any ecosystem are nothing but photosynthetic organisms such as plants, phytoplanktons and photosynthetic bacteria.
5. Energy enters the producers in the ecosystem from the sun in the form of solar energy during photosynthesis.
6. From the producers, the chemical energy passes to the consumers from one tropic level to the next through food.
7. For example, in a grassland ecosystem, grass traps the solar energy and stores in its body.
8. When this grass is eaten and assimilated by insects this stored energy enters into the body of insects.
9. From the insects it will pass to frog, from them to snake and so on to eagle.
10. Thus, all the energy in the ecosystem is ultimately derived from sunlight.

Question 11.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:

1. The tendency of pollutants to concentrate as they move from one trophic level to the next trophic level is known as Biomagnification.
2. Plants absorb pesticides, heavy metals from the soil.
3. The primary consumers when eat these plants the remaining of pesticides and heavy metals enter their bodies.
4. As these chemicals are not degradable, they accumulate in the bodies of organisms of all trophic levels in the food chain.
5. Most of the plants products which we eat are grown in fields in which pesticides and fertilisers have been used.
6. These are absorbed by the plants and cannot be removed by washing or other means.
7. Human beings are at the top level of the food chain these chemicals get accumulated in our bodies and cause various disorders.
8. Levels of biological magnification would increase as the trophic level increases.

Question 12.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels ? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:

1. If we remove producers from ecosystem, herbivores will not survive and the entire ecosystem collapse.
2. Removing herbivores result in increase number of producers and carnivores would not get food.
3. Removing carnivores result in increase of herbivores to unsustainable levels.
4. If we remove decomposers from ecosystem waste material and animal dead remains would pile up and nutrients would not be available to the producers.
5. Some or the other damage would be caused to the ecosystem if the organisms of any trophic level is removed.
6. However impact of removing producers or decomposers would be serve as the whole ecosystem would collapse.
7. Without plants sun’s energy cannot be converted to chemical energy which is the basis of life on earth.
8. Without decomposers the nutrients cannot be recycled and made available to producers.

Question 13.
Every organism has got the right to live on this planet. Write slogans to motivate the people on preservation of biodiversity.
Answer:

1. Live and let live.
2. Conserve nature – conserve life.
3. Clean the environment, live happily.
4. Think eco-friendly and live eco-friendly.
5. If we protect the environment, it protect us.
6. Reduce pollution – conserve the biodiversity.

Question 14.
Write some friendly ecosystem activities you will conduct in your school.
Answer:

1. Forming eco-clubs: These clubs consists of student representatives from each class. They will take up the eco-friendly activities and encourage the people of that village to follow environment friendly activities.
2. Setting up garden at school: This ensures the school and its premises green through planting of flowering plants, vegetables and fruit trees. It is a symbol of biodiversity because various plants and animals inhabit the garden.
3. Electricity conservation programme: To save energy the school implements certain hours to be switched off habit. This switching off programme for one hour from 3.30 p.m. to 4.30 p.m. help conserve electricity in every classroom.
4. Pollution prevention programme: A ‘no burning of trash policy should be implemented in the school. Waste materials are recycled and properly disposed to ensure a clean, waste-free environment.
5. Making compost by organic wastes: By digging a pit at one corner of the school and throwing the organic waste particularly of mid day meal waste into pit and covering with soil layers prepares compost which can be used as manure for plants. This creates a clean environment in the school.
6. Using cloth bags instead of polythene bags by pupil.
7. Collection of solid waste materials and proper management of its helps in reducing soil pollution.
8. Children should be encouraged to follow ‘3R’ system i.e. Reduce, Re use and Recycle different substances.

Question 15.
What is Ecological pyramid? Describe different types of Ecological pyramids.
Answer:

1. The graphical representation of the feeding level structure of an ecosystem by taking the shape of a pyramid is called ecological pyramid.
   
2. There are three types of ecological pyramids. They are
   1) Pyramid of number, 2) Pyramid of biomass and 3) Pyramid of energy.
3. Pyramid of number is a graphical representation designed to show the number of organisms at each tropic level in a given ecosystem.
4. The shape of this pyramid varies from ecosystem to ecosystem.
5. In forest ecosystem the pyramid structure is partly upright and in parasitic food chain is inverted.
6. Pyramid of biomass is a graphical representation designed to show the quantity of living matter (bio mass) at each trophic level in a given ecosystem.
7. In terrestrial ecosystems, the biomass progressively decreases from producers to top carnivores hence the pyramid structure is upright whereas in aquatic ecosystem it is inverted.
8. Pyramid of energy is a graphical representation designed to show the quantity of energy present at each tropic level in a given ecosystem. The pyramid of energy is always upright.

Question 16.
Collect information regarding pesticides commonly used in your area and prepare a chart showing pesticide and common name and on which crop and pest it is commonly used.
Answer:

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 10th Lesson Chemical Bonding

10th Class Chemistry 10th Lesson Chemical Bonding Textbook Questions and Answers

Improve Your Learning

Question 1.
List the factors that determine the type of bond that will be formed between two atoms. (AS1)
(OR)
How can you identify the type of bond formation between two atoms?
Answer:

• The strength of attraction or repulsion between atoms.
• Electrons in valence shell (valence electrons).

Question 2.
Explain the difference between the valence electrons and the covalency of an element. (AS1)
(OR)
How are valence electrons different from the covalency of element? Explain with examples.
Answer:
Valence electrons :

• Number of electrons in the outermost orbit or an atom is called its valence electrons.
• Ex: Na (Z = 11). It has 2e^– in I orbit, 8e^– in II orbit and 1e^– in III orbit.
• So number of valence electrons in Na atom are ‘l’.

Covalency of an element:

• Number of valance electrons which are taking part in covalent bond is called covalency.
• The electron configuration of Boran is 1s² 2s² 2p¹.
• It has three valance electrons.
• So its covalency is 3.

Question 3.
A chemical compound has the following Lewis notation : (AS1)
a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the elements X and Y.

Answer:
a) 6
b) 2
c) 1
d) two
e) X – is hydrogen and Y – is oxygen. The formed molecule is H₂O.

Question 4.
Why do only valence electrons involve in bond formation? Why not electron of inner shells? Explain. (AS1)
(OR)
Which shell electrons involve in bond formation? Explain. What is the reason behind it?
Answer:

• The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
• But the electrons in the outermost shell (valence shell) of atoms get affected.
• The inner shell electrons are strongly attracted by the nucleus when compared to the valence electrons.
• So electrons in valence shell (valence electrons) are responsible for the formation of bond between atoms.

Question 5.
Explain the formation of sodium chloride and calcium oxide on the basis of the concept of electron transfer from one atom to another atom. (AS1)
(OR)
Explain the formation of any two compounds according to Kossel’s theory.
Answer:
I. Formation of sodium chloride (NaCl) :
1) Sodium chloride is formed from the elements sodium (Na) and chlorine (Cl).

2) Cation formation:
i) When sodium (Na) atom loses one electron to get octet electron configuration, it forms a cation (Na⁺).
ii) Now Na⁺ gets electron configuration that of Neon (Ne) atom.

3) Anion Formation :
i) Chlorine has shortage of one electron to get octet in its valence shell.
ii) So it gains the electron that was lost by Na to form anion and gets electron configuration of Argon (Ar).

4) Formation of NaCl :
i) Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–’ ions.
ii) These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–(g) → Na⁺Cl^–_(s) or NaCl

II. Formation of calcium oxide (CaO) :
1. Calcium (Ca) reacts with oxygen (0) to form an ionic compound calcium oxide (CaO).

2. Atomic number of Calcium is 20. Its electronic configuration is 2, 8, 8, 2.
3.

4. Atomic number of Oxygen is 8. Its electronic configuration is 2, 6.

5.

6. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound calcium oxide (CaO).
Ca²⁺ + O^2- → Ca²⁺O^2- (or) CaO

Question 6.
A, B and C are three elements with atomic number 6, 11 and 17 respectively.
i) Which of these cannot form ionic bond? Why? (AS1)
ii) Which of these cannot form covalent bond? Why? (AS1)
iii) Which of these can form ionic as well as covalent bonds? (AS1)
Answer:
i) ‘A’ cannot form ionic bond. Its valence electrons are 4. It is difficult to lose or gain 4e^– to get octet configuration. So it forms covalent bond [Z of A is 6 so it is carbon (C)].

ii) ‘B’ cannot form covalent bond. Its valence electrons are 1 only. So it is easy to donate for other atom and become an ion. So it can form ionic bond [Z of B is 11, so it is sodium (Na)].

iii) Element C can form ionic as well as covalent bonds. Atomic number of Cl is 17. It is able to participate with Na in ionic bond and with hydrogen in HCl molecule as covalent bond.

Question 7.
How do bond energies and bond lengths of molecule help us in predicting their chemical properties? Explain with examples. (AS1)
(OR)
How can you explain with examples that bond energies and bond lengths are used to recognise chemical properties?
Answer:
1. Bond length :
Bond length or bond distance is the equilibrium distance between the nuclei of two atoms which form a covalent bond.

2. Bond energy :
Bond energy or bond dissociation energy is the energy needed to break a covalent bond between two atoms of a diatomic covalent compound in its gaseous state.

3. If the nature of the bond between the same two atoms changes the bond length also changes. For example, the bond lengths between two carbon atoms are C – C > C = C > C = C.

4. Thus the various bond lengths between the two carbon atoms are in ethane 1.54 Å, ethylene 1.34 Å, acetylene 1.20 Å.

5. The bond lengths between two oxygen atoms are in H₂O₂ (O – O) is 1.48 Å and in O₂ (O = O) is 1.21 Å.
6. Observe the table.

7. When bond length decreases, then bond dissociation energy increases.

8. When bond length increases, then bond dissociation energy decreases.

9. Bond length of H – H in H₂ molecule is 0.74 Å and its bond dissociation energy is 436 KJ/mol, whereas bond length of F – F in F₂ molecule is 1.44 Å and its bond dissociation energy is 159 KJ/mol.

10. Melting and boiling points of substances also can be determined by this bond energies and bond lengths.

Question 8.
Predict the reasons for low melting point for covalent compounds when compared with ionic compounds. (AS2)
(OR)
“Covalent compounds have low melting point.” What Is the reason for this statement? Explain.
Answer:
They are covalent compounds.

• The melting point is low due to the weak Vander Waal’s forces of attractions between the covalent molecules.
• The force of attraction between the molecules of a covalent compound is very weak.
• Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compounds have low melting points and low boiling points.
• Please note that some of the covalent solids like diamond and graphite have, however very high melting points and boiling points.

Question 9.
Collect the information about properties and uses of covalent compounds and prepare a report. (AS4)
(OR)
Generally these compounds are non-polar in nature. What are those compounds? Explain their properties and uses.
(OR)
Write any two uses and two properties of covalent compounds.
Answer:
The compounds are covalent.
Properties of covalent compounds :

1. Covalent compounds are usually liquids or gases, only some of them are solids.
2. The covalent compounds are usually liquids or gases due to the weak force of attraction between their molecules.
3. Covalent compounds have usually low melting and low boiling points.
4. Covalent compounds are usually in soluble in water but they are soluble in organic solvents.
5. Covalent compounds do not conduct electricity.

Uses of covalent compounds :

1. Covalent compounds form 99% of our body.
2. Water is a covalent compound. We know its many uses.
3. Sugars, food substances, tea and coffee are all covalent compounds.
4. Air we breathe in contains covalent molecules of oxygen and nitrogen.
5. Almost everything on earth other than most simple in organic salts are covalent.

Question 10.
Draw simple diagrams to show how electrons are arranged in the following covalent molecules : (AS5)
a) Calcium oxide (CaO)
b) Water (H₂O)
c) Chlorine (Cl₂)
Answer:
a) Calcium oxide (CaO) :

b) Water (H₂O):

The formation of water molecule can be shown like this also

c) Chlorine (Cl₂):

We can explain the formation of Cl₂ molecule in this way also.

Question 11.
Represent the molecule H₂O using Lewis notation. (AS5)
(OR)
How can you explain the formation of H₂O molecule using dot structure?
Answer:
One atom of oxygen shares its two electrons with two hydrogen atoms to form a water molecule.

Question 12.
Represent each of the following atoms using Lewis notation : (AS5)
a) Beryllium
b) Calcium
c) Lithium
Answer:

Question 13.
Represent each of the following molecules using Lewis notation : (AS5)
a) Bromine gas (Br2)
b) Calcium chloride (CaCl₂)
c) Carbon dioxide (CO₂)
d) Which of the three molecules listed above contains a double bond?
Answer:
a) Bromine gas (Br₂) :

b) Calcium chloride (CaCl₂)

c) Carbon dioxide (CO₂) :

d) CO₂, contains double bond in above list. Its structure is like this : O = C = O.

Question 14.
Two chemical reactions are described below. (AS5)
♦ Nitrogen and hydrogen react to form ammonia (NH₃).
♦ Carbon and hydrogen bond to form a molecule of methane (CH₄).
For each reaction give :
a) The valency of each of the atoms involved in the reaction.
b) The Lewis structure of the product that is formed.
Answer:
a) ♦ Nitrogen and hydrogen react to form ammonia (NH₃):
i) The valency of nitrogen is 3 and hydrogen is 1.
ii) The chemical formula of the product is NH₃

♦ Carbon and hydrogen bond to form a molecule of methane (CH₄):
i) The valency of carbon is 4 and hydrogen is 1.
ii) The chemical formula of the product is CH₄.

b) ♦ The Lewis structure of the product that is formed (: NH₃)

♦ The Lewis structure of the product that is formed (CH₄)

Question 15.
How does Lewis dot structure help in understanding bond formation between atoms? (AS6)
(OR)
What is the use of Lewis dot structure in bond formation? Explain.
Answer:

1. Only the outermost electrons of an atom take part in chemical bonding.
2. They are known as valence electrons.
3. The valence electrons in an atom are represented by putting dots (•) on the symbol of the element, one dot for each valence electron.
4. For example, sodium atom has 1 valence electron in its outermost shell, so we put 1 dot with the symbol of sodium and write Na• for it.
5. Sodium atom loses this 1 electron to form sodium ion.
6. By knowing the valence electrons of two different atoms by Lewis dot structure, we can understand which type of bond is going to establish between them and forms corresponding molecule.

Question 16.
What is octet rule? How do you appreciate role of the ‘octet rule’ in explaining the chemical properties of elements? (AS6)
(OR)
Which rule decides whether given element is chemically stable or not? Appreciate that rule.
Answer:
Octet rule decides whether given element is stable or not.
Octet rule :

• ‘The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer shell electrons.”
• It was found that the elements which participate in chemical reaction get octet (or) ns2 np6 configuration similar to that of noble gas elements.

Role of octet in chemical properties of elements :

1. Except He remaining inert gas elements have 8 electrons in their outermost orbit. Since these elements are having stable octet configuration in their outermost orbit, they are very stable.
2. They do not allow the outermost electrons to take part in chemical reactions.
3. So by having octet configuration for these elements we can conclude these are chemically inertial.
4. If any group of elements (take halogens) which contain 7 electrons in their outermost orbit, they require only 1 e^– to get octet configuration.
5. So they try to participate in chemical reaction to get that 1 difference electron for octet configuration.
6. Similarly, Na contains 2, 8, 1 as its electronic configuration.
7. So it loses le^– from its outermost shell; it should have 8e^– in its outer shell and get the octet configuration.
8. Thus the octet rule helps in explaining the chemical properties of elements.

Question 17.
Explain the formation of the following molecules using valence bond theory.
a) N₂ molecule
b) O₂ molecule
(OR)
Write the formation of double bond and triple bond according to valence bond theory.
(OR)
Who proposed Valence Bond Theory? Explain the formation of N₂ molecule by using this theory.
Answer:
Linus Pauling was proposed valence bond theory.
Formation of N₂ molecule :

1. Electronic configuration of Nitrogen is 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹.
2. Suppose that p_x orbital of one Nitrogen atom overlaps the p_x orbital of other ‘N’ atom giving σ p_x – p_x bond along the inter nuclear axis.
3. The p_y and p_z orbitals of one ‘N’ atom overlaps with the p_y and p_z orbital of other ‘N’ atom laterally giving π p_y – p_y and π p_z – p_z bonds.
4. Therefore, N₂ molecule has a triple bond between two Nitrogen atoms.

