Category: Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}\)
= \(\frac{2}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) – \(\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\)
= \(\frac{8}{4}\)
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)
= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)
= \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\)
= \(\frac{4}{4}\) = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)
√3 = \(\frac{RQ}{6}\)
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = \(\sqrt{3 x^{2}}\) = √3x
Now, from the △XYZ
tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)
tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.1

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.1 Textbook Questions and Answers

Question 1.
A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower?
Answer:
Let the height of the tower = h m
Distance of the point of observation from the foot of the tower =15 cm.
Angle of elevation of the top of the tower = 45°
From the figure tan θ = \(\frac{\text { opp. side }}{\text { adj. side }}\)
tan 45° = \(\frac{h}{15}\)
⇒ 1 = \(\frac{h}{15}\)
∴ h = 1 × 15 = 15 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.
Answer:
Distance between the foot of tree and the point of contact of the top of the tree on the ground = 6 cm.
Let the length of the remaining part be = h m.
Let the length of the broken part be = x m.
Angle made by the broken part with the ground = 30°.
From the figure
tan 30° = \(\frac{h}{6}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{6}\)
∴ h = \(\frac{6}{\sqrt{3}}=\frac{3 \times 2}{\sqrt{3}}\) = 2√3 m
Also cos 30° = \(\frac{6}{x}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}\) = \(\frac{3 \times 2 \times 2}{\sqrt{3}}\) = 4√3
∴ Height of the tree = broken part + remaining part
= x + h
= 2√3 + 4√3 = 6√3 m
= 6 × 1.732
≃ 10.392 m.

Question 3.
A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2 m and by making an angle of 30° with the ground. What should be the length of the slide?
Answer:
Height of slide = 2 m
Let the length of the slide = x m.
Angle made by the slide with the ground = 30°
From the figure
sin 30° = \(\frac{2}{x}\)
⇒ \(\frac{1}{2}\) = \(\frac{2}{x}\)
⇒ x = 2 × 2 = 4 m
Length of the slide = 4 m.

Question 4.
Length of the shadow of a 15 meter high pole is 5√3 meters at 7 o’clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?
Answer:
Height of the pole = 15 m
Length of the shadow = 5√3 m
Let the angle of elevation be ‘θ’.
Then from the figure
tan θ = \(\frac{15}{5 \sqrt{3}}=\frac{5 \times \sqrt{3} \times \sqrt{3}}{5 \times \sqrt{3}}\) = √3
tan θ = √3 = tan 60°
∴ θ = 60°
∴ Angle of elevation of Sun rays with the ground = 60°.

Question 5.
You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30° with the pole. What should be the length of the rope?
Answer:
Height of the pole = 10 m
Let the length of each rope = x
Angle made by the rope with the pole = 30°

From the figure
cos 30° = \(\frac{10}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{10}{x}\)
⇒ x = \(\frac{10 \times 2}{\sqrt{3}}=\frac{20}{\sqrt{3}}\)
∴ Length of each rope = \(\frac{20}{\sqrt{3}}\)m
= 11.546 m.

∴ Total length of the rope = 3 × \(\frac{20}{\sqrt{3}}\)
= 20√3
= 20 × 1.732
≃ 34.64 m.

Question 6.
Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance between you and the object?
Answer:
Height of the building = 6 m
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{6}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}=\frac{2 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\) = 4√3
∴ Distance = 4√3 m or
4 × 1.732 = 6.928 m.

Question 7.
An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground? What will be the distance between foot of the ladder and foot of the pole?
Answer:
Height of the pole = 9m
Height of the point from the ground where he reaches the pole = 9 – 1.8 = 7.2 m
Angle of elevation = 60°
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{7.2}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{7.2}{x}\)
⇒ x = \(\frac{7.2 \times 2}{\sqrt{3}}=\frac{3 \times 2.4 \times 2}{\sqrt{3}}=\frac{\sqrt{3} \times \sqrt{3} \times 4.8}{\sqrt{3}}\)
⇒ x = 1.732 × 4.8
≃ 8.31 m
Also tan 60° = \(\frac{7.2}{d}\)
√3 = \(\frac{7.2}{d}\)
⇒ d = \(\frac{7.2}{\sqrt{3}}=\frac{2.4 \times 3}{\sqrt{3}}\) = 2.4 × √3 = 2.4 × 1.732
∴ d ≃ 4.1568 m