Formation of O₂ molecule :

1. Electronic configuration of ‘O’ is 1s² 2s² 2p_x² 2p_y¹ 2p_z¹.
2. If the P_y orbital of one ‘O’ atom overlaps the p_y orbital of other ‘O’ atom along inter- nuclear axis, a σ p_y – p_y bond is formed.
3. p_z orbital of oxygen atom overlaps laterally, perpendicular to inter nuclear axis giving a π p_y – p_z bond.
4. So O₂ molecule has a double bond between the two oxygen atoms.

Question 18.
What is hybridisation? Explain the formation of the following molecules using hybridisation.
a) BeCl₂
b) BF₃
Explain the formation of sp and sp² hybridisation using examples.
(OR)
What is the name given to inter mixing of atomic orbitals to form new orbitals. Explain the formation of following molecules by using that process,
a) BeCl₂
b) BF₃
Answer:
This process is called hybridisation.
Hybridisation :
It is a phenomenon of inter mixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

a) Formation of BeCl₂ (Beryllium chloride) molecule :

1. ₄Be has electronic configuration 1s² 2s².
2. It has no unpaired electrons.
3. It is expected not to form covalent bonds, but informs two covalent bonds one each with two chlorine atoms. „
4. To explain this, an excited state is suggested for Beryllium in which an electron from ‘2s’ shifts to 2p_x level.
5. Electronic configuration of ₄Be is 1s² 2s¹ 2p_x¹].
6. Electronic configuration of ₁₇Cl is 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹.
7. If Be forms two covalent bonds with two chlorine atoms, one bond should be σ 2s-3p due to the overlap of ‘2s’ orbital of Be, the ‘3p_z‘ orbital of one chlorine atom.
8. The other bond should be σ 2p-3p due to the overlap of ‘2p_x’ orbital of Be atom the 3p orbital of the other chlorine atom.
9. As the orbitals overlapping are different, the bond strengths of two Be-Cl must be different.
10. But, both bonds are of same strength and Cl\(\hat{\mathrm{Be}}\) Cl is 180°.

The Hybridisation of BeCl₂ can be explained in this way also :
a) Be atom in its excited state allows its 2s orbital and 2p_x orbital which contain unpaired electrons to intermix and redistribute to two identical orbitals.
b) As per Hund’s rule each orbital gets one electron.
c) The new orbitals based on the types of orbitals that have undergone hybridisation are called sp orbitals.
d) The two sp orbitals of Be get separated by 180°.
e) Now each chlorine atom comes with its 3p_z¹ orbital and overlaps it the sp orbitals of Be forming two identical Be-Cl bonds (σ sp-p bonds).
Cl\(\hat{\mathrm{Be}}\) Cl = 180°.

f) Both the bonds are of same strength.

b) Formation of BF₃ molecule :

1. ₅B has electronic configuration 1s² 2s² 2pxh
2. The excited electronic configuration of ₅B is 1s² 2s¹ 2p_x¹ 2p_y¹
3. As it forms three identical B-F bonds in BF₃.
4. It is suggested that excited ‘B’ atom undergoes hybridisation.
5. There is an intermixing of 2s, 2p_x, 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one election.
7. Now three fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (F₉ 1s² 2s² 2p_x² 2p_y² 2p_z¹) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three σsp²-p bonds.

Fill in the Blanks

1. Electrons in the outermost orbit are called …………………… .
2. Except …………………… gas all other noble gases have octet in their valence shell.
3. Covalency of elements explains about member of …………………… formed by the atom.
4. Valence bond theory was proposed by …………………… .
5. In …………………… bonding the valence electrons are shared among all the atoms of the metallic elements.
Answer:

1. valence electrons
2. Helium
3. covalent bonds
4. Linus Pauling
5. covalent

Multiple Choice Questions

1. Which of the following elements is electronegative?
A) Sodium
B) Oxygen
C) Magnesium
D) Calcium
Answer:
B) Oxygen

2. An element ₁₁X²³ forms an ionic compound with another element ‘Y’. Then the charge on the ion formed by X is
A) +1
B) +2
C) -l
D) – 2
Answer:
A) +1

3. An element ‘A’ forms a chloride ACl₄. The number of electrons in the valence shell of ‘A’
A) 1
B) 2
C) 3
D) 4
Answer:
D) 4

10th Class Chemistry 10th Lesson Chemical Bonding InText Questions and Answers

10th Class Chemistry Textbook Page No. 153

Question 1.
How do elements usually exist?
Answer:
They may exist as a single atom or as a group of atoms.

Question 2.
Do atoms exist as a single atom or as a group of atoms?
Answer:
Atoms exist as a single atom, sometimes as a group of atoms also.

Question 3.
Are there elements which exist as atoms?
Answer:
Yes. There are elements which exist as atoms.

Question 4.
Why do some elements exist as molecules and some as atoms?
Answer:
By following different laws of chemical combination the chemical compounds take place as a result of combination of atoms of various elements in different ways.

Question 5.
Why do some elements and compounds react vigorously while others are inert?
Answer:
1) Number of electrons in their outermost shell.
2) Bond strength between the atoms in compound.

Question 6.
Why is the chemical formula for water H₂O and for sodium chloride NaCl, why not HO₂ and NaCl₂?
Answee:
Valencies of the atoms participating in the molecules.

Question 7.
Why do some atoms combine dille tl do not?
Answer:
1) Atoms which have 8e“ in their outer shell will not combine.
2) Atoms which have more than or less than 8e“ in their outer shell will combine.

Question 8.
Are elements and compounds simply made up of separate atoms Individually arranged?
Answer:
No. Elements and compounds are not simply made up of separate atoms individually arranged.

Question 9.
Is there any attraction between atoms?
Answer:
Yes. There is some attraction betwen atoms.

Question 10.
What is that holding them together?
Answer:
Force of attraction between them.

10th Class Chemistry Textbook Page No. 155

Question 11.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
Answer:’
Because of bond energy between the atoms in a molecule.

Question 12.
Where does the absorbed energy go?
Answer:
For breaking chemical bonds between atoms in a molecule.

Question 13.
Is there any relation to energy and bond formation between atoms?
Answer:
Yes. There is some relation to energy and bond formation between atoms.

Question 14.
What could be the reason for the change in reactivity of elements?
Answer:
Number of electrons in their outermost orbit.

Question 15.
What could be the reason for this?
Answer:
They have 8 (e^–) electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 157

Question 16.
What did you notice in Lewis dot structure of noble gases and electronic configurations of the atoms of these elements shown in table – 1?
Answer:
Except He remaining Ne, Ar, Kr have 8 electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 158

Question 17.
What have you observed from the above conclusions about the main groups?
Answer:

1. Number of gained electrons of non-metals in their valency.
2. Number of lost electrons of metals in their valency.

Question 18.
Why do atoms of elements try to combine and form molecules?
Answer:
To attain stable electronic configuration.

10th Class Chemistry Textbook Page No. 159

Question 19.
Is it accidental that IA to VIIA main group elements during chemical reactions get eight electrons in the outermost shells of their ions, similar to noble gas atoms?
Answer:
No, it cannot be simply accidental.

Question 20.
Explain the formation of ionic compounds NaCl, MgCl₂, Na₂O and AlCl₃ through Lewis electron dot symbols (formulae).
Answer:
1) Lewis electron dot symbol for NaCl:

Formation of sodium chloride (NaCl) :
Sodium chloride is formed from the elements sodium and chlorine. It can be explained as follows.
Na_(s) + ½Cl_2(g) → NaCl₂

Cation formation :
When sodium (Na) atom loses one electron to get octet electron configuration it forms a cation (Na+) and gets electron configuration that of Neon (Ne) atom.

Anion formation :
Chlorine has shortage of one electron to get octet in its valence shell. So it gains the electron from Na atom to form anion and gets electron configuration as that of argon (Ar).

Formation of the compound NaCl from its ions :
Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–‘ ions. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–_(g) → Na⁺Cl^–_(s) or NaCl

2) Lewis electron dot symbol for MgCl₂:
MgCl₂

Formation of magnesium chloride (MgCl₂):
Magnesium chloride is formed from the elements magnesium and chlorine. The bond formation MgCl₂ in brief using chemical equation is as follows :
Mg_(s) + Cl_2(g) → MgCl_2(g)
Cation formation:

Anion formation :

The compound MgCl₂ formation from its ions :
Mg²⁺ gets ‘Ne’ configuration and
Each Cl^– gets ‘Ar’ configuration
Mg²⁺_(g) + 2 Cl^–_(g) → MgCl_2(s)
One ‘Mg’ atom transfers two electrons one each to two ‘Cl’ atoms and so formed Mg²⁺ and 2Cl^– attract to form MgCl₂.

3) Lewis electron dot symbol for (Na₂O) :

Formation of di sodium monoxide (Na₂O):
Di sodium monoxide formation can be explained as follows:
Cation formation (Na⁺ formation):

Two ‘Na’ atoms transfer one electron each to one oxygen atom to form 2 Na⁺ and O^2-
Each Na⁺ gets ‘Ne’ configuration and O^2- gets ‘Ne’ configuration.
These ions (2Na⁺ and O^2-) attract to form Na₂O.

4) Lewis electron dot symbol for (AlCl₃):

Formation of aluminium chloride (AlCl₃):
Aluminium chloride formation can be explained as follows:
Formation of aluminium ion (Al³⁺), the cation:

Each aluminium atom loses three electrons and three chlorine atoms gain them, one electron each.
The compound AlCl₃ is formed from its component ions by the electrostatic forces of attractions.
Al³⁺_(g) + 3 Cl^–_(g) → AlCl_3(s)

10th Class Chemistry Textbook Page No. 163

Question 21.
How do cations and anions of an ionic compound exist in its solid state?
Answer:
Ionic compounds exist in crystalline state.

Question 22.
Do you think that pairs of Na⁺ Cl^– as units would be present in the solid crystal?
Answer:
Yes. I think that pairs of Na⁺ Cl^– as units would be present in the solid crystal.

10th Class Chemistry Textbook Page No. 164

Question 23.
Can you explain the reasons for all these?
Answer:
Ionic bond is formed between atoms of elements with electronegativity, difference equal to or greater than 1.9.

10th Class Chemistry Textbook Page No. 165

Question 24.
Can you say what type of bond exists between atoms of nitrogen molecule?
Answer:
Triple bond exists between atoms of nitrogen molecule.

10th Class Chemistry Textbook Page No. 168

Question 25.
What do you understand from bond lengths and bond energies?
Answer:
Bond lengths and bond energies are not same when the atoms that form the bond are different.

Question 26.
Are the values not different for the bonds between different types of atoms?
Answer:
Yes. The values are not different for the bonds between different types of atoms.

10th Class Chemistry Textbook Page No. 170

Question 27.
What is the bond angle in a molecule?
Answer:
It is the angle subtended by two imaginary lines that pass from the nuclei of two atoms which form the covalent bonds with the central atom through the nucleus of the central atom at the central atom.

10th Class Chemistry Textbook Page No. 172

Question 28.
How is MCI molecule formed?
Answer:
The ‘1s’ orbital of ‘H’ atom containing unpaired electron overlaps the ‘3p’ orbital of chlorine atom containing unpaired electron of opposite spin.

10th Class Chemistry 10th Lesson Chemical Bonding Activities

Activity – 1

1. Write the Lewis structures of the given elements in the table. Also, consult the periodic table and fill in the group number of the element.
Answer:

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 12 Electromagnetism Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 12th Lesson Electromagnetism

10th Class Physics 12th Lesson Electromagnetism Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
How do electric appliances work?
Answer:
Electrical appliances work due to the electric force. Electrical force works in displacing the charges. Electric force is independent of the state of rest or the motion of the charged particle. Electric motor, washing machine are some of the examples of electric appliances.

Question 2.
How do electromagnets work?
Answer:
An electromagnet acquires the magnetic properties only when electric current is passed through the solenoid. Once the current is switched off, it almost loses its magnetic properties as retentivity of soft iron is very low. The strength of the electromagnet depends upon number of turns per unit length of the solenoid and the current through the solenoid.

Question 3.
Is there any relation between electricity and magnetism?
Answer:
The first evidence that there exists such a relationship between electricity and magnetism was observed by Oersted. When current carrying conductor was parallel to the axis of the needle, and the needle was deflected. This was much against his expectations. On reversing the direction of the current the needle moved in opposite direction.

Question 4.
Can we produce magnetism from electricity?
Answer:
We can produce magnetism from electronic current. Ampere with his Ampere’s swimming rule explained the direction of electric current and the deflection of magnetic needle.

Improve Your Learning

Question 1.
Are the magnetic field lines closed? Explain. (AS1)
Answer:

• Magnetic field lines are closed.
• If we observe the field lines formed by a current carrying straight wire, circular field lines are formed. They are closed circles.
• If we observe the field lines by a current carrying solenoid the field lines out side the solenoid are continuous with those inside.
• Thus the magnetic field lines are closed loops.

Question 2.
See figure, magnetic lines are shown. What is the direction of the current flowing through the wire? (AS1)

Answer:
If field lines are in anti-clockwise direction as shown in the diagram, the direction of current is vertically upwards. This can be demonstrated with right hand thumb rule.

Question 3.
A bar magnet with north pole facing towards a coil moves as shown in figure. What happens to the magnetic flux passing through the coil? (AS1)

(OR)
Why would induced current be generated in the coil when a north pole of a bar magnet pushed into it ?
Answer:
If north pole of the magnet moves towards the coil, there is a continuous change of magnetic flux linked with closed coil, then current is generated in the coil.

Question 4.
A coil is kept perpendicular to page. At P, current flows into the page and at Q it comes out of the page as shown in figure. What is the direction of magnetic field due to the coil? (AS1)

Answer:
At the top, anti-clockwise direction.
At the bottom, clockwise direction.
Try This:

Take a test tube and wound minimum 50 turns of 24 guage insulated copper wire with 2cms length at the centre of test tube as shown in figure, ‘l Now solenoid is ready. Take 3cms length of iron nail and make it floats on water with appropriate foam (thermocol) on the water. Now connect the j two ends of solenoid two 3-6 volts battery eliminator and switch on the eliminator. You can observe the motion of the nail towards the solenoid. (If not move decrease the water level or increase the potential).
Try to explain motion of the nail into the water using solenoid concept.

Question 5.
The direction of current flowing in a coil is shown in the figure. What type of magnetic pole is formed at the face that has flow of current as shown in the figure? (AS1)

Answer:
North. Since the current in the coil flows in anti-clockwise direction, north pole is formed at the face we are watching. 

Question 6.
Why does the picture appear distorted when a bar magnet is brought close to the screen of a television? Explain. (AS1)
(OR)
Explain magnetic force on moving charge and current carrying wire.
(OR)
What happens when you bring a bar magnet near a picture of TV screen ? What inference do you conclude from this activity?
Answer:
This is due to the fact that magnetic field exerts a force on the moving charge.

TV screen Activity :

• Take a bar magnet and bring it near the TV screen.
• Then the picture on the screen is distorted.
• Here the distortion is due to the motion of the electrons reaching the screen are affected by the magnetic field.
• Now move the bar magnet away from the screen.
• Then the picture on the screen stabilizes.
• This must be due to the fact that the magnetic field exerts a force on the moving charges. This force is called magnetic force.
• The magnitude of the force is F = Bqv where B is magnetic induction, ‘q’ is the charge and v is the velocity of the charged particle.

Question 7.
Symbol ‘X’ indicates the direction of a magnetic held into the page. A straight long wire carrying current along its length is kept perpendicular to the magnetic field. What is the magnitude of force experienced by the wire? In what direction does it act? (AS1)

Answer:
1) Magnetic force (F) experienced by the wire with the magnitude of ILB :
Here I = Current, L = Length of the wire B = Magnetic field

2) Direction of the magnetic force (F) :

1) The direction of force can be find by using Right hand rule.
2) Fore finger → i (North)
Middle finger → B (into the page)
Thumb → F (Towards west parallel to the paper)

Question 8.
Explain the working of electric motor with a neat diagram. (AS1)
(OR)
Which device converts electrical energy into mechanical energy? Explain the working of that device with a neat diagram.
Answer:
Electric motor converts electrical energy into mechanical energy.
Electric motor:
It is a device which converts electrical energy into mechanical energy.

Principle :
It is based on the principle that a current carrying conductor placed perpendicular to the magnetic field experiences a force.

Construction :
a) Armature coil:
It contains a single loop of an insulated copper wire in the form of a rectangle.

b) Strong magnetic field :
Armature coil is placed between two permanent poles (N & S) of a strong magnet.

c) Slip-ring Commutator:
It consists of two halves (C₁ and C₂) of a metallic ring. The two ends of the armature coil are connected to these two halves of the ring. Commutator reverses the direction of current in the armature coil.

d) Brushes:
Two carbon brushes B₁ and B₂ press against the commutator. These brushes act as the contact between the commutator and terminals of the battery.

e) Battery :
A battery is connected across the carbon brushes. The battery supplies the current to the armature coil.

Working and Theory :

1. When current flows through the coil, AB and CD experience magnetic force.
2. In the arm, AB of the coil experiences a force in one direction, similarly, in CD it experiences in opposite direction.
3. These two equal and opposite forces constitute a couple; which rotates the coil.
4. At this position, the supply of current to the coil is cut off because contacts of commutator and brushes break.
5. Hence no force acts on the arms of the coil.
6. The coil will not come to rest because of rotational inertia of motion, till the commutator again comes in contact with the brushes B₁ and B₂.
7. Now the direction of the current in the arms AB and CD is reversed.
8. Then the couple again rotates in opposite direction.
9. The coil of DC motor continues to rotate in the same direction. Hence electrical energy is converted into mechanical energy.
10. The speed of rotation of the motor depends on
    a) current through the armature
    b) number of turns of the coil
    c) area of the coil
    d) magnetic induction.

Question 9.
Derive Faraday’slaw of induction from law of conservation of energy. (AS1)
Answer:
Faraday’s law :
Whenever there is a continuous change in magnetic flux linked with coil closed the current is generated in the coil.

• Consider a pair of parallel bare conductors which are separated by 7′ meters.
• They are placed in uniform magnetic field of induction ‘B’ supplied by ‘N’ and ‘S’ poles of the magnet.
• A galvanometer is connected to the ends of the parallel conductors.
• We can close the circuit by touching the parallel conductor with another bare conductor which is taken in our hand.
• If we move our hand to the left, the galvanometer needle will deflect in one direction.
• If we move our hand to the right, the needle in the galvanometer moves in opposite direction.
• A current will be set up in the circuit only when there is an EMF in the circuit. Let EMF be ‘ε’.
• The principle of conservation energy tells us that this electric energy must come from the work that we have done in moving the cross wire.

Question 10.
The value of magnetic field induction which is uniform is 2T. What is the flux passing through a surface of area 1.5 m2 perpendicular to the field? (AS1)
Answer:
B = 2T ; Φ == ? ; A = 1.5 m²
We know B = \(\frac{\phi}{\mathrm{A}}\)
or Φ = BA = 2 × 1.5 =3 Webers

Question 11.
An 8N force acts on a rectangular conductor 20 cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40 A. (AS1)
Answer:
F = 8N ; l = 20 cm or 20 × 10^-2 m ; B = ? ; i = 40 Amp

Question 12.
Explain with the help of two activities that current carrying wire produces magne tic field. (AS1)
(OR)
How can you verify that a current carrying wire produces a magnetic field with the help of experiment?
Answer:
Activity – 1

• Take a thermocole sheet and fix two thin sticks ol height 1cm.
• Join the two sticks with the help of copper wire.
• Take a battery, tap key and connect them in series with the copper wire which is thin.
• Keep a marine compass needle beneath the wire.
• If you press the tap key, current flows in the copper wire.
• Immediately the magnetic needle gets deflected.
• This indicates that the magnetic field is increased when current flows through the conductor.

Activity – 2

• Take a wooden plank and make a hole.
• Place the plank on the table.
• Place the retort stand on it.
• Pass copper wire through the hole.
• Connect the two ends of the wire with battery through switch.
• Place some compass needle around the hole.
• When the current flows the magnetic needle deflects.
• We can verify this by changing the direction of current.
• So we can conclude the magnetic field surrounds a current carrying conductor.

Question 13.
How do you verify experimentally that the current carrying conductor experiences a force when it is kept in magnetic field? (AS1)
Answer:

1. A copper wire is passed through splits of wooden sticks.
2. Connect the wire to 3 volts battery.
3. Close the switch of the battery and pass the current.
4. Bring the horse-shoe magnet near the wire.
5. Then a force is experienced on the wire.
6. Reverse the polarities of the magnet, then the direction of the force is also reversed.
7. The right hand rule helps the direction of flow of current and the direction of current.

Question 14.
Explain Faraday’s law of induction with the help of an activity. (AS1)
(OR)
Write an activity which proves changing magnetic flux produces induced current in the circuit.
Answer:

• Connect the terminals of a coil to a sensitive ammeter.
• Push a bar magnet towards the coil, with its north pole facing the coil, the needle in the galvanometer deflects.
• It shows that a current is set up in the coil.
• The galvanometer does not deflect if the magnet is at rest.
• If the magnet is moved away from the coil, the needle in the galvanometer again deflects in opposite direction.
• Further this experiment enables us to understand that the relative motion of the magnet and coil set up a current in the coil. It makes no difference whether the magnet is moved towards the coil. This is one form of Faraday’s law.

Question 15.
Explain the working of AC electric generator with a neat diagram. (AS1)
(OR)
Which device converts mechanical energy into electrical energy? Explain the working of that device with a neat diagram.
Answer:
Generator converts mechanical energy into electrical energy.

• As armature is rotated about an axis, the magnetic flux linked with armature changes. Therefore, an induced current is produced in the armature.
• If the armature rotates in anti-clockwise direction, from Flemming’s right hand rule the direction of current and deflection of the coil are noted.
• Alter armature has turned through 180°, it occupies another position.
• By applying Flemming’s right hand rule we can find the direction of current and deflection of the needle.
• Hence we can conclude the induced current is alternating in nature.

Question 16.
Explain the working of DC generator with a neat diagram. (AS1)
Answer:

• The principle and working of D.C generator is same as that of AC generator except that in place of slip – rings as sliding contacts, we have a slip-ring or a commutator.
• In a slip ring, there are two half rings.
• The ends of armature coil are connected to these rings and these rings rotate the armature.
• By using slip-ring, the direction of induced current does not change in the external circuit throughout the complete rotation of the armature. In other words, the current in the external circuit always flows in the same direction. Hence the induced current is unidirectional.

Question 17.
Rajkumar said to you that the magnetic field lines are open and they start at north pole of bar magnet and end at south pole. What questions do you ask Rajkumar to correct him by saying “field lines are closed”? (AS2)
Answer:

• If the magnetic field lines start at north pole and end at south pole, where do the lines go from south pole?
• What is happening within the bar magnet?
• Are the magnetic field lines passing through bar magnet?
• What is the direction of magnetic field lines inside the bar magnet? (Recall the solenoid activity).
• Can you say now, that the magnetic field lines are open?

Question 18.
As shown in figure, both coil and bar magnet move in the same direction. Your friend is arguing that there is no change in flux. Do you agree with his statement? If not, what doubts do you have? Frame questions about the doubts you have regarding change in flux. (AS2)

Answer:

• What happens if both magnet and coil move in same direction?
• What happens if both magnet and coil move in opposite direction?
• What is the direction of the current in the coil?
• If both move in same direction, is there any linkage of flux with the coil?
• When ‘N’ pole is moved towards the coil what is the direction of current?
• If magnet is reversed, what is the direction of current in the coil?

Question 19.
What experiment do you suggest to understand Faraday’s law? What items are required? What suggestions do you give to get good results of the experiment? Give precautions also. (AS3)
Answer:
Aim :
To understand Faraday’s law of induction.

Materials required :
A coil of copper wire, a bar magnet, Galvanometer, etc.

Procedure :

1. Connect the terminal of a coil to a sensitive galvanometer as shown in the figure.
2. Normally, we would not expect any deflections of needle in the galvanometer because there is to be no electromotive force in this circuit.
3. Now if we push a bar magnet towards the coil, with its north pole facing the coil, we observe the needle in the galvanometer deflects, showing that a current is set up in the coil.
4. The galvanometer does not deflect if the magnet is at rest.
5. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction.
6. If we use the end of south pole of a magnet instead of north pole in the above activity, the deflections are exactly reversed.
7. This experiment proves “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Precautions :

1. The coil should be kept on an insulating surface.
2. Bar magnet should be of good magnetic moment.
3. The centre of the galvanometer scale must be zero.
4. The deflections in the galvanometer must be observed while introducing the bar magnet into the coil and also while withdrawing it.

Question 20.
How can you verify that a current carrying wire produces a magnetic field with the help of an experiment? (AS3)
Answer:
Experiment:

• Take a thermocole sheet and fix two thin wooden sticks of height 1cm.
• These sticks are joined with the help of a copper wire.
• Connect battery and tap key to this copper wire.
• Place a magnetic compass beneath the wire.
• Now press the tap key and allow the current through the wire. It is observed that magnetic needle deflects.
• If you change the direction of the current, the direction of deflection of needle also changes.
• So we can say current carrying conductor produces magnetic field.

Question 21.
Collect information about generation of current by using Faraday’s law. (AS4)
Answer:
Faraday’s law is useful in generation of current.

1. According to this law, the change in magnetic flux induces EMF in the coil.
2. Fie also proposed electromagnetic induction.
3. Electromagnetic induction is a base for generator, which produces electric current.
4. Transformer also works on the principle of electromagnetic induction, which is helpful in transmission of electricity.
5. Hence Faraday’s law is used in the generation and transmission of current.

Question 22.
Collect information about material required and procedure of making a simple electric motor from internet and make a simple ntotor on your own. (AS4)
Answer:
Aim :
Preparation of a simple electric motor.

Material requried :
A wire of nearly 15 cm, 1.5v Battery, Iron nail, strong magnet, paper clip.

Procedure:

1. Attach the magnet to the head of the iron nail.
2. Attach a paper clip to the open end of the magnet.
3. Now attach the other end of the nail (Free end) to the cap (positive terminal) of the battery.
4. Now connect the negative terminal of the battery and the head of the iron nail through a wire.
5. We observe that the paper clip rotates.

Another model:
Materials required :
1.5 m enamelled copper wire (about 25 gauge), 2 safety pins,
1.5 v battery, magnets, rubber bands or bands cut from cycle tube.

Procedure :

1. Wind copper wire on the battery nearly 10 – 15 turns to make a coil.
2. Remove the coil and fix the ends as shown in the figure.
3. Scrape the insulation com¬pletely on one end of the coil.
4. Scrape the insulation on top, left and right of the other end. The bottom should be insulated.
5. Now complete the electric mo¬tor as shown in the figure. “5

Question 23.
Collect information of experiments done by Faraday. (AS4)
Answer:
Experiment – 1

1. Connect the terminals of a coil to a sensitive galvanometer as shown in the figure.
2. Normally, we would not expect any deflection of needle in the galvanometer because there is no EMF in the circuit.
3. Now, if we push a bar magnet towards the coil, with its north pole facing the coil, the needle in the galvanometer deflects, showing that a current has been set up in the coil, the galvanometer does not deflect if the magnet is at rest.
4. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction.
5. If we use the end of south pole of a magnet instead of north pole, the results i.e., the deflections in galvanometer are exactly opposite to the previous one.
6. This activity proves that the change in magnetic flux linked with a closed coil, produces current.
7. From this Faraday’s law of induction can be stated as “whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”. This induced EMF is equal to the rate of change of magnetic flux passing through it.

Experiment – 2

1. Prepare a coil of copper wire C₁ and connect the two ends of the coil to a galvanometer.
2. Prepare another coil of copper wire similar to C₂ and connect the two ends of the coil to a battery via switch.
3. Now arrange the two coils C₁ and C₂ nearby as shown in the figure.
4. Now switch on the coil C₂. We observe a deflection in the galvanometer connected to the coil C₁.
5. The steady current in C₂ produces steady magnetic field. As coil C₂ is moved towards the coil C₁ the galvanometer shows a deflection.
6. This indicates that electric current is induced in coil C₁.
7. When C₂ is moved away, the galvanometer shows a deflection again, but this time in the opposite direction.
8. The deflection lasts as long as coil C₂ is in motion.
9. When C₂ is fixed and C₁ is moved, the same effects are observed.
10. This shows the induced EMF due to relative motion between two coils.

Question 24.
Draw a neat diagram of electric motor. Name the parts. (AS5)
Answer:

Question 25.
Draw a neat diagram of an AC generator. (AS5)
(OR)
Draw the diagram of electric generator and label its parts. A.
Answer:

Question 26.
How do you appreciate the Faraday’s law, which is the consequence of conservation of energy? (AS6)
Answer:

• Law of conservation of energy says energy neither be created nor be destroyed, but can be converted from one form to another.
• Faraday’s law says whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil. This induced EMF is equal to the rate of change of magnetic flux passing through it.
• We have to do some work to move the magnet through a coil. This work produces energy.
• This energy is converted into electrical energy in the coil.
• In this way conservation of energy takes place in electromagnetic induction.

Question 27.
How do you appreciate the relation between magnetic field and electricity that changed the lifestyle of mankind? (AS6)
Answer:

• Changing life style of mankind is a result of many inventions, utilising a lot of scientific principles.
• Scientists all ways going on searching for new principles and new applications to make our life more comfortable.
• If you consider electricity, right from amber stone to nuclear power, so many changes have been incorporated.
• The idea of Oersted and Faraday that current carrying wire produces electricity and electromagnetic induction, enable us to use electric motors, generators, fans, mixers, grinders, induction stoves, etc.
• All these appliances makes our life more comfortable. Hence Faraday and Oersted rendered a lot of servies in this field.
• Hence, I appreciate the relation between magnetic field and electricity that changed the life style of mankind.
  So if current is more, induction is also more.

Question 28.
Give a few applications of Faraday’s law of induction in daily life. (AS7)
Answer:
Applications:
The daily life applications of Faraday’s law of induction are

1. Generation of electricity
2. Transmission of electricity
3. Metal detectors in security checking
4. The tape recorder
5. Use of ATM cards
6. Induction stoves
7. Transformers
8. Induction coils (spark plugs in automobiles)
9. Break system in railway wheels
10. AC and DC generators
11. Windmills, etc.

Question 29.
Which of the various methods of current generation protects the nature well? Give examples to support your answer. (AS7)
Answer:
Windmill :

• Electricity is produced when an armature of a generator rotates between two poles of a strong magnet.
• Whereas when wind falls on the wheel of a windmill, it rotates. So the armature of the generator rotates between two poles of a magnet along with the rotation of the wheel of the windmill.
• Thus electric current is produced.
• This is how, KE of the wind is converted into electric energy.

Advantages :
Wind energy produces no smoke and no harmful gases. So this form of energy is pollution free or environment-friendly.

Fill in The Blanks

1. The SI unit of magnetic field induction is ………………….
2. Magnetic flux is the product of magnetic field induction and …………………
3. The charge is moving along the direction of magnetic field. Then force acting on it is ………………..
4. A current carrying wire of length L is placed perpendicular to a uniform magnetic field B. Then the force acting on the wire with current I is ……………..
5. Faraday’s law of induction is the consequence of …………………
Answer:

1. weber/m² (or) Tesla
2. area
3. zero
4. ILB
5. Law of conservation of energy

Multiple Choice Questions

1. Which converts electrical energy into mechanical energy?
A) motor
B) battery
C) generator
D) switch
Answer:
A) motor

2. Waich converts mechanical energy into electrical energy?
(OR)
The device used to convert mechanical energy into electrical energy among the following is
A) motor
B) battery
C) generator
D) switch
Answer:
C) generator

3. The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field, is
A) 0
B) ILB
C) 2ILB
D) ILB/2
Answer:
B) ILB

10th Class Physics 12th Lesson Electromagnetism InText Questions and Answers

10th Class Physics Textbook Page No. 221

Question 1.
Why does the needle get deflected by the magnet?
Answer:
Because of strength of the magnetic field of the magnet, the needle gets deflected since it is in the field.

10th Class Physics Textbook Page No. 213

Question 2.
How can we find the strength of the field and direction of the field?
Answer:
We can find the strength of the field with magnetic flux and the direction of the field from the tangent drawn to the line of force.

10th Class Physics Textbook Page No. 214

Question 3.
Can we give certain values to magnitude of the field at every point in the magnetic field?
Answer:
In uniform magnetic field it is same whereas in non-uniform magnetic field it is different.

10th Class Physics Textbook Page No. 215

Question 4.
What is the flux through unit area perpendicular to the field?
Answer:
Flux density or magnetic induction.

Question 5.
Can we generalize the formula of flux for any orientation of the plane taken in the field?
Answer:
Yes, Φ = BA cos θ

Question 6.
What is the flux through the plane taken parallel to the field?
Answer:
Magnetic flux (or) Magnetic field.

Question 7.
What is the use of introducing the ideas of magnetic flux and magnetic flux density?
Answer:
Magnetic flux and flux density help in understanding the concept of electromagnetic induction and relation between electricity and magnetism.

Question 8.
Are there any sources of magnetic field other than magnets?
Answer:
Current carrying straight wires and loops act as sources of magnetic filed.

Question 9.
Do you know how old electric calling bells work?
Answer:
Yes. They work on the principle of magnetic effect of electric currents.

10th Class Physics Textbook Page No. 218

Question 10.
What happens when a current carrying wire is kept in a magnetic field?
Answer:

• Magnetic field applies force on current carrying wire.
• So it gets deflected and the direction of deflection is given by right hand rule.
• Or there will be no force acting on the wire when wire is in the direction of the field.

10th Class Physics Textbook Page No. 219

Question 11.
Do you feel any sensation on your skin?
Answer:
Yes. The hair on my skin rises up when I stand near TV screen.

Question 12.
What could be the reason for that?
Answer:
It is due to the magnetic field produced by electric charges in motion.

Question 13.
Why does the picture get distorted?
Answer:
Due to motion of electrons that form the picture is affected by the magnetic field of bar magnet.

Question 14.
Is the motion of electrons reaching the screen affected by the magnetic field of the bar magnet?
Answer:
Yes. The motion of electrons reaching the screen is affected by the magnetic field of the bar magnet.

Question 15.
Can we calculate the force experienced by a charge moving in a magnetic field?
Answer:
Yes, If the force is F, it is given by the expression F = qvB.

Question 16.
Can we generalize the equation for magnetic force on charge when there is an angle ‘0’ between the directions of field “B” and velocity “v”?
Answer:
No, Then force F is given by the formula F = qvB sin θ.

Question 17.
What is the magnetic force on the charge moving parallel to a magnetic field?
Answer:
When the charge moves parallel to the magnetic field the value of “θ” becomes zero. In the equation F = qvB sin θ, since θ = θ, the value of force F also becomes zero.

Question 18.
What is the direction of magnetic force acting on a moving charge?
Answer:
By applying right hand rule we can guess the direction of magnetic force acting on a moving charge is the “thumb” direction.

10th Class Physics Textbook Page No. 221

Question 19.
Can you determine the magnetic force on a current carrying wire which is placed along a magnetic field?
Answer:
F = BIl sin θ. If the current carrying wire is placed along direction field θ = 0.
∴ F = 0

Question 20.
What is the force on the wire if its length makes an angle ‘θ’ with the magnetic field?
Answer:
F = Bqv sin θ or F = Bil sin θ, where ‘i’ is current. WorhA
Here B = magnetic induction, q = charge, v = velocity of the charge and ‘θ’ is the angle between direction of field and velocity.

10th Class Physics Textbook Page No. 222

Question 21.
How could you find its (current carrying wire) direction?
Answer:
We can find by using right hand rule.

Question 22.
Is the direction of deflection observed experimentally same as that of the theoretically expected one?
Answer:
Yes. But it depends on polarities of the horse shoe magnet.

Question 23.
Does the right hand rule give the explanation for the direction of magnetic force exerted by magnetic field on the wire?
Answer:
The right hand rule does not help us to explain the reason for deflection of wire.

Question 24.
Can you give a reason for it (deflection of wire)?
Answer:
There exists only magnetic field due to external source. When there is a current in the wire, it also produces a magnetic field. These fields overlap and give non-uniform field. This is the reason for it.

10th Class Physics Textbook Page No. 223

Question 25.
Does this deflection fit with the direction of magnetic force found by right hand rule?
Answer:
Yes. This deflection fits with the direction of magnetic force found by right hand rule.

Question 26.
What happens when a current carrying coil is placed in a uniform magnetic field?
Answer:
It gets deflected since magnetic lines of force are perpendicular to the length of the coil.

Question 27.
Can we use this knowledge to construct an electric motor?
Answer:
Yes. This is the principle of electric motor.

Question 28.
What is the angle made by AB and CD with magnetic field?
Answer:

AB and CD are at right angles to the magnetic field.

10th Class Physics Textbook Page No. 224

Question 29.
Can you draw the direction of magnetic force on sides AB and CD?
Answer:
Yes. The direction of magnetic force on sides AB and CD can be determined by applying right hand rule.

Question 30.
What are the directions of forces on BC and DA?
Answer:

ADBC, magnetic force pulls the coil up and at DA magnetic force pulls it down.

Question 31.
What is the net force on the rectangular coil?
Answer:
Net force on the rectangular coil is zero.

Question 32.
Why does the coil rotate?
Answer:

The rectangular coil rotates in clockwise direction because of equal and opposite pan’ of forces acting on the two sides of the coil.

Question 33.
What happens to the rotation of the coil if the direction of current in the coil remains unchanged?
Answer:
The coil comes to halt and rotates in anti-clockwise direction.

Question 34.
How could you make the coil rotate continuously?
Answer:
If the direction of current in coil, after the first half rotation, is reversed, the coil will continue to rotate in the same direction.

10th Class Physics Textbook Page No. 225

Question 35.
How can we achieve this (convertion of electrical energy to mechanical energy)?
Answer:
Brushes B₁ and B₂ are used to achieve this.

Question 36.
What happens when a coil without current is made to rotate in magnetic field?
Answer:
When the coil rotated due to the change in magnetic flux electricity is generated.

Question 37.
How is current produced?
Answer:
The current is produced from the battery to the coil.

Question 38.
Why is there a difference in behaviour in these two cases?
Answer:
The A.C. supply changes its direction a number of times in a second. But D.C. is unidirectional current. So there is a difference in the behaviour of the metal ring in these two cases.

Question 39.
What force supports the ring against gravity when it is being levitated?
Answer:
The magnetic force developed in the coil of copper wire supports the ring against gravity when it is being levitated.

10th Class Physics Textbook Page No. 226

Question 40.
Could the ring be levitated if DC is used?
Answer:
The metal ring is levitated because the net force on it should be zero according to Newton’s second law.

Question 41.
What is this unknown force acting on the metal ring?
Answer:
The change in polarities at certain intervals at the ends of the solenoid causes the unknown force acting on the metal ring.

Question 42.
What is responsible for the current in the metal ring?
Answer:
The field through the metal ring changes so that flux linked with the metal ring changes and this is responsible for the current in metal ring.

Question 43.
If DC is used, the metal ring lifts up and falls down immediately. Why?
Answer:
The flux linked with metal ring is zero. When the switch is on, at that instant there should be a change in the flux linked with ring. So the ring rises up and falls down. If the switch is off, the metal ring again raises up and falls down. There is no change in flux linked with ring when the switch is off.

10th Class Physics Textbook Page No. 227

Question 44.
What could you conclude from the above analysis (metal ring lifts up and falls down)?
Answer:
The relative motion of the magnet and coil sets up a current in the coil.

10th Class Physics Textbook Page No. 228

Question 45.
What is the direction of induced current?
Answer:
The direction of the induced current is such that it opposes the charge that produced it.

Question 46.
Can you apply conservation of energy for electromagnetic induction?
Answer:
Yes, we can apply. The mechanical energy is converted into electrical energy.

10th Class Physics Textbook Page No. 229

Question 47.
Can you guess what could be the direction of induced current in the coil in such case?
Answer:
The direction of the induced current in the coil must be in anti-clockwise direction.

Question 48.
Could we get Faraday’s law of induction from conservation of energy?
Answer:
Yes, we can get. Here we have to ignore the friction everywhere.

10th Class Physics Textbook Page No. 230

Question 49.
Can you derive an expression for the force applied on crosswire by the field “B”?
Answer:
Yes. The force applied F = BIl.

10th Class Physics Textbook Page No. 232

Question 50.
How could we use the principle of electromagnetic induction in the case of using ATM card when its magnetic strip is swiped through a scanner? Discuss with your friend or teacher.
Answer:
If the card is moved through a card reader, then a change in magnetic flux is produced in one direction, which induced potential or EMF. The current received by the pickup coil goes through signal amplification and translated into binary code, so that it can be read by computer.

Question 51.
What happens when a coil is continuously rotated in a uniform magnetic field?
Answer:
An induced current is generated in the coil.

Question 52.
Does it (continuous rotation of coil) help us to generate electric current?
Answer:
Yes. Continuous rotation of coil helps us to generate electric current.

10th Class Physics Textbook Page No. 233

Question 53.
Is the direction of current induced in the coil constant? Does it change?
Answer:
Yes, it changes. When the coil is at rest in vertical position, with side (A) of coil at top position side (B) at bottom position, no current will be induced in it.

Question 54.
Can you guess the reason for variation of current from zero to maximum and vice-versa during the rotation of coil?
Answer:
The reason for variation of current from zero to maximum and vice-versa during the rotation of coil current generated follows the same pattern so that in first half except that the direction of current is reversed.

Question 55.
Can we make use of this current? If so, how?
Answer:
Two carbon brushes are arranged in such a way that they press the slip rings to obtain current from the coil. When these brushes are connected to external devices like TV, Radio we can make them work with current supplied from ends of carbon brushes.

10th Class Physics Textbook Page No. 234

Question 56.
How can we get DC current using a generator?
Answer:
By connecting two half-slip rings instead of a slip ring commutator on either side to the ends of the coil we can get D.C. current.

Question 57.
What changes do we need to make in an AC generator to be converted into a DC generator?
Answer:
Instead of two slip rings, we have to use a slip ring commutator to change A.C. generator into a D.C. generator.

10th Class Physics 12th Lesson Electromagnetism Activities

Activity – 2

Question 1.
Show that the magnetic field around a bar magnet is three dimensional and its strength and direction varies from place to place.
Answer:

• Take a sheet of white paper and place it on the horizontal table.
• Place a bar magnet in the middle of the sheet.
• Place a magnetic compass near the magnet it settles to a certain direction.
• Use a pencil and put dots on the sheet on either side of the needle. Remove the compass. Draw a small line segment connecting the two dots. Draw an arrow on it from south pole of the needle to north pole of the needle.
• Repeat the same by placing the compass needle at various positions on the paper. The compass needle settles in different directions at different positions.
• This shows that the direction of magnetic field due to a bar magnet varies from place to place.
• Now take the compass needle to places far away from magnet, on the sheet and observe the orientation of the compass needle in each case.
• The compass needle shows almost the same direction along north and soiath at places far from the magnet.
• This shows that the strength of the field varies with distance from the bar magnet.
• Now hold the compass a little above the table and at the top of the bar magnet.
• We observe the deflection in compass needle. Hence we can say that the mag¬netic field is three dimensional i.e., magnetic field surrounds its source.
• From the above activities we can generalize that a magnetic field exists in the region surrounding a bar magnet and is characterized by strength and direction.

Activity – 3

Question 2.
Explain how you draw magnetic lines of force in the magnetic field.
(OR)
What is the name given to the imaginary lines joining from north pole to south pole of a bar magnet called? Explain how you can draw those lines around a bar magnet.
Answer;
These lines are called magnetic lines of forces.

Procedure:

1. Take a white drawing sheet.
2. Place a marine compass at the centre of the sheet.
3. Draw a line which shows north and south of the earth on the drawing sheet.
4. Now remove the compass needle and place a bar magnet at the centre of the sheet showing north of the bar magnet pointing north of the earth.
5. Place the magnetic compass near the bar magnet without contact. The needle comes to rest after oscillations.
6. Locate the end of the pointer with pencil. Now place the compass needle at this point and once again notice the end of the pointer.
7. We can repeat the same around the magnet, and draw all the points with the help of the pencil.
8. We can draw the lines taking the needle too far to the magnet and we can observe the orientation of needle of compass.
9. So we can conclude that the strength of field varies with distance from the bar magnet.
10. These lines of force are from north of the bar magnet to south of the bar magnet.

Activity – 4

Question 3.
Explain the direction of magnetic field around the straight conductor carrying current.
(OR)
What field would be formed around straight conductor carrying current? How do you find the direction of that field experimentally?
Answer:
Magnetic field would be formed around current carrying conductor.

Procedure:

1. Take a wooden plank and make a hole and place il on the table.
2. Place a stand on the plants, and suspend a c opper wire from the stand and see that it passes through the hole made to the plank.
3. Connect a battery and switch to this wire in series. Place some magnetic needle at the hole.
4. If the current is passed through the wire, the magnetic needle deflects and it is directed as the tangent to the circle
5. If the current flows in downward direction, the field lines are in anti-clockwise direction and if the current flows in upward direction, the field lines are in clockwise direction.
6. The direction ol the current and magnetic fines of force can be easily explained with the help of right hand thu mb rule. If you hold the current carrying conductor with your right hand grip stretching the thumb, the direction of the I humb shows the direction of the current, the direction of the other four fingers shows direction of magnetic lines of force.

Activity 5

Question 4.
Explain the direction of magnetic field due to circular coil.
Answer:

Procedure :

1. Take a thin wooden plank and cover it with whitepaper.
2. Make two holes to the plank and pass insulated copper wire through the holes and wind the wire 4 to 5 times through the holes such that it looks like a coil.
3. The ends of the wire are connected to the battery terminals.
4. Now place a compass needle at the centre of the coil.
5. Put dots on either side of the compass. Repeat this by keeping at the dots. We can observe that field lines are circular.
6. Here the direction of the field is perpendicular to the plane of the coil.
7. The direction of the magnetic field due to coil points towards you when the current in the coil is in anti-clockwise direction.
8. When you curl your right hand fingers in the direction of current, thumb gives the direction of magnetic field.

Activity – 6

Question 5.
Explain the magnetic field due to solenoid.
(OR)
What is the name given to the device which is a long wire wound in a close pack helix? Find the direction of magnetic field around that device.
Answer:
It is called solenoid.

Procedure :

1. Take a wooden plank covered with white paper.
2. Make holes on its surface.
3. Pass copper wire through the holes.
4. Join the ends of the coil to a battery through a switch.
5. Current passes through the coil, when we switch on the circuit.
6. Now sprinkle iron filings on the surface of the plank, around the coil. Then orderly pattern of iron filings is seen on the paper.
7. The iron filings arrange themselves in orderly way and look like lines of force.
8. The long coil is known as solenoid. The direction of the field due to solenoid is determined by using right hand rule.
9. One end of the solenoid behaves like a north pole and the other behaves like south pole.
10. Outside the solenoid the direction of the field lines of force is from north to south while inside the direction is from south to north. Thus the magnetic field lines are closed loops.
11. Hence electric charges in motion produce magnetic field.

Activity – 8

Question 6.
Explain the field lines due to horse-shoe magnet between its poles.
(OR)
Which field is set up between poles of a horse-shoe magnet? Explain the field lines due to horse magnet between its poles.
Answer:
Non-uniform magnet is set up between poles of a horse shoe magnet.

Procedure :

1. The field in between north and south pole of horse-shoe magnet are straight and parallel.
2. If the wire is passing perpendicular to the paper, the magnetic lines of force are concentric circles, when the current is passed.
3. The direction of field lines due to the wire in upper part coincides with the direction of field lines of horse-shoe magnet.
4. The direction of field lines by the wire in lower part is opposite to the direction of field lines of horse-shoe magnet.
5. Hence the net field in upper part is strong and in the lower part is weak.
6. Hence a non-uniform field is created around the wire.

Activity – 9

Question 7.
Explain electromagnetic induction.
(OR)
Which current will levitate the ring in the following figure? Explain the experimental activity.
Answer:
AC will levitate the ring.

Procedure :

1. Fix a soft iron cylinder on the wooden base vertically.
2. Wind copper wme around the soft iron.
3. Take a metal ring which is slightly greater in radius than the radius of soft iron cylinder and insert it through the soft iron cylinder.
4. Connect the ends of the coil to an AC source and switch on the current.
5. Here metal ring levitates on the coil (appears to rise and floats in the air).
6. In this experiment we can conclude that if AC current is used, the magnetic induction changes in both magnitude and direction in the solenoid and in the ring. The field through the metal ring changes, so that flux linked with the metal ring changes.
7. If DC current is used, the metal ring lifts up and falls down immediately.

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 10 Natural Resources.

AP State Syllabus SSC 10th Class Biology Important Questions 10th Lesson Our Environment

10th Class Biology 10th Lesson Natural Resources 1 Mark Important Questions and Answers

Question 1.
Recently a new programme was launched in our state known as “Vanam – Manain”. Prepare any two slogans to promote the programme.
Answer:
a) Save forest, forest will save you.
b) Grow the plants and get the fresh air.

Question 2.
Suggest any two practices suitable to farmers with less water resources.
Answer:

1. Construction of percolation tanks (or) Soak pits
2. Irrigation techniques like drip irrigation and usage of sprinklers.

Question 3.
Why should we conserve forests? Give two reasons.
Answer:
a) Forests serve as lungs for the world. They purify the air and protect the earth from greenhouse effect and global warming,
b) Forests are rich habitats for plants

Question 4.
Ravi observed symbol on the plastic water bottle purchased by him. What does this symbol indicate? and animals.
Answer:
The symbol on the plastic bottle indicates that the bottle is made from recycled plastic and after its use it can be recycled.

Question 5.
Write any two suggestions for the conservation of biodiversity at your village.
Answer:

1. Protecting and preserving the natural habitats of birds and animals.
2. Replace the wood products with alternative sources.
3. Using Recycled products and following the 4’R Principle in day to day life.

Question 6.
Suggest any two activities to check soil erosion in your school.
Answer:

1. Observe the school ground after the rain.
2. Conduct a field project on soil erosion.

Question 7.
To create awareness on “Water conservation” in your locality, what slogan you will suggest?
Prepare two slogans on ‘Save Water’ propaganda.
Answer:
“Don’t Waste Water”.
“Save every drop”.
“Water is life”.

Question 8.
The symbol is there on the item you bought. What it instructs? (OR)
What does the given logo indicate? What does it mean?

Answer:
It is the Recycle logo. It indicates that the item we bought is prepared from recycled materials or the item can be recycled after use.

Question 9.
What happens if the forest area decreases rapidly?
Answer:
If the forest area decreases

1. It destroys wildlife habitat.
2. It increases soil erosion.
3. It releases green house gases into the atmosphere contributing to global warming.
4. It also harms people who relay on forest for their survival, hunting and gathering, harvesting forest products or using timber and firewood.

Question 10.
Write two activities which you are performing to save electricity.
(OR)
Write any two measures vou take in your home to reduce the consumption of electricity.
Answer:

1. We can reduce the consumption of electricity by putting off the fans and lights when there is no need.
2. We can use LED (Lighting Emitting Device) bulbs to save electricity.
3. To shut down laptops and computers when they are not in use.

Question 11.
Prepare two slogans on protecting non-renewable resources.
Answer:

1. Use Biofuel – Reduce Fossil Fuel.
2. Use alternative resources – Save the environment.

Question 12.
Write two examples for non-renewable resources.
Answer:
Examples for non-renewable resources are coal, petroleum and natural gas.

Question 13.
What is sustainable development? Is it needful for us?
Answer:
When we use the environment In ways that ensure we have resources for the future, It Is called sustainable development. It Is needed because development and conservation can coexist In harmony.

Question 14.
What are examples for natural resources?
Answer:
Examples for natural resources are water, soil, forests, flora, fauna, etc.

Question 15.
What are percolation tanks?
Answer:
Percolation tanks are normally earthen dams with masonry structures where water may overflow.

Question 16.
What are Kharif crops?
Answer:
Crops grown In the rainy season are termed as Kharif crops, e.g: Paddy, maize, millet and cotton crops.

Question 17.
What are Rabi crops?
Answer:
The crops that are grown only in winter season are generally called Rabi crops, e.g.: Wheat, Gram and Mustard.

Question 18.
What is the average fall of ground water level in Andhra Pradesh state during the period of 1998 – 2002?
Answer:
The average fall of ground water level In Andhra Pradesh state during the period of 1998 – 2002 Is 3 meters.

Question 19.
Which agency in villages of Warangal district helped in recharging wells that were being dried up?
Answer:
Centre for water solidarity (Secundrabad, T.S.) helped In recharging wells that were drying up In the villages.

Question 20.
Give examples for micro irrigation techniques.
Answer:
Drip irrigation, sprinklers are the examples for micro irrigation techniques.

Question 21.
Mow did the boundaries between the villages were fixed in ancient times?
Answer:
In ancient times village boundaries were decided upon a water shed (Land between water sources usually of two rivers or streams) basis fixed at the common point of the drainage system In between two villages by the expert farmers In the village.

Question 22.
Expand ICRISAT.
Answer:
International Crop Research Institute for Semi-Arid Tropics.

Question 23.
What is the other name for Sri Rama Sagar Project?
Answer:
Sri Rama Sagar Project also known as the Pochampadu project on the Godavari river,

Question 24.
What is the use qf planting Gliricidia on field bunds?
Answer:
Planting Gliricidia on field bunds help In strengthen them and make the soli nitrogen-rich.

Question 25.
What is the micro irrigation system that can reduce water consumption by 70%?
Answer:
Drip irrigation can reduce water consumption by 70%.

Question 26.
Who predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity ?
Answer:
The Food and Agriculture Organisation (FAO) of the united nations has predicted that by 2025, 1.8 billion people will be living in countries or regions with absolute water scarcity.

Question 27.
What happens if we use resources wisely?
Answer:
If resources are used wisely and efficiently they will last much longer. Through conservation people can reduce waste and manage natural resources wisely.

Question 28.
Give an example of country where restrictions on water usage were imposed.
Answer:
In Australia restrictions were imposed on activities like, watering lawns by using sprinkler systems, washing vehicles, using house pipes to clean paved areas, and refilling swimming pools.

Question 29.
Why are the natural resources used up quickly?
Answer:
The population of human beings has grown enormously in the past two centuries. Billions of people use up resources quickly as they eat food, build houses, produce goods and burn fuel for transportation and electricity.

Question 30.
What happens if we damage a forest resource?
Answer:
Harm to animals that may be forced to find new habitats. If we damage a forest resource indiscriminately the depletion of resources occur and we may have to face problem for water and timber in future.

Question 31.
What are the results of deforestation?
Answer:
Deforestation destroys wild life habitats and increases soil erosion and also releases green house gases into atmosphere, contributing to global warming.

Question 32.
How the people in China and Mexico recycle paper? (OR)
Give an example of recycling paper by the people. What is the use of recycling paper?
Answer:
People in China and Mexico reuse much of their waste paper, including writing paper, wrapping paper and card board.

Question 33.
How the soil is important for us ? How the soil is importane for us?
Answer:
Soil is vital to food production and also important to plants that grow in the wild.

Question 34.
What are the reasons for depletion of nutrients in soil?
Answer:
Poor farming methods, such as repeatedly planting the same type of crop in the same place cause depletion of nutrients in the soil.

Question 35.
What is biodiversity?
Answer:
Biodiversity is the variety of living things that populate the Earth.

Question 36.
How are people speeding up the loss of biodiversity?
Answer:
Through hunting, pollution, habitat destruction, people are speeding up the loss of biodiversity.

Question 37.
How many plant species are being used by us for medicines world wide?
Answer:
We use between 50,000 to 70,000 plant species for medicines world wide.

Question 38.
What is selective harvesting?
Answer:
The practice or removing individual plants or small groups of plants leaving other plants standing to anchor the soil is called selective harvesting.

Question 39.
How are fossil fuels produced?
Answer:
The fuels that are produced from the remains of ancient plants and animals are called fossil fuels. They include coal, petroleum and natural gas.

Question 40.
What are the alternate sources of energy?
Answer:
The alternate sources of energy are sun, wind and water.

Question 41.
What are the other products made from petroleum?
Answer:
Plastic, synthetic, rubber, fabrics like nylon, medicines, cosmetics, waxes, cleaning products, medical devices, etc., are the other products made from petroleum.

Question 42.
Which plant’s seeds are used for the production of bio-fuel?
Answer:
Seeds from the Jatropa Curcas plant are used for the production of bio-fuel.

Question 43.
How does the mining method, Mountain Top Removal mining (MTR) devastate the environment?
Answer:
The mining method Mountain Top Removal mining devastate the environment. They destroy soil, plants and animal habitats.

Question 44.
In which country car manufacturers recycle many raw materials used in making automobiles?
Answer:
In Japan car manufacturers recycle many raw materials used in making automobiles.

Question 45.
In which country nearly one third of the iron produced comes from recycled automobiles?
A. In the United States, nearly one-third of the iron produced comes from recycled automobiles.

Question 46.
What does the Indian tradition teach us?
Answer:
The Indian tradition teaches us that all forms of life – human, animal and plant are so closely inter linked that disturbance of one gives rise to imbalance in the other.

Question 47.
Expand IUCN.
Answer:
IUCN stands for International Union for the Conservation of Nature.

Question 48.
How is IUCN planning to protect wild life and habitats?
Answer:
IUCN monitors the status of endangered wild life, threatened national parks and preserves.

Question 49.
What are the four R’s to save the environment?
Answer:
Reduce, Reuse, Recycle, Recover are the four R’s to save the environment.

Question 50.
How did Amritha Devi and her daughters protest against the king’s order?
Answer:

1. Amritha Devi and her daughters, followed by villagers, who clung to trees in the forest surrounding their village and laid down their lives to save them.
2. They protested against the king’s order to collect wood for the construction of his palace.

Question 51.
Write a method of soil conservation.
Answer:
One soil conservating method is called contour strip cropping. Several crops such as corn, wheat and clover are planted to alternating strips across a slope or across the path of the prevailing wind.

Question 52.
What is the rate of extinction by the estimation of the scientists?
Answer:

1. Scientists estimate that the current rate of extinction is 1,000 times the natural rate through hunting, pollution, habitat destruction.
2. Based on various estimates of the number of species on Earth, we could be losing anywhere from 200 to 1,00,000 species each year.

Question 53.
What is the need to protect biodiversity?
Answer:
We need to protect biodiversity to ensure plentiful and varied food sources. Biodiversity is important for more than just food because many plant species are being used for medicines.

Question 54.
Mention two ways in which water harvesting can be undertaken?
Answer:
The two ways by which water harvesting can be undertaken are

1. Capturing run off water from, rooftops.
2. Capturing run off water from local catchments.

Question 55.
On the basis of the issues raised in the chapter management of natural resources, what changes you in corporate in your lifestyle in a move towards a sustainable use of our resources?
Answer:
I would incorporate the maximum of four R’s i.e., reduce, recycle, reuse and recover in my lifestyle in a move towards a sustainable use of our resources.

10th Class Biology 10th Lesson Natural Resources 2 Marks Important Questions and Answers

Question 1.
Rahul remarked that different human activities are responsible for global warming.
What might be the reasons for his statement?
Answer:

1. Deforestation.
2. Industrialisation and urbanization.
3. Conversion of agriculture lands into residential areas.
4. Home appliances like A/C, refrigerators, vehicle pollution.
5. Population explosion.

Question 2.
What steps do you take to improve natural resources?
Answer:

1. Motivate the people to conserve water.
2. Try to avoid wastage of water whenever possible.
3. Plantation in the vacant lands.
4. Educating the farmer regarding proper utilization of water for irrigation.
5. Encourage the people to recycle the water wherever possible.

Question 3.
Proper utilisation of natural resources is the way to show gratitude to our nation.
Can you support this statement? Give your argument.
Answer:

1. Natural resources of a nation influence its economical and social development.
2. Natural resources are freely available in nature and help in many activities and development of people.
3. The generation of natural resources take a lot of time.
4. They disappear by indiscriminate usage.
5. So proper utilization of natural resources is the way to show gratitude to our nation.

Question 4.
The humans are utilising natural resources indiscriminately. These resources are decreasing more rapidly. Guess what will be the consequences in future?
Answer:
Indiscriminate usage of natural resources causes the following consequences.

1. Reduction in rainfall
2. Drought will occur.
3. Atmospheric temperature becomes increase.
4. The rare species become extinct.

Question 5.
Write any four slogans on the conservation of natural resources.
Answer:
Slogans:

1. Waste water today – live in desert tomorrow
2. Practice eco-friendly methods.
3. Use natural resources judiciously.
4. Save nature – Save future.

Question 6.
There is an increase in the atmospheric temperature year by year. If it continues, guess and write what would be the consequences?
Answer:
If the temperature on earth increases, the consequences would be as follows.
a) All the glaciers and the frozen ice in the polar region start melting leading to rise in the sea water levels.
b) It results in the submergence of low lying coastal areas throughout the world. Millions of people of those areas would lost their homes.
c) Changes in rainfall patterns take place and it result in the occurance of droughts and decrease in crop production.
d) Global warming results in climate change which cause the breakout of climatic sensitive diseases like Malaria, Dengue, Diarrhoea, etc.

Question 7.
There is water scarcity in Ravi’s village during summer. He wants to conduct a rally to create awareness regarding conservation of water. Write any four slogans required to conduct this rally.
Answer:

1. Water is life.
2. Save water – Save a life.
3. Today’s rain water is tomorrow’s life saver.
4. No matter your occupation, water conservation is your obligation.

Question 8.
What steps you take to conserve the biofuels in your daily life?
Answer:

1. Development and usage of alternative energy resources in place of bio-fuels.
2. Minimise the usage of bio-fuels whenever possible.
3. Use public transport, ride by bicycle and walking regularly.
4. Use and purchase energy efficient appliances to save bio-fuels.

Question 9.
Why do we use fossil fuels judiciously?
Answer:

1. Fossil fuels were produced from the remains of ancient plants and animals.
2. They include coal, petroleum (oil) and natural gas.
3. We need to use fossil fuels judiciously because they are non – renewable resources.
4. We need to conserve fossil fuels so we don’t run out of them.
5. The pollution caused by them when burnt, to limit our fossil fuel use.
6. Future generations may not get these resources.
7. Balance in the nature will be disturbed.
8. Electricity production will be stopped.
9. Vehicles running with fossil fuels become useless.

Question 10.
Write two suggestions to create awareness on groundwater conservation.
Answer:

1. We need to adapt different methods to Improve the quality and increase the quantity of groundwater.
2. We should dig water harvesting pits for every house.
3. We should clean the silt, mud fill in tanks and ponds.
4. We should prohibit the establishment of borewells for extraction of groundwater for agricultural and Industrial use.
5. These measures will improve quality and quantity of groundwater.

Question 11.
What is the importance of 4R’s in achieving the goal of “Swachh Bharat”?
Answer:

1. Reduce the production of garbage.
2. Reuse the garbage for the production of manure and electricity.
3. Recycle the garbage by separating It as dry and wet garbage.
4. Recover the plants.

Question 12.
Suggest four measures to conserve fossil fuels.
Answer:
Measures to conserve fossil fuels:

1. Usage of alternatives to fossil fuel.
2. Minimise the usage of fossil fuel.
3. Walk, ride by bicycle and use public transportation whenever possible.
4. Purchase energy efficient appliances.
5. Turn off light and other electronics when you are not using them.

Question 13.
The indiscriminate digging of Borewells may result in what type of consequences in future?
Answer:

1. Due to over drilling of borewells and pulling out water by electric motors, the ground water level Is decreasing day by day.
2. It Is goes on without recharging, ground water becomes scarce.
3. It shows impact on agriculture and the productivity will decrease.
4. Fluorine levels In ground water will increase.
5. Sometimes, saline water may intrude Into the interior places of land and water becomes unfit for consumption.
6. Farmers have to drill the bore wells to more depths which Increase the losses for them.

Question 14.
Ramaiah made broad bed furrow around his field under employment guarantee scheme. Guess the reasons for if. If all the farmers of your village work together, will their water scarcity meet?
Answer:
The reason for Ramaiah making broad bed furrow around his field was, it is useful to conserve soil and water, fertilizer application weeding operations. It also conserves rain water.

The farmers are over coming the water scarcity by sharing water available in the village. They formed groups of farmer including large and small ones who would use the same water resource. Farmers were also motivated to use irrigation techniques like drip irrigation.

Question 15.
What are renewable sources and non-renewable resources?
Answer:
Renewable resources: Resources that can be replaced after they are used are called renewable resources.
Ex: Air, water and soil.
Non-renewable resources: Some other resources, cannot be replaced at all: Once they are used up they are gone forever and are called non renewable resources.
Ex: Coal, Petroleum, Natural gas (fossil fuels).

Question 16.
How do people waste natural resources?
Answer:

1. People often waste natural resources.
2. Animals are over hunted, forests are cleared, exposing land to wind and water damage.
3. Fertile soil is exhausted and lost to erosion because of poor farming practices.
4. Fuel supplies are depleted.
5. Water and air are polluted.
6. Water resources is indiscriminately used for crop growth.

Question 17.
How do people use the forest resources differently?
Answer:

1. The need to conserve resources often conflicts with other needs.
2. For some people, a forest area may be a good place to put a farm.
3. A timber company may want to harvest the area’s trees for construction materials.
4. A business company may want to build a factory or a shopping mall on the land.

Question 18.
What are die effects of deforestation?
Answer:

1. Deforestation destroys wild life habitats and increases soil erosion.
2. It also releases green house gases into the atmosphere, contributing to global warming,
3. Deforestation accounts for 15 per cent of the world’s green house gas emissions.
4. Deforestation also harms the people who rely on forests for their survival, hunting and gathering, harvesting forest products, or using the timber for firewood.

Question 19.
In your opinion What are the causes for soil erosion?
Answer:

1. Soil erosion is caused by poor farming methods such as repeatedly planting the same type of crop in the same place.
2. These methods deplete nutrients in the soil.
3. Soil erosion Is also caused by water and wind currents.
4. When farmers plough up and down hills, soil erosion occurs.
5. Overgrazing by cattle also causes soil erosion.
6. Natural floods causes the extensive damage to the top layer of the soil.

Question 20.
What is Biodiversity? Explain.
(OR)
What is the importance of biodiversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. The products and benefits we get from nature rely on biodiversity.
3. We need to protect biodiversity to ensure plentiful and varied food sources.
4. Biodiversity is important for more than just food. For instance we use between 50,000 to 70,000 plant species for medicines world wide.

Question 21.
How can we use the fossil fuels carefully?
Answer:
We can use the fossil fuels carefully by taking the following measures.

1. Turn off lights and other electronics when we are not using them.
2. Purchase energy-efficient appliances.
3. Walk, ride a bicycle, if the distance is less.
4. Use public transportation whenever possible.
5. It is better to prefer public transport system like bus or train, instead of travel in personal vehicles.

Question 22.
Why the prices of aluminium and iron are expensive?
Answer:
Earth’s supply of raw material resources is in danger. Many mineral deposits that have been located and mapped have been depleted. As the ores for minerals like aluminium and iron become harder to find and extract, their prices go up.
This makes tools and machinery more expensive to purchase and operate.

Question 23.
What are the effects of mining?
Answer:

1. Many mining methods such as Mountain Top Removal mining (MTR) devastate the environment.
2. They destroy soil, plants and animal habitats.
3. Many mining methods also pollute water and air, as toxic chemicals leak into the surrounding ecosystem.

Question 24.
What did Smt. Indira Gandhi said, while launching the world conservation strategy in India on 6th March 1980?
Answer:
“The interest in conservation is not a sentimental one but the discovery of a truth well known to our ancient stages. The Indian tradition teaches us that all forms of life- human, animal and plant – are so closely inter-linked that disturbance in one gives rise to imbalance in the other” said by Smt. Indira Gandhi.

Question 25.
What are the steps taken by the government to conserve resources?
Answer:

1. Government enacts laws defining how land should be used and which areas should be set aside as parks and wild life preserves.
2. The government enforces laws designed to protect the environment from pollution, such as requiring factories to install pollution control devices and also provide incentives for conserving resources.

Question 26.
What is the necessity of sustainable management of natural resources? Out of the two methods reuse and recycle which one would you suggest to practice and why?
Answer:

1. Sustainable management of natural resources is necessary to Increase the over all life of natural resources specially non renewable resources and also to control the environmental pollution.
2. Both reuse and recycle are the good choice.
3. Reuse: If we reuse something then the cost of recycle will be saved.
4. Recycle: It is not necessary that each and everything can be reused, so after getting recycled the life of the resource will be enhanced.

Question 27.
“Burning fossil fuels is a cause of global warming”. Justify this statement?
Answer:

1. Fossil fuels are composed of carbon, hydrogen, nitrogen and sulphur.
2. When these are burnt they produce CO₂, H₂O, Oxides of Nitrogen and Sulphur.
3. Incomplete combustion of fossil fuels produces green house gases such as CO₂,
4. If huge amount of fossil fuels are burnt, It would produce high amount of CO₂ resulting intense global warming.

Question 28.
Can you suggest some changes in your school which would make it environment friendly?
Answer:
The changes that would make my school environment friendly are

1. Save energy by turning off lights that we are not using.
2. I will suggest to buy recycled paper for decoration and other purposes.
3. Use writing paper on both the sides.
4. Growing trees and plants all around the play ground.

Question 29.
What is the necessity of replenishment of forest? State four reasons.
Answer:
The replenishment of forest is necessary because of the following reasons.

1. It is used to conserve soil.
2. It provides shelter to wild animals.
3. It reduces atmospheric pollution.
4. It controls flood and increases frequency of rainfall.

10th Class Biology 10th Lesson Natural Resources 4 Marks Important Questions and Answers

Question 1.
Forest is renewable resource. But, each year, the Earth loses about 36 million acres of forest. In this type of situation, what suggestions do you give to save forests from turning into non-renewable resources ?
Answer:

1. Forests are the lungs of the world. So I will suggest the following measures to save forests from turning into non-renewable resources.
2. Sustainable forestry practices for ensuing resources into the future.
3. Low impact logging practices, harvesting with natural regeneration in mind. Prevention of removing all the high value trees or all the largest trees from the forests. Recycling methods should be adopted.
4. Replace wood products with alternative sources.
5. Preventing forest fires.
6. Implementing methods like agro forestry, social forestry crop rotation, green plantation, etc. are essential.

Question 2.
What are four R’s? Explain how they help to conserve the environment?
(OR)
Write about the 4 ‘R’s needed for the protection and conservation of environment.
Answer:
By pursuing the maximum of four R’s i.e., Reduce, Reuse, Recycle and Recover, we can save the environment in an effective way.

1. Reduce : It means to use less, I would save electricity by switching off unnecessary lights and fans, prefer walking or cycling than using a vechicle, turn off the engine of car at red light, repair leaky taps and would not waste food.
2. Recycle: It means to collect used things like plastic, paper, galss and metal items and recycle these materials to make required things instead of synthesising or extracting fresh plastic, paper, glass or metal.
3. Reuse: It refers to use things again and again. For example instead of throwing away used envelops, they can be used by pasting new labels.
4. Recover: We should implement ‘recover’ to prevent environmental threat. For example when we cut trees to construct industries or roads for transportation, it is important to grow trees in another areas.

Question 3.
What steps you would like to follow on your part to conserve bio-diversity?
Answer:

1. Biodiversity is the variety of living things that populate the earth.
2. To conserve biodiversity we should avoid hunting.
3. Sustainable forest conservation methods should be followed.
4. I will actively participate Vana Mahosthavam programmes.
5. I will educate and encourage people and make them participate in conservation programmes.
6. Create awareness programmes in and around school.
7. Writing slogans and also make some posters about conservation of biodiversity
8. Judicious use of electricity wherever possible.
9. Finding out of various alternative sources of energy.
10. Plant the saplings in the habitat.
11. Encouraging of social forestry.

Question 4.
Observe the pie diagram showing water resources available in our state for agriculture and answer the given questions.
A) Which water resource is using more for agriculture?
B) What are the consequences of excess utilization of underground water?
C) Which water resource should be utilized for agriculture?
D) What are the alternative ways to increase the underground water resources?

Answer:
A) Ground water.
B) Underground water table will be depleted and scarcity of drinking water will arise.
C)

1. Tanks should be constructed to harvest with rain water.
2. Projects should be constructed across the rivers to store water that can be utilized for agriculture.

D)

1. Construction of rain water storage structures on large scale.
2. Constructing soaking or percolation pits.
3. Contour field bunding.
4. Recharge of wells by building dykes or barriers in the nalla.
5. Plantation in waste lands.
6. Adapting micro irrigation techniques.
   (Any two points you can write)

Question 5.
Forests are renewable resource. Write four sentences supporting this.
(OR)
“Forest is a renewable resource”. Do you agree? Justify.
Answer:

1. Forests are rich habitat for plants and animals. Forests serve as lungs for the world and a bed of nutrients for new fife to prosper.
2. Forest’s pure air protects the earth from green house effect by removing carbon dioxide from the atmosphere and converts it into oxygen.
3. Many fruits, medicines, dyes, sandle wood and bamboo is obtained from forest by local people.
4. Forest provide employment to large number of people and also help in generating revenue.

Question 6.
Observe the above table and answer the following questions.

1. Which crop is most suitable to cultivate for small farmer in both the villages?
Answer:
Cotton, paddy

2. If you are a large farmer, which crop do you select to cultivate?
Answer:
Cotton, paddy, mirchi

3. What similarities you have identified in village A and village B?
Answer:
Small and large fanners cultivated same type of crop in both villages. Large farmer gets more income per acre on crops than small farmer in both the villages.

4. Which is the lowest income crop ?
Answer:
Mirchi.

5. Is there any relationship between production of crops and income ? How ?
Answer:
Commercial crops are good for income. Income may or may not related to production of crop. It depends upon demand of the market.

Question 7.
Read the given information and answer the following questions.

A survey was conducted in two villages, Vanaparthy and Vaddicherla of Warangal district in Telangana State. The first with no scarcity and the second with scarce groundwater. Well census was carried out in the villages in order to get a complete picture of well irrigation and its status as well as availability of water. There are no alternative sources of supply as against wells in Vaddicherla, where there is an existing tank that has been converted into a percolation tank, so that the water situation is much better in Vanaparthy.

i) Why did they conduct survey?
Answer:
A compartive study on available water resources irigation method in the Vaddicharla and Wanaparthi of Warangal Dist of Telangana State.

ii) What are irrigation resources in Telangana State?
Answer:
Lakes, wells, canals and ground water etc…,

iii) In which village, do you suggest drip irrigation?
Answer:
Vaddicherla.

iv) Why the water situation is much better in Vanaparthy village compared to Vaddicherla?
Answer:
Existing tank has been converted into a percolation tank.

Question 8.
Observe the Pie diagram and answer the following questions.
i) Identify the fossil fuels from the above diagram.
Answer:
Coal, natural gas, oils are fossil fuels.

ii) Why wastes should be considered as primary energy source in future?
Answer:
The fossil fuels may be exhausted in future. So we may be considered that wastes are primary alternative energy resources.

iii) Why can’t we depend on fossil fuels forever?
Answer:
We can’t depend on fossil fuels forever because fossil fuels are non-renewable resources.

iv) What are the alternatives for fossil fuels?
Answer:
Solar energy, wind energy, tide energy, nuclear energy, energy from waste materials.

Question 9.
Explain the importance and implementation of community based interventions and farmer based interventions for water management.
Answer:
Community based interventions:

1. For water harvesting, there is an urgent need to construct earthen and masonry dams. They help us to store rain water during rainy seasons. They are help in increasing the ground water table.
2. Construction of percolation pits and field bunding are very helpful in the harvesting every rain drop.
3. Open dry wells near nalla canal were recharged by building dykes or barriers in the nalla and maintaining the run – off rain water. The ground water is recharged by these community based interventions.
   Farmer based interventions:
4. Broad Bed Furrow (BBF) land form and contour planting methods are very useful to conserve soil, water and fertilizer application and weeding operations.
5. Planting Gliricidia, a leguminous plant adapted to grow in dry areas on field bunds to strengthen them and make the soil nitrogen rich.
6. Farmers were encouraged to use water resource jointly and irrigate land using micro irrigation methods like sprinklers and drip irrigation.

Question 10.
Explain the farmer based and community based interventions to conserve soil and water resources.
Answer:

Question 11.
“The humans who were developed by using the natural resources, today has become the reason for destroying them”. Explain analytically.
Answer:
“The humans who were developed by using the natural resources, today has become the reason for destroying them” – This statement is absolutely true.

1. Primitive man lived in forests and hills. He used the natural resources for his livelyhood. He worshipped nature and used them wisely for his development.
2. After his development, he becomes greedy and using the natural resources indiscriminately and held responsible for their destruction.
3. To meet the needs of growing population, industrialization, urbanization, and huge constructive activities, man utilised natural resources Indiscriminately. At the same time, he did not planned for their revival.
4. But now he realised the importance of natural resources and taken up steps for their conservation. The concept of “Sustainable development” is being implemented in natural resource management.
5. He focussed on development of alternatives for fossil fuels, conservation of water and soil at community level and farmer based interventions.
6. Now he is so keen on conserving forests, wild life and biodiversity.
7. He is so cautious in minimising the utilization of natural resources by following 4’R principle in the day to day life [R – Reduce, R – Reuse, R – Recycle, R – Recover]
8. Now, he is adopting micro-irrigation methods like sprinklers and drip Irrigation to minimise the water usage in low water available areas.
9. He is very interested in following eco-friendly techniques, natural farming methods, using biofertilizers, vermicompost and natural pest control methods in place of toxic chemical pesticides.

Question 12.
The wells and tanks in your village become dry. Ground water levels decreased. Assume the causes for this. Will there be no water scarcity if all the farmers of your village work collectively?
Answer:
Causes for decreasing ground water levels:

1. Varying monsoon behaviour in recent years, there is a pressure on ground water utilization.
2. Indiscriminate tapping of ground water in our village by too much drilling and construction of deep tube wells and bore wells have resulted in over exploitation and depletion of ground water resources.
3. There will be no water scarcity if all the farmers of our village work collectively. Farmers in our village were encouraged to use water resource jointly and irrigate land using micro irrigation techniques. By using micro irrigation techniques farmers in our village obtained more crop yield. Farmers in our village follow the micro irrigation method i.e. drip irrigation and can reduce water consumption by 70% in our village.

Question 13.
Whom do you meet to collect the information of the methods of farmer based, community based water management? Prepare information table to note down your observation.
Answer:
I will meet officials of International Crop Research Institute for Semi – Arid Tropics (ICRISAT) located at Hyderabad to collect information of the methods of farmer based and community based water management.
I also collect information from Central Research Institute for Dry Land Agriculture (CRIDA), National Remote Sensing Agency (NRSA), District Water Management Agency (DWMA) and M Venkatarangaiah Foundation (MVF) and NGO.
The information I gathered from these institutions is summarised below.
Information table:
For Information table See Q.No. 10 in 4 Marks.

Question 14.
Think that there is much scarcity of water for drinking and cultivation in your village. What advice do you give to prevent this?
(OR)
How do you overcome the problem of water scarcity in your village?
Answer:

1. Motivate the people to conserve water.
2. I will educate the people to avoid wastage of water whenever possible.
3. Construction of recharge pits in the house, school and in the open areas to increase the underground water level.
4. Planting trees wherever possible in the village particularly in the vacant lands.
5. Educate the farmers about the micro irrigation system like drip irrigation, sprin¬klers, etc.
6. Encourage the farmers to form groups to share available water among themselves.
7. Construction of percolation tanks in the low lying areas of the village.

Question 15.
What type of fossil fuels are used in your house? What measures do you take to conserve them?
Answer:
Fossil fuels are sources of energy for cooking, heating and burning in our households. Petrol and diesel are being used in our house for transport and running generators and water pumps.
Measures to be taken to conserve fossil fuels in my house :

1. I will put the food material to be cooked on the stove only after arranging all the things which are necessary for cooking.
2. By using pressure cookers 20% gas on rice and 41.5% on meat would be saved when compared to Other cooking means.
3. We must reduce the flame as soon as the boiling process starts in a pressure cooker. This process saves nearly 35% of fuel.
4. I will soak the food material before cooking. It saves 22% of fuel.
5. I will cook food in broad and low depth vessel.
6. I will keep lid on the cooking vessel. If not, it takes more time to cook.
7. For short distances to travel I will go by walk to save fuel for longer distance. I use public transport.
8. Encourage people to use solar water heater and solar cooker.

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 1 Nutrition – Food Supplying System Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 1st Lesson Nutrition – Food Supplying System

10th Class Biology 1st Lesson Nutrition – Food Supplying System Textbook Questions and Answers

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Question 1.
Write differences between
(a) Autotrophic nutrition – Heterotrophic nutrition:
Answer:

(b) Ingestion – Digestion :
Answer:

(c) Light reaction – Dark reaction :
(OR)
Differentiate the reactions that take place in presence of light and the reactions which do not require light in photosynthesis.
Answer:

(d) Chlorophyll – Chloroplast:
Answer:

Question 2.
Give reasons.
a) Why photosynthesis is considered as the basic energy source for most of living world?
(OR)
Why can we say that photosynthesis is the basic energy source for the living world?
Answer:

1. All living organisms constantly need energy to be alive.
2. They get energy from the food they eat.
3. The food directly or indirectly comes from the green plants through photosynthesis.
4. Hence photosynthesis can be considered as the basic energy source for most of living world.

b) Why is it better to call the dark phase of photosynthesis as a light independent phase?
Answer:

1. The term dark reaction or light independent does not mean that they occur when it is dark at night.
2. It only means that the reactions are not depend on light.
3. Hence we call the dark phase of photosynthesis as a light independent phase.

c) Why is it necessary to destarch a plant before performing any experiment on photosynthesis?
Answer:
1) To get better results, it is necessary to destarch a plant before performing any experiment on photosynthesis.
2) Because if starch is present it may interfere with the result of the experiment.

d) Why is it not possible to demonstrate respiration in green plants kept in sunlight?
Answer:

1. We cannot demonstrate an experiment of respiration in green plants kept in sunlight.
2. Because if sunlight is present, the C02 produced in respiration will be used in photosynthesis.
3. So we must conduct an experiment on respiration in green plants in a dark room.

Question 3.
Give examples.
a) Digestive enzymes
Answer:
The digestive enzymes are:

1. Salivary Amylase (Ptyalin),
2. Pepsin,
3. Trypsin,
4. Lipase,
5. Peptidases,
6. Sucrase,
7. Amylase (Pancreatic juice)

b) Organisms having heterotrophic nutrition is seen in organisms like:
Answer:
Heterotrophic nutrition is seen in organisms like:

1. All animals and human beings.
2. Some protozoans Ex: Amoeba.
3. Some parasitic plants Ex: Cuscuta
4. Saprophytes Ex: Bread moulds, yeast, mushrooms, etc.

c) Vitamins
Answer:
Water soluble vitamins:
B complex (B₁) Thiamine, (B₂) Riboflavin, (B₃) Niacin, (B₆) Pyridoxine,
(B₁₂) Cyanocobalamine, Folic acid, Pantothenic acid, Biotin, (C) Ascorbic Acid.
Fat soluble: (A) Retinol, (D) Calciferol, (E) Tocoferol, (K) Phylloquinine.

d) Nutritional deficiency diseases
Answer:
Eg: Kwashiorkor, Marasmus etc.

Question 4.
Where do plants get each of the raw materials required for photosynthesis?
Answer:

Question 5.
Explain the necessary conditions for autotrophic nutrition and what are its by products.
Answer:
A. Necessary conditions:

1. Autotrophic nutrition takes place through the process of photosynthesis.
2. Carbon dioxide, water, chlorophyll pigment and sunlight are the necessary conditions required for autotrophic nutrition.
3. The rate of photosynthesis depends on availability of sunlight.

B. By products:

1. Photosynthesis is the main process for autotrophic nutrition.
2. Carbohydrates and oxygen are the by products of photosynthesis.

Question 6.
With the help of chemical equation explain the process of photosynthesis In detail with the help of a flow chart.
Answer:
Process of photosynthesis:

1. The chemical equation representing the process of photosynthesis is
   
2. Definition: Photosynthesis is a photochemical reaction during carbohydrates are formed using carbon dioxide and water in the chloroplasts of the green plants in the presence of sunlight.
3. CO₂ water, sunlight and chlorophyll are the requirements of photosynthesis.
4. Glucose, O and water are the end products of the reaction.
5. Photosynthesis have two phases.
   1) Light reaction 2) Dark reaction
6. Light reaction have three steps, i) Oxidation of chlorophyll ii) Photolysis iii) Formation of ATP, NADPH and O₂
7. In dark reaction CO₂ is utilized and finally glucose is formed which is converted and stored as starch.

FLOW CHART:

Question 7.
Name the three end products of photosynthesis.
Answer:
Glucose, oxygen and water are the three end products of photosynthesis.

Question 8.
What is the connecting substance between light reaction and dark reaction?
Answer:
The hydrogen of NADPH present in the stroma is the connecting substance between light reaction and dark reaction.

Question 9.
Most leaves have the upper surface is more green and shiny than the lower ones. Why?
(OR)
In most of leaves the upper surface will be more green and shiny than the lower surface. Why?
Answer:

1. The upper surface comprising of the palisade parenchyma.
2. The lower surface comprising of the spongy parenchyma.
3. Palisade parenchyma contains more number of chloroplasts than the spongy parenchyma.
4. Thus the upper surface is more green and shiny than the lower ones.

Question 10.
Explain the structure of chloroplast with a neatly labelled sketch.
(OR)
Explain the structure of a chloroplast with the help of a rough diagram.
Answer:

1. Chloroplast is a membranous structure consisting of 3 membranes.
2. The third layer forms stacked sac like structures called granum.
3. The intermediatery fluid filled colourless portion is called stroma.
4. It is responsible for enzymatic reaction leading to the synthesis of glucose in plants.
5. Substances found in chloroplast, capture sunlight are called photosynthetic pigments.
6. Chlorophyll pigment contain one atom of magnesium.
7. Two major kinds of chlorophylls are associated with thyakoid membranes.
8. Chlorophyll-a is blue-green in colour and chlorophyll-b is yellow-green colour.
9.  Around 250-400 pigments molecules are grouped as light harvesting complex units in granum.
10. Some of the events occur in chloroplast are :
    a) Conversion of light energy to chemical energy.
    b) Splitting of water molecule.
    c) Reduction of carbondioxide to carbohydrates.

Question 11.
What is the role of acid in stomach?
Answer:

1. The internal walls of stomach has number of gastric glands. They secret gastric juice.
2. It contains HCl and enzymes.
3. HCl kills the bacteria present in the food and protects us from their harmful effects.
4. And also denatures the proteins so that enzymes can act easily on them.

Question 12.
What is the function of digestive enzyme?
Answer:

1. The function of digestive enzyme is to increase the process of breaking up of complex molecules into simpler and absorb molecules.
2. This makes easy for the body to absorb food.

Question 13.
How is the small intestine designed to absorb digested food? Explain.
(OR)
How is food absorbed by villi in small intestine?
Answer:

1. Small intestine is the largest part in digestive system.
2. Absorption is its main function including last stage of digestion.
3. The inner surface of small intestine has millions of tiny finger-like projections called villi.
4. Due to the presence of villi, the absorbing surface area of small intestine increases.
5. And the large surface area of small intestine helps in the rapid absorption of digested food.
6. The digested food which is absorbed through the walls of the small intestine goes into our blood.
7. Long and folding structure increase the ability of small intestine.

Question 14.
How are fats digested in our bodies? Where does this process take place?
What is emulsification? How it helps in digestion of fats? (OR)
How are fats digested? Where do they get digested?
Answer:

1. Bile juice and lipase enzymes helps in fat digestion.
2. Bile juice is secreted by liver.
3. Fats are digested by converting them into small globules like forms by the help of the bile juice.
4. This process is called emulsification.
5. Lipase enzyme is secreted by pancreas.
6. It converts emulsified fats into fatty acids and glycerol.
7. 
8. This process takes place in duodenum and small intestine.

Question 15.
What is the role of saliva in the digestion of food ?
(OR)
How does saliva digest food ?
Answer:

1. Saliva is secreted by three pairs of salivary glands present in the mouth.
2. Human saliva contains an enzyme called amylase (ptyalin).
3. It converts starch into maltose (a sugar).
4. The food is mixed thoroughly with saliva and becomes wet and slippery.
5. Saliva helps in the smooth passage of food in the food pipe.

Question 16.
What will happen to protein digestion as the medium of intestine is gradually rendered alkaline ?
Answer:

1. The food coming from the stomach to intestine is acidic in nature.
2. Bile and pancreatic juices render the internal condition of the intestine gradually to a basic or alkaline one.
3. Protein digestion continues even if the medium of intestine is gradually changed to alkaline.
4. In the alkaline medium pancreatic enzyme trypsin can act on the food and digests the proteins.
   
5. The enzymes present in the intestinal juice like peptidases complete the digestion of proteins into amino acids.
   

Question 17.
What is the role of roughages in the alimentary tract?
Answer:

1. Roughages are the fibres of either carbohydrates or proteins.
2. Plenty of roughages in the diet avoid constipation.
3. Roughages help in the easy movement of faeces in the large intestine.
4. They help in the easy digestion of food and keep the alimentary canal clean and healthy.

Question 18.
What is malnutrition? Explain some nutrition deficiency diseases.
Answer:
Malnutrition: Eating of food that does not have one or more than one nutrients in required amount is known as malnutrition.
Malnutrition is of three types:

1. Calorie malnutrition,
2. Protein malnutrition,
3. Protein calorie malnutrition.

Nutrition deficiency diseases:

1. Kwashiorkor disease: This is due to protein deficiency in diet.
   Symptoms:
   i) Body parts becomes swollen due to accumulation of water in the intercellular spaces,
   ii) Very poor muscle development,
   iii) Swollen legs,
   iv) Fluffy face,
   v) difficult to eat,
   vi) diarrhoea,
   vii) Dry skin.
2. Marasmus: This is due to deficiency of both protein and calories. Generally this disease occurs when there is an immediate pregnancy or repeated child births.
   Symptoms:
   i) Lean and weak,
   ii) Swelling in joints of limbs,
   iii) Less developed muscles,
   iv) Dry skin,
   v) diarrhoea.

Question 19.
How do non-green plants such as fungi and bacteria obtain their nourishment?
Answer:

1. Bacteria and fungi are non-green plants. So they cannot prepare their own food materials.
2. They are saprophytes which feed on dead and decaying plant and animal bodies.
3. The fungi and bacteria breakdown the complex organic molecules present in dead and decaying matter by releasing chemical substances into simple substances out¬side the body.
4. These simpler substances are then absorbed by fungi and bacteria as their food.

Question 20.
If we keep on increasing CO₂ concentration in the air, what will be the rate of photosynthesis?
Answer:

1. If the CO₂ concentration in the air increases, the rate of photosynthesis also increases.
2. If the CO₂ concentration raises above 5% then the rate of photosynthesis reduces.
3. At certain CO₂ concentration the rate of photosynthesis is constant.
4. Here a rise in CO₂ levels has no affect on the rate of photosynthesis as the other factors such as light intensity become limited.

Question 21.
What happens to plant if the rate of respiration becomes more than the rate of photosynthesis ?
Answer:

1. Respiration is a catabolic (destructive) process and photosynthesis is an anabolic (constructive) process.
2. If the rate of respiration becomes more than the rate of photosynthesis, the amount of food oxidised will be more than the food produced.
3. This affects the growth and development of plants and may even results in the death of the plant.

Question 22.
Why do you think that carbohydrates are not digested in the stomach?
(OR)
Where are carbohydrates digested in alimentary canal?
Answer:

1. For the digestion of carbohydrates enzyme ptyalin or amylase are required.
2. The gastric juice produced by stomach do not contain the enzyme ptyalin or amylase, it contains only pepsin which digests proteins.
3. Hence carbohydrates are not digested in the stomach.
4. Carbohydrates are partially digested in the mouth and completely in small intestine.

Question 23.
What process do you follow in your laboratory to study the presence of starch in leaves?
(OR)
(Activity – 1)
How do you test the presence of starch in leaves ? (OR)
Mention the materials required and explain the experiment to prove the presence of starch in leaves. What inference do you draw from this experiment?
Answer:
Aim: To study the presence of starch in leaves.
Apparatus: Beaker, test tube, bunsen burner, tripod stand, asbestos gauze, ethanol, leaf, petridish, iodine solution.
Procedure:

1. Select a leaf of a potted plant with soft thin leaves.
2. Boil the leaf in methylated spirit over a water bath till it becomes pale white due to the removal of chlorophyll.
3. Observe the leaf.
   
4. Spread the leaf in a dish and add a few drops of tincture iodine / betadine solution on it. Again observe the leaf.

Observation: The presence of starch will be indicated by a blue-black colour in leaf. Result: The experiment proves that starch is present in leaves. It is formed by Photo-synthesis.
Precautions:

1. Do not boil the methylated spirit test tube directly on flame.
2. Boil the water bath with low flame.

Question 24.
How would you demonstrate that green plants release oxygen when exposed to light? (OR) (Lab Activity)
Write the experimental procedure to prove that oxygen is produced during photosynthesis in the presence of light. (OR)
What materials are required to prove that oxygen is produced during photosynthesis in the presence of light? What procedure we need to follow to perform the above experiment?
We have conducted experiment that prove the release of oxygen when photosynthesis happens?
i) What are the plants used for this experiment? Where do they grow?
ii) How did you conduct the above experiment? In which context large number of air bubbles released? Do you noticed?
Answer:
i) Hydrilla plants are used for this experiment they grow in water.
ii) Experiment to demonstrate the release of oxygen during photosynthesis.
Aim: To prove that oxygen is produced during photosynthesis by hydrilla funnel experiment.
Apparatus: Beaker with water, test tube, funnel, hydrilla twigs, glowing splinter.
Procedure:

1. Arrange the apparatus as shown in the figure.
2. Place some water plant hydrilla in a beaker containing pond water, and cover these by a short stemmed funnel.
3. Invert a test – tube full of water over the stem of the funnel.
4. Ensure that the level of water in the beaker is above the level of stem of the inverted funnel.
5. Place the apparatus in the sun for at least 2 or 3 hours.
6. After sometime it is observed that gas bubbles come from the hydrilla plant. These bubbles are collected at the end of the test tube pushing the water into the beaker.
7. After sufficient gas is collected test – tube is taken out of the beaker carefully by closing it with thumb.