Question 8.
A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river?
Answer:
Let the width of the river = AB = x m
Angle made by the boat = 60°
Distance travelled = AC = 600 m
From the figure
cos 60° = \(\frac{x}{600}\)
\(\frac{1}{2}\) = \(\frac{x}{600}\)
⇒ x = \(\frac{600}{2}\) = 300 m.
In the figure
A = Boat’s place
C = Reach place of another side (or) Point of observation.
AC = Travelling distance of the boat ∠AC = 60°
AB = width of the river AB
In △ABC, sin 60° = \(\frac{AB}{AC}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{AB}{600}\)
⇒ AB = 600 × \(\frac{\sqrt{3}}{2}\) = 300√3

Question 9.
An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45°. What is the height of the palm tree?
Answer:
Height of the observer = 1.8 m
Distance of the observer from the palm tree = 13.2 m
From the figure
tan 45° = \(\frac{x}{13.2}\)
⇒ 1 = \(\frac{x}{13.2}\)
⇒ x = 13.2 m
∴ Height of the palm tree = 13.2 + 1.8 = 15 m.

Question 10.
In the given figure, AC = 6 cm, AB = 5 cm and ∠BAC = 30°. Find the area of the triangle.
Answer:
Draw a perpendicular BD to AC
∴ BD ⊥ AC
Now let AD = 6 – x and DC = x
Given AB = 5 cm and ∠BAD = 30° then in △ABD
sin 30° = \(\frac{BD}{AB}\) = \(\frac{BD}{5}\) = \(\frac{1}{2}\)
⇒ BD = \(\frac{5}{2}\) = 2.5 cm
and cos 30° = \(\frac{AD}{AB}\) = \(\frac{6-x}{5}\) = \(\frac{\sqrt{3}}{2}\)
⇒ 6 – x = \(\frac{5 \sqrt{3}}{2}\)
⇒ x = 6 – \(\frac{5 \sqrt{3}}{2}\) = 6 – \(\frac{5(1.732)}{2}\)
∴ x = 1.67
∴ Area of △ABC = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) × AC × BD
= \(\frac{1}{2}\) × 6 × 2.5
= 7.5 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.3

10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Answer:

Sum of the frequencies = 68
∴ \(\frac{n}{2}\) = \(\frac{68}{2}\) = 34
Hence median class = 125 – 145
Lower boundary of the median class, l = 125
cf – cumulative frequency of the class preceding the median class = 22
f – frequency of the median class = 20
h = class size = 20
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 125 + \(\frac{[34-22]}{20}\) × 20
∴ Median = 125 + 12 = 137
Maximum number of consumers lie in the class 125 – 145
Modal class is 125 -145
l – lower limit of the modal class =125
f₁ – frequency of the modal class = 20
f₀ – frequency of the class preceding the modal class =13
f₂ – frequency of the class succeeding the modal class =14
h – size of the class = 20
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
Mode (Z) = 125 + \(\frac{20-13}{(20-13)+(20-14)} \times 20\)
= 125 + \(\frac{7}{7+6}\) × 20
= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
∴ Mode = 135.769
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 = 135
∴ \(\overline{\mathbf{x}}\) = 135 + \(\frac{7}{68}\)
= 135 + 0.102941
≃ 135.1
Mean, Median and Mode are approximately same in this case.

Question 2.
If the median of 60 observations, given below is 28.5, find the values of x and y.

Answer:

Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
It is given that ∑f = n = 60
So, 45 + x + y = 60
x + y = 60 – 45 = 15
x + y = 15 ….. (1)
The median is 28.5 which lies be-tween 20 and 30.
Median class = 20 – 30
Lower boundary of the median class ‘l’ = 20
\(\frac{N}{2}\) = \(\frac{60}{2}\) = 30
cf – cumulative frequency = 5 + x
h = 10
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
⇒ 28.5 = 20 + \(\frac{30-5-x}{20}\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
\(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
25 – x = 2 × 8.5
x = 25- 17 = 8
also from (1); x + y = 15
8 + y = 15
y = 7
∴ x = 8; y = 7.

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.

Answer:

The given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60.
Observe that from the given distribution 2 persons with age less than 20.
i.e., frequency of the class below 20 is 2.
Now there are 6 persons with age less than 25 and 2 persons with age less than 20.
∴ The number of persons with age in the interval 20 – 25 is 6 – 2 = 4.
Similarly, the frequencies can be calculated as shown in table.
Number of observations = 100
n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50, which lies in the class 35-40
∴ 35 – 40 is the median class and lower boundary l = 35
cf = 45;
h = 5;
f = 33
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 35 + \(\frac{50-45}{33}\) × 5
= 35 + \(\frac{5}{33}\) × 5
= 35 + 0.7575
= 35.7575
∴ Median ≃ 35.76

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5,…, 171.5 – 180.5.)
Answer:
Since the formula, Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\) assumes continuous classes assumes continuous class, the data needs to be converted to continuous classes.
The classes then changes to 117.5 – 126.5; 126.5 – 133.5, …… 171.5 – 180.5.