Observation: Test the gas in the test – tube by inserting a glowing incense stick which would burst into flames. This shows the presence of oxygen.
Result: This shows that oxygen is produced during photosynthesis.
Precautions:

1. Funnel should be smaller than the beaker.
2. Necessary care is to be taken while removing the test tube from the stem of the funnel.

Question 25.
Collect information from your primary health centre of malnutrition child at various ages and make a table your own and display in the classroom.
Answer:

Question 26.
If there were no green plants, all life on the earth would come to an end ! Comment.
(OR)
The survival of organisms would become difficult, if there are no green plants on the earth. How do you support?
Answer:

1. Plants play the most important part in the cycle of nature.
2. Without plants there could be no life on earth.
3. Plants are the only organisms that can make their own food and all other living beings directly or indirectly depend on plants for their food.
4. Moreover plants release oxygen into the atmosphere through photosynthesis.
5. Oxygen is essential for the organisms to respire.
6. Hence without green plants, all life on the earth would come to an end.

Question 27.
Draw a neat labelled diagram of chloroplast found in leaf, and its role in photosynthesis.
Answer:
Role of Chloroplast in photosynthesis:

1. Chloroplasts trap solar energy.
2. They convert that solar energy into chemical energy.
3. They help in the formation of glucose.

Question 28.
Draw the label diagram of human digestive system. List out the parts where peristalsis takes place.
Answer:
Parts where peristalsis takes place: Oesophagus, stomach, small intestine and large intestine.

Question 29.
Raheem prepared a model showing the passage of the food through different parts of the alimentary canal. Observe this and label its parts.
Answer:

Question 30.
Observe the following diagram and write a note on light dependent, light independent reactions.
Answer:
Note on light dependent reactions:

1. Light dependent reactions are also called as photochemical phase.
2. The light dependent reaction takes place in chlorophyll containing thylakoids called grana of chloroplasts.
3. Several steps occur in the light dependent reaction.
4. Step – 1: The chlorophyll on exposure to light energy becomes activated by absorbing photons.
5. Step – II: The energy is used in splitting the water molecule into two component ions named hydrogen (H⁺), hydroxyl ion (OH^–). This reaction is known as photolysis.
6. Step – III: OH^– ions through a series of steps produce water (H₂O) and O₂.
7. The end products of light reaction are ATP, NADPH and O₂.

Note on dark reaction or light independent reaction:

1. In light independent phase the hydrogen of the NADPH is used to combine it with CO₂, by utilizing ATP energy and to produce glucose.
2. This synthesis occurs in a number of steps using certain special intermediate compounds (mainly RUBP – Ribulose hi phosphate) and enzymes. Finally glucose is converted to starch.
3. All these reactions occur in the stroma region of the chloroplast.

Question 31.
Almost all the living world depends on plants for food material. How do you appreciate the process of making food by the green plants?
(OR)
What facts about the green plants do you appreciate?
Answer:
Leaf is a wonderful machine to synthesize food:

1. The leaf is the important site of photosynthesis and is called as food factory of the plant.
2. This plant organ can be treated as a wonderful natural machine which converts solar energy into useful chemical energy.
3. With all his scientific knowledge and technical skills, man has not produced anything similar leaf for utilization of solar energy without polluting the atmosphere.
4. This machine provides food and supports the life by providing oxygen for all the organisms including man on this planet.
5. Nature has given us such a wonderful machine free !!

Question 32.
Even a hard solid food also becomes smooth slurry in the digestive system by the enzymes released at a particular time. This mechanism is an amazing fact. Prepare a cartoon on it.
Answer:

Question 33.
What are good food habits?
Answer:
The food habits I am going to follow after reading this chapter are:

1. I take balanced diet which contains proper amounts of carbohydrates, proteins, fats, vitamins and minerals.
2. I avoid taking food containing high proportion of fat.
3. I eat food as much required by my body. I do not over eat.
4. I will not eat rich meals over several days.
5. I eat simple balanced meals, eat it leisurely and thoroughly masticating the food.
6. I avoid doing violent exercise soon after eating food.
7. I empty the bowels regularly avoiding constipation.
8. I will see to have plenty of roughages in the diet.

(OR)
After reading the chapter nutrition, I would like to follow the following food habits.

1. Having simple, well balanced meals.
2. Eating them in a leisurely manner.
3. Thoroughly masticating the food.
4. Avoiding strenuous exercise soon after eating food.
5. Drinking plenty of water and having regular bowel movement.
6. Decreasing consumption of coffee or tea per day.
7. Taking leafy vegetables at least 3 times a week and taking of fruits and vegetables plenty.
8. Maintaining regular timings for daily food consumption.

Fill in the blanks.

1. The food synthesized by the plant is stored as ———–.
2. ———– are the sites of photosynthesis.
3. Pancreatic juice contains enzymes for carrying the process of digestion of ———– and ———–.
4. The finger-like projections which increases the surface area in small intestine are called ———–.
5. The gastric juice contains ———– acid.
6. ———– vitamin is synthesized by bacteria present in intestine.

Answer:

1. carbohydrates
2. Chloroplasts
3. proteins, fats
4. Villi
5. HCl
6. Cyanocobalamin

Choose the correct answer.

1. Which of the following plant take the food by parasitic nutrition? [ ]
   A) Yeast
   B) Mushrooms
   C) Cuscuta
   D) Leeches
   Answer: C & D
2. The rate of photosynthesis is not affected by [ ]
   A) Light intensity
   B) Humidity
   C) Temperature
   D) Carbon dioxide concentration
   Answer: B
3. A plant is kept in dark cupboard for about forty eight hours before conducting any experiment on photosynthesis in order to [ ]
   A) Remove chlorophyll from leaves
   B) Remove starch from leaves
   C) Ensure that no photosynthesis occurred
   D) Ensure that leaves are free from the starch
   Answer: B
4. The digestive juice without enzyme is [ ]
   A) Bile
   B) Gastric juice
   C) Pancreatic juice
   D) Saliva
   Answer: A
5. In single-celled animals, the food is taken by [ ]
   A) the entire body surface
   B) mouth
   C) teeth
   D) vacuoles
   Answer: A
6. Which part of the plant takes in carbon dioxide from the air for photosynthesis? [ ]
   A) Root hair
   B) Stomata
   C) Leaf veins
   D) Sepals
   Answer: B

10th Class Biology 1st Lesson Nutrition – Food Supplying System Activities

Activity – 1

How do you prove experimentally that carbon dioxide is necessary for photosynthesis by Mohl’s half leaf experiment?
(OR)
List out the materials required and the procedure to be followed to prove that ‘carbon dioxide’ is essential for photosynthesis.
(OR)
You know that the factors like CO₂, Light and Chlorophyll are essential for photosynthesis. Write any one of experiment related to the factors essential for photosynthesis.
Answer:
Aim:
To prove-that carbon dioxide is essential for photosynthesis by Mohl’s half leaf experiment.
Apparatus:
Wide mouthed transparent bottle, KOH solution, potted plant, vertically split cork, Iodine solution.
Procedure:
Arrange the apparatus as shown in the figure.

1. Take a healthy potted plant and keep it in the dark for nearly a week for the removal of starch from the leaves.
2. A wide mouthed transparent bottle is taken.
3. Put potassium hydroxide pellets or potassium hydroxide solution (KOH) in the bottle.
4. This KOH absorbs CO₂ present in the bottle.
5. Insert splitted cork in the mouth of the bottle.
6. Insert one of the leaves of destarched plant through a split cork into transparent bottle.
7. Arrange half of the leaf is inside bottle and the remaining half outside.
8. Leave the plant in the sunlight for 2-3 hours.
9. After a few hours, test this leaf and other leaf of this plant for starch.

Observation :

1. The part of the leaf outside the bottle turns blue-black because starch is formed in this part due to photosynthesis.
2. The part of the leaf inside the bottle does not turn blue-black because the carbon dioxide present inside the bottle is absorbed by potassium hydroxide solution.
3. All the other factors water, sunlight and chlorophyll are available but not CO₂. Hence starch is not formed in the leaf part which is inside the bottle.

Result: This experiment proves that CO₂ is necessary for photosynthesis. Precautions:

1. The part of the leaf kept inside the bottle should not touch potassium hydroxide solution.
2. The apparatus should be kept air tight by applying grease or vaseline.

Activity – 2

Sunlight is necessary to form starch in green leaves.
(OR)
Write the materials required and the procedure to prove that light is essential for Photosynthesis.
(OR)
Write the procedure, precautions and observations in the lab activity, “Sunlight is necessary for photosynthesis”.
Answer:
Aim:
To prove that light is necessary for photosynthesis to form starch.
Apparatus:
Potted plant, light screen, iodine solution.

1. Keep potted plant in dark for one week to remove starch.
2. Take one black paper and cut it with your own design.
3. Keep design paper properly on the both sides with the help of clips.
4. Ensure that light does not pass through the covered area with black paper.
5. Keep the arranged apparatus at sunlight available area.
6. After few hours of exposure to bright sunlight detach the leaf.
7. Boil the leaf in methylated spirit over water bath. It becomes pale white due to the removal of chlorophyll. Take the leaf from test tube and spread the leaf in a petridish.
8. Add few drops of Iodine on leaf. The parts of the leaf, which could get light through the cut out design, turns blue-black colour.
9. The parts of the leaf which could not get light are not turned into blue – black colour.
   

Observation:
It is observed that only the parts of the leaf, which could get light through the cut out design, turn blue black, showing the presence of starch.

Result:
This experiment proves that light is necessary to form starch in the process of photosynthesis.

Activity – 3

Demonstrate litmus paper test on salivary amylase in the mouth.
Answer:

1. Before taking food into the mouth, take a litmus indicator paper and touch it to the tongue.
2. We observe no colour change in litmus paper.
3. Perform the litmus test again after chewing the food and swallowing it.
4. The red litmus paper turns to blue colour.
5. The blue litmus paper do not turns to red colour.
6. This demonstrates that amylase converts complex carbohydrates to simple sugar.
7. Amylase is alkaline in nature. This turns litmus paper blue when touches glucose at the second time.
   

Activity – 4

Observe different digestive enzymes and their role in digesting food in a tabular form.
Answer:

10th Class Biology 1st Lesson Nutrition – Food Supplying System InText Questions and Answers

10th Class Biology Textbook Page No. 2

Question 1.
Can you think of some raw materials needed for photosynthesis?
Answer:
Yes. Photosynthesis needs the following raw materials.

1. Sunlight, CO₂ water are external factors.
2. Chlorophyll and enzymes are internal factors.

Question 2.
What could be the end products of the process of photosynthesis?
Answer:
Glucose, water and oxygen are the end products of photosynthesis.

10th Class Biology Textbook Page No. 4

Question 3.
Do you think solar energy transforms into chemical energy by the process of photosynthesis?
Answer:
Yes, solar energy transforms into chemical energy by the process of photosynthesis.

Question 4.
What are the materials that you think would be essential for the synthesis of carbohydrates in the process of photosynthesis?
Answer:
The materials essential for the synthesis of carbohydrates in the process of photosynthesis are carbon dioxide, water, sunlight and chlorophyll.

Question 5.
Do you think the equation tells us about all the materials involved?
Answer:
Yes, the materials which are essential for photosynthesis and the products formed are involved in the equation.

10th Class Biology Textbook Page No. 5

Question 6.
What had Priestly done to introduce the mint plant without disturbing the experimental setup?
Answer:
Priestly should have tilted the bell jar to one side and introduced the mint plant without disturbing the experimental set up.

Question 7.
How did Priestly light the candle from outside?
Answer:
Priestly might have used convex lens through which beam of sun rays can light the candle from outside or he might have used long burning stick, to light the candle by lifting the jar partially.

Question 8.
Do you find any relationship between candle, rat, mint plant? Discuss.
Answer:
Priestly’s experiment confirmed that gaseous exchange was going on and plants were giving out a gas that supported burning and was essential for the survival of animals.
By combustion process candle releases carbondioxide. By respiration process rat also releases carbondioxide. During photosynthesis process mint plant uses this carbondioxide and releases oxygen. This oxygen will be used by rat to stay alive and for the candle to burn.
So there is a relationship between respiration and photosynthesis by candle, rat and mint plant.

10th Class Biology Textbook Page No. 6

Question 9.
Why was the plant kept in dark and then in sunlight?
Answer:

1. The plant is kept in the dark for nearly a week to remove the starch from the leaves.
2. Then only we can understand that the starch is formed in the leaves or not after the experiment when the plant is kept in the sunlight.

Question 10.
Why did we study two leaves in the Mohl’s half leaf experiment?
Answer:

1. To test CO₂ is essential or not for photosynthesis, two leaves are used in the experiment.
2. One leaf with the plant and another one used in the experiment.
3. The leaf which is exposed to the atmospheric air becomes bluish-black. It proves that starch is prepared in the leaf by using CO₂ from atmosphere.
4. The leaf inside the flask containing potassium hydroxide, which absorbs CO₂ present in the bottle does not become bluish black. It shows that CO₂ is necessary for photosynthesis.

10th Class Biology Textbook Page No. 7

Question 11.
What precautions do you need while removing test tube from the beaker? Discuss with your teacher.
Answer:

1. When sufficient gas is collected lift the test tube carefully from the beaker by closing its mouth with the thumb.
2. Because of that the gas present in the test tube cannot escape into the atmosphere.

10th Class Biology Textbook Page No. 8

Question 12.
Which part of leaf turns blue black? What about the remaining part?
Answer:

1. The part of the leaf, which could get light through the cut design turns to blue black showing the presence of starch.
2. The remaining part of the leaf which did not get light, do not turn blue, indicating that starch is not prepared.

Question 13.
Observe the colour of the leaf stained with iodine. Can you tell why it is stained differently?
Answer:

1. Some parts of the leaf prepared starch.
2. Some parts of the leaf does not prepared starch.
3. So it is stained differently.

Question 14.
What about plants having coloured leaves?
Answer:
Plants having coloured leaves also carry out photosynthesis. The coloured leaves containing pigments pass on the energy of sunlight trapped by them to chlorophyll.

Question 15.
How is that new leaves which look dark red in colour in several plants turn green?
Answer:

1. The new leaves which look dark red in colour contain coloured chromoplasts.
2. As the leaf grows the chromoplasts turns to chloroplasts and the leaf appears green in colour.

Question 16.
Do plants having reddish or yellowish leaves also carry out photosynthesis?
Answer:

1. Yes. Plants having reddish or yellowish leaves also carry out photosynthesis.
2. The pigments present in reddish or yellowish leaves pass on the energy of sunlight trapped by them to chlorophyll.

Question 17.
What made plants carry out photosynthesis while even green coloured animals (like some birds) could not?
Answer:

1. Chlorophyll and other pigment molecules trap (harvest) solar energy, convert it into chemical energy in the thylakoid membranes of the chloroplast.
2. But animals having green colour on their body cannot trap solar energy and cannot perform photosynthesis. Photosynthesis is possible only in plants but not in animals.

10th Class Biology Textbook Page No. 9

Question 18.
Where is chlorophyll and other pigments present in the plant?
Answer:
Chlorophyll and other pigments are present in the grana thylakoids of chloroplast in leaf.

Question 19.
Do you think the new reddish leaves of plants also carry out photosynthesis? What could be the role of their colour?
Answer:

1. Yes. New reddish leaves of plants also carry out photosynthesis.
2. Chromoplasts are responsible for the reddish colour of leaves.
3. They also pass on the energy of sunlight which they trap to the photosystems.

10th Class Biology Textbook Page No. 10

Question 20.
What makes chloroplast appear completely different from other cell organelles?
Answer:

1. Substances found in chloroplast, which capture sunlight are called photosynthetic pigments.
2. Two major kinds of chlorophyll are associated with thylakoid membranes.
3. Chlorophyll – a (blue – green in colour) and chlorophyll – b (yellow – green); around 250 to 400 pigment molecules are grouped as light harvesting units in each granum.
4. Such innumerable units in chloroplasts make them appear completely different from other organelles.

10th Class Biology Textbook Page No. 13