∑f_i = n = 40
\(\frac{n}{2}\) = \(\frac{40}{2}\) = 20
\(\frac{n}{2}\)th observation lie in the class 144.5- 153.5
∴ Median class = 144.5 – 153.5
Lower boundary, l = 144.5
Frequency of the median class, f = 12
c.f. = 17
h = 9
∴ Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 144.5 + \(\frac{20-17}{12}\) × 9
= 144.5 + \(\frac{3}{12}\) × 9
= 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
∴ Median length = 146.75 mm.

Question 5.
The following table gives the distribution of the life-time of 400 neon lamps.

Find the median life-time of a lamp.
Answer:

Total observations are n = 400
\(\frac{n}{2}\)th observation i.e \(\frac{400}{2}\) = 200
200 lies in the class 3000 – 3500
∴ Median class = 3000 – 3500
Lower boundary l = 3000
frequency of the median class f = 86
c.f = 130
Class size, h = 500
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 3000 + \(\frac{200-130}{86}\) × 500
= 3000 + \(\frac{70}{86}\) × 500
= 3000 + 406.977
= 3406.98
∴ Median life ≃ 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows.

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
Answer:
Number of letters in the surnames.
Also find the modal size of the surnames.

Total observations are n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50
50 lies in the class 7 – 10
∴ Median class = 7 – 10
l – lower boundary = 7
f – frequency of the median class = 40
cf = 36
Class size h = 3
Median:
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 7 + \(\frac{50-36}{40}\) × 3
= 7 + \(\frac{14}{40}\) × 3
= 7 + \(\frac{42}{40}\)
= 7 + 1.05
= 8.05
∴ Median = 8.05.

Mean:
Assumed mean, a = 8.5
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 8.5 + \(\frac{(-18)}{100}\)
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32.

Mode:
Maximum number of surnames = 40
∴ Modal class = 7-10
l – lower boundary of the modal class = 7
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
l = 7; f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode (Z) = 7 + \(\frac{40-30}{(40-30)+(40-16)}\) × 3
= 7 + \(\frac{10}{10+24}\) × 3
= 7 + \(\frac{30}{34}\)
= 7 + 0.882
= 7.882

Median = 8.0.5; Mean = 8.32; Modal size = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

Number of observations (n) = ∑f_i
\(\frac{n}{2}\) = \(\frac{30}{2}\) = 15
15 lies in the class 50 – 55
∴ Median class = 50-55
l – lower boundary of the median class = 55
f – frequency of the median class = 8
c.f = 5
Class size h = 6
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 50 + \(\frac{15-5}{8}\) × 6
= 50 + 7.5
= 57.5
= 50 + 7.5 = 57.5
∴ Median weight = 57.5 kg.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.1

10th Class Maths 14th Lesson Statistics Ex 14.1 Textbook Questions and Answers

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Answer:

Since f_i and x_i are of small values we use direct method.
∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{162}{20}\)
= 8.1

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:

Here the x_i are of large numerical values.
So we use deviation method then,
\(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
Here the assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 275 + \(\frac{1900}{50}\)
= 275 + 38
= 313.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Answer:

\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
x_i = 18 (given)
\(\Rightarrow 18=\frac{752+20 \mathrm{f}}{(44+\mathrm{f})}\)
18 (44 + f) = 752 + 20 f
⇒ 20f- 18f= 792-752
⇒ 2f = 40
∴ f = \(\frac{40}{20}\) = 20.

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Answer:

\(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
75.5 + \(\frac{12}{30}\)
= 75.5 + 0.4
= 75.9.

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?
Answer:

Here we use step deviation method where a = 135, h = 5,a multiple of all d_i
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 22 + \(\frac{25}{400}\) × 5
= 22 + 0.31
= 22.31

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
Answer:

Here a = 125, h = 50, ∑f_iu_i = 43
Now
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 125 + \(\frac{43}{25}\) × 50
= 125 + (43 × 2)
= 125 + 86
= 211.
NOTE: If we consider first value as “a” then we dont get negative values in u_i, f_iu_i columns. Then it becomes easy for calculation.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Answer:

∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\)
= 0.00986666…….
≃ 0.099

Question 8.
A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Answer:

Here, a = 51.5
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
= 51.5 – \(\frac{99}{40}\)
= 51.5 – 2.475
= 49.025
≃ 49 days

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

a = 70; h = 10
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}} \times \mathbf{h}\)
\(\Rightarrow \bar{x}=70-\frac{2}{35} \times 10\)
= 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\)
= 70 – 0.57142
= 69.4285
≃ 69.43%

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: