Category: Class 10

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 4 Acids, Bases and Salts Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 4th Lesson Acids, Bases and Salts

10th Class Chemistry 4th Lesson Acids, Bases and Salts Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Which property do you think of while suggesting the remedy from a problem of acidity?
Answer:
Neutralization Property. Antacid tablets neutralise acidity.

Improve your learning

Question 1.
Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11,7 and 9 respectively, which solution is : (AS1)
a) neutral
b) strongly alkaline
c) strongly acidic
d) weakly acidic
e) weakly alkaline
Arrange the pH in increasing order of hydrogen ion concentration.
Answer:
Solution – pH Value
A → 4
B → 1
C → 11
D → 7
E → 9
a) Solution ‘D’ is neutral
b) Solution ‘C’ is strongly alkaline
c) Solution ‘B’ is strongly acidic
d) Solution ‘A’ is weakly acidic
e) Solution ‘E1 is weakly alkaline
∴ Increasing order of Hydrogen ion concentration : C < E < D < A < B.

Question 2.
What is a neutralization reaction? Give two examples. (AS1)
Answer:
Neutralization reaction : When acid reacts with base, forms its salt and water. This reaction is called a neutralization reaction.
Examples :

Equation: HCl + NaOH → NaCl + H₂O
ii) Acetic Acid + Sodium Hydroxide → Sodium Acetate + Water
Equation : CH₃COOH + NaOH → CH₃COONa + H₂O
Formula : Acid + Base > Salt + Water

Question 3.
What happens when an acid or base is mixed with water? (AS1)
Answer:
When an acid or base is mixed with water it changes into dilute acid or dilute base.
(OR)
Dilute acid or dilute base will be formed when an acid or base is mixed with water. Mixing an acid or base with water results in decrease in the concentration of ions (H30+/ OH-) per unit volume. Such a process is called dilution and the acid or base is said to be diluted.

Question 4.
Why does tooth decay start when the pH of mouth is lower than 5.5? (AS1)
(OR)
Does the pH change tooth decay? Explain.
Answer:

1. Tooth enamel is the hardest substance in the body.
2.  It doesn’t dissolve in water but corroded when the pH in the mouth is below 5.5.
3. It happens due to the bacteria which produce acids by degradation of sugar and food particles remaining in the mouth.

Question 5.
Why does not distilled water conduct electricit? (AS2)
Answer:

1. Distilled water does not contain impurities.
2. It is also extremely weak electrolyte.
3. So it does not dissociate into ions.
4. It does not have charge carriers.
5. Because of that it does not conduct electricity.

Question 6.
Dry hydrogen chloride gas does not turn blue litmus to red whereas hydrochloric acid does. Why? (AS1)
Answer:
1. Dry hydrogen chloride gas is not an acid. Because it does not produce H⁺_(aq) ions. Hence it can’t turn blue litmus into red.
2. Hydrochloric acid is an aqueous solution. So it can produce H⁺_(aq) ions. Hence it can turn blue litmus into red.

Question 7.
Why pure acetic acid does not conduct electricity? (AS1)
Answer:
The reasons for pure acetic acid does not conduct electricity are :
i) Acetic acid is a weak acid.
ii) It gives fewer H₃O⁺ ions.

Question 8.
A milkman adds a very small amount of baking soda to fresh milk. (AS2)
a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
Answer:
1. By adding a very small amount of baking soda to fresh milk, the milkman keeps the milk unspoiled for little more time than usual time.
2. As the pH value increases the milk turns to slightly alkaline.

b) Why does this milk take a long time to set as curd?
Answer:

1. Curd form from the milk by the action of Lactic acid produced by bacteria in the milk.
2. If milk man add Baking soda (NaHCO₃) to the milk it neutralise acid, which is produced by the bacteria.
3. Excess acid is required to change the milk as curd.
4.  It takes long time.

Question 9.
Plaster of Paris should be stored in a moisture-proof container. Explain why? (AS2)
Answer:
Storing of Plaster of Paris :

1. Plaster of Paris is a white powder.
2. It easily absorbs water in air and forms hard gypsum.
3. So, it should be stored in a moisture-proof container.

Question 10.
Fresh milk has a pH of 6. Explain why the pH changes as it turns into curd.
Answer:
1. Fresh milk has a pH of 6. Hence it is a weak acid.
2. To turn the milk as curd, we have to add yeast in the form of some curd. The fermentation takes place during this process and lactose changes in lactic acid and the pH decreases as milk sets as curd.

Question 11.
Compounds such as alcohols and glucose contain hydrogen but are not categorized as acids. Describe an activity to prove it. (AS3)
(OR) (Activity – 7)
Write an activity to show that the solutions of compounds like alcohol and glucose do not show acidic character even though they are having Hydrogen.
(OR)
Write an activity which proves acids are good conductors of electricity.
(OR)
The acidity of acids is attributed to the H+ ions produced by them in solution explain the above statement with an activity.
List out the material for the experiment to investigate whether all compounds containing Hydrogen are acids or not and write the experimental procedure.
Answer:
List of the material required :

1. Glucose
2. Alcohol
3. Dil. HCl
4. Dil-H₂SO₄
5. Beaker
6. Connecting wires
7. 230 voltage AC supply
8. Bulb
9. Graphite rods.

Procedure :

1. Prepare glucose, alcohol, hydrochloric acid and sulphuric acid solutions.
2. Connect two different coloured electrical wires to graphite rods separately as shown in figure.
3. Connect free ends of the wire to 230 volts AC plug.
4. Complete the circuit as shown in the figure by connecting a bulb to one of the wires.
5. Now pour some dilute HCl in the beaker and switch on the current.

Observation :
The bulb starts glowing.

Repetition:
Repeat activity with dilute sulphuric acid, glucose and alcohol solutions separately.

Observation :

1. We will notice that the bulb glows only in acid solutions.
2. But the bulb does not glow in glucose and alcohol solutions.

Result:

1. Glowing of bulb indicates that there is flow of electric current through the solution.
2. Acid solutions have ions and the movement of these ions in solution helps for flow of electric current through the solution.

Conclusion :

1. The positive ion (cation) present in HCl solution is H⁺.
2. This suggests that acids produce hydrogen ions H⁺ in solution, which are, responsible for their acidic properties.
3. In glucose and alcohol solution the bulb did not glow indicating the absence of H⁺ ions in these solutions.
4. The acidity of acids is attributed to the H⁺ ions produced by them in solutions.

Question 12.
What is meant by “water of crystallization” of a substance? Describe an activity to show the water of crystallisation. (Activity – 16) (AS3)
Answer:
Water of Crystallization : Water of crystallization is the fixed number of water molecules present in one formula unit of a salt in its crystaline form.
Ex : CuSO₄ • 5H₂O.
It means that five water molecules are present in one formula unit of copper sulphate.

Activity to show the water of crystallization :

1. Take a few crystals of copper sulphate in a dry test tube and heat the test tube.
2. We observe water droplets on the walls of the test tube and salt turns white.
3. Add 2 – 3 drops of water on the sample of copper sulphate obtained after heating.
4. We observe, the blue colour of copper sulphate crystals is restored.

Reason :
1. In the above activity copper sulphate crystals which seem to be dry contain the water of crystallization, when these crystals are heated, water present in crystals is evaporated and the salt turns white.

2. When the crystals are moistened with water, the blue colour reappears.
Removing water of crystallization

Question 13.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. Amount and concentration of both the acids are same. In which test tube will the fizzing occur more vigorously and why? (AS4)
Answer:
1. The volatility of acetic acid (CH₃COOH) is more than that of hydrochloric acid.
2. But HCl solution has more strength than acetic acid.
3. Hence magnesium ribbon in test tube A will react more vigorously than in B.
4. So fizzing occurs more vigorously in test tube ‘A’.

Question 14.
Draw a neat diagram showing acid solution in water conducts electricity. (AS5)
(OR)
Draw a neat diagram which shows acids contains H⁺ ions.
(OR)
Draw a neat diagram showing how does dilute HCl solution conduct electricity.
Answer:

Question 15.
How do you prepare your own indicator using beetroot ? Explain. (AS5)
Aim : To prepare own indicator.
Materials required :
1) Beetroots-2 or 3
2) Knife
3) Bowls
4) Water
5) Spoon
6) Mixy
7) Orange juice

Procedure:
1) Take the beetroots and peel them with the help of a knife. (Firstly wash them).
2) Chop them into pieces.
3) Put those pieces into a mixy jar and make a paste.
4) Add some water to the paste. Now filter this and collect only juice from this.

Observation and Result:
1) Now add 5 to 6 drops of this juice, (beetroot juice (indicator)) to orange juice (5 to 6 drops) and mix it.
2) We can see the colour changes. This indicates the presence of acidic nature in orange juice.

Question 16.
How does the flow of acid rain into a river make the survival of aquatic life in a river difficult? (AS7)
(OR)
What are the harmful effects of acid rain?
Answer:
1) Acid rains are combination of carbonic acid, sulphuric acid and nitric acid with rain water.
2) The pH of acid rain is less than 5.6.
3) Living organisms can survive only in a narrow range of pH change.
4) When acid rain with pH value less than 5.6, flows into rivers, it lowers the pH of river water.
5) Due to less pH, the river water becomes acidic and hence the aquatic life in such rivers becomes difficult.

Question 17.
What is baking powder? How does it make the cake soft and spongy? (AS7)
Answer:
1) Baking Powder:
Baking powder is a mixture of baking soda (NaHCO₃) and a mild edible acid such as tartaric acid. COOH (CHOH)₂ COOH

2) Chemical reaction :
When baking powder is heated or mixed in water, the following reaction takes place.
NaHCO₃ + H⁺ → CO₂ + H₂O + Sodium salt of acid.

3) Carbondioxide produced during the reaction causes bread or cake to rise making them soft and spongy.

Question 18.
Give two important uses of washing soda and baking soda. (AS7)
(OR)
Write the chemical formulae for washing soda and Baking soda and give their uses.
(OR)
Write any four uses of washing soda.
Answer:
Uses of washing soda (Na₂CO₃.10H₂O) :
1) Washing soda (sodium carbonate) is used in glass, soap and paper industries.
2) It is used in the manufacture of sodium compounds such as borax.
3) Sodium carbonate can be used as a cleaning agent for domestic purposes.
4) It is used for removing permanent hardness of water.

Uses of baking soda (NaHCO₃ 10H₂O) :
1) Baking soda (Sodium hydrogen carbonate) is used for faster cooking.
2) Baking powder (a mixture of baking soda and a mild acid) is used in preparation of cakes.
3) Sodium hydrogen carbonate is also an ingredient in antacids.
4) It is also used in soda – acid, fire extinguishers.
5) It acts as a mild antiseptic.

Fill in the Blanks

1. i) ………………….. taste is a characteristic property of all acids in aqueous solution.
ii) Acids react with some metals to produce ………………….. gas.
iii) Because aqueous acid solutions conduct electricity, they are identified as …………………..
iv) Acids react with bases to produce a ………………….. and water.
v) Acids turn methyle orange into ………………….. colour.
Answer:
1. i) Sour
ii) hydrogen
iii) electrolytes or conductors
iv) salt
v) red

2. i) Bases tend to taste ………………….. and feel ………………….. .
ii) Like acids, aqueous basic solutions conduct ………………….., and are identified as ………………….. .
iii) Bases react with ………………….. to produce a salt and
iv) Bases turn phenophthalein into ………………….. colour.
Answer:
2. i) bitter, soapy (slippery) to touch
ii) electricity, electrolytes
iii) acids, water
iv) pink

Match the following :

Answer:
3. a – 4,
b – 5,
c – 1,
d – 2,
e – 3.

Multiple Choice Questions

1. The colour of methyl orange indicator in acidic medium is
A) yellow
B) green
C) orange
D) red
Answer:
D) red

2. The colour of phenolphthalein indicator in basic solution is
A) yellow
B) green
C) pink
D) orange
Answer:
C) pink

3. Colour of methyl orange in alkali conditions
A) orange
B) yellow
C) red
D) blue
Answer:
B) yellow

4. A solution turns red litmus blue, its pH is likely to be
A) 1
B) 4
C) 5
D) 10
(OR)
If a solution converts red litmus into blue colour, then its pH value is …………….. .
A) 1
B) 4
C) 5
D) 10
Answer:
D) 10

5. A solution reacts with crushed egg-shells to give a gas that turns lime-water milky, the solution contains …………….. .
A) NaCl
B) HCl
C) LiCl
D) KCl
Answer:
B) HCl

6. If a base dissolves in water, by what name is it better known?
A) neutralization
B) basic
C) acid
D) alkali
Answer:
D) alkali

7. Which of the following substances when mixed together will produce table salt?
A) Sodium thiosulphate and sulphur dioxide
B) Hydrochloric acid and sodium hydroxide
C) Chlorine and oxygen
D) Nitric acid and sodium hydrogen carbonate
Answer:
B) Hydrochloric acid and sodium hydroxide

8. What colour would hydrochloric acid (pH = 1) turn universal indicator?
A) Orange
B) Purple
C) Yellow
D) Red
Answer:
D) Red

9. Which one of the following types of medicines is used for treating indigestion?
A) antibiotic
B) analgesic
C) antacid
D) antiseptic
Answer:
C) antacid

10. What gas is produced when magnesium is made to react with hydrochloric acid?
A) hydrogen
B) oxygen
C) carbon dioxide
D) no gas is produced
Answer:
A) hydrogen

11. Which of the following is the most accurate way of showing neutralization?
A) Acid + base → acid-base solution
B) Acid + base → salt + water
C) Acid + base → sodium chloride + hydrogen
D) Acid + base → neutral solution
Answer:
B) Acid + base → salt + water

10th Class Chemistry 1st Lesson Acids, Bases and Salts InText Questions and Answers

10th Class Chemistry Textbook Page No. 25

Question 1.
Is the substance present in antacid tablet acidic or basic?
A. The substance present in antacid tablet is basic.

Question 2.
What type of reaction takes place in stomach when an antacid tablet is consumed?
mrearx
A. Neutralization reaction takes place in stomach when an antacid tablet is consumed.

10th Class Chemistry Textbook Page No. 26

Question 3.
You are provided with three test tubes containing distilled water, an acid and a base solution respectively. If you are given only blue litmus paper, how do you identify the contents of each test tube?
Answer:
I know that acid turns blue litmus to red. With the help of this test I can find the acid. Distilled water and base don’t do so. Thus I identify each.

Question 4.
Which gas is usually liberated when an acid reacts with a metal? How will you test for the presence of this gas?
Answer:
Usually acids generate hydrogen gas on reacting with metals.

Test: When a burning splinder is brought near to the collected gas (H₂), it puts off with a pop sound.
This test proves that the gas is H₂.

Question 5.
A compound of a calcium reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle ; turns lime water milky. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:
Equation is : CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O

10th Class Chemistry Textbook Page No. 30

Question 6.
Why do HCl, HNO₂ etc. show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
HCl, HNO₃, etc. show acidic characters in aqueous solutions as they liberate H⁺ ions. But alcohol and glucose don’t liberate H⁺ ions. So, they do not show acidic character.

Question 7.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
Answer:
1) If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns.
2) The glass container may also break due to excessive local heating.

10th Class Chemistry Textbook Page No. 33

Question 8.
What will happen if the pH value of chemicals in our body increases?
Answer:
When pH value of chemicals in our body increases then the body will effect by some problems. They are
1) Digestion problems raise in the stomach.
2) pH changes as the cause of tooth decay.

Question 9.
Why do living organism have narrow pH range?
Answer:
Because increasing acidity is thought to have a range of possibly harmful consequences such as depressing metabolic rate and immune response in some organisms and causing coral bleaching

10th Class Chemistry 4th Lesson Acids, Bases and Salts Activities

Activity – 1

Question 1.
Observe the change in colour in each case and tabulate the results in the table.
Answer:
Procedure:
1) Collect the following samples from the science laboratory ;
i) Hydrochloric acid (HCl)
ii)Sulphuric acid (H₂SO₄)
iii) Nitric acid (HNO₃)
iv) Acetic acid (CH₃COOH)
v) Sodium hydroxide (NaOH)
vi) Calcium hydroxide[Ca(OH)₂]
vii) Magnesium hydroxide [Mg(OH)₂]
viii) Ammonium hydroxide(NH₄OH)
ix) Potassium hydroxide (KOH)
2) Prepare dilute solutions of the respective substances.
3) Take four watch glasses.
4) Put one drop of the first solution in each one of them and test the solution as follows.
i) Dip the blue litmus paper in the first watch glass.
ii) Dip the red litmus paper in the second watch glass.
iii) Add a drop of methyl orange to the third watch glass.
iv) Add a drop of phenolphthalein to the fourth watch glass.

Observation :
Observe the respective colour changes and note down in the chart below.

1. What do you conclude from the observations noted in the above table? (AS1)
Answer:
Conclusion : Acids turn blue litmus to red and bases turn red litmus to blue. Acids turn phenolphthalein to colourless and bases turn pink. Acids turn methyl orange to red and bases turn methyl orange to yellow.

2. Identify the above sample as acidic or basic solution. (AS4)
Answer:
Acids : HCl, H₂SO₄, HNO₃, CH₃COOH.
Bases : NaOH, KOH, Mg(OH)₂, NH₄OH, Ca(OH)₂.

Activity -2

Question 2.
What are Olfactory indicators? Write an activity to prove them.
(OR)
What is the name given to a substance which identifies an acid or base by virtue of smell? Write an activity to prove the fact with an example.
Answer:
Olfactory Indicators : There are some substances whose odour changes in acidic or basic media. These are called olfactory indicators.
Activity :
Aim : To check the olfactory indicator.
Required materials :
1) Onions
2) Knife
3) Plastic bag
4) Clean clothes.

Procedure :
1) Take some onions and finely chop them.
2) Put the chopped onions in a plastic bag along with some clean cloth.
3) Tie up the bag tightly and keep it overnight in the fridge.
4) Then remove onions from fridge and add some base. We observe it loses its smell. Observation : Check the odour of the cloth strips.

Result: It is used as the basic indicator.

LAB ACTIVITY Reaction of Acids with metals

Question 3.
Write an experiment showing the reaction of acids with metals. (AS3)
(OR)
Ramu added acid to active metal then what is the gas which has been liberated. What are the apparatus required to prove the experiment. Write the experimental acitivity.
(OR)
Write the required material and experimental procedure for the experiment, “Hydrochloric acid reacts with ‘Zn’ pieces and liberates H₂“.
Answer:
Aim : To show the reaction of acids with metals.

Required Materials :

1. Test tube
2. Delivery tube
3. Glass trough
4. Candle,
5. Soap water
6. Dil. HCl
7. Zinc granules
8. One holed rubber stopper
9. Retard stand

Experimental procedure :

1. Take some zinc granules in a test tube and arrange the test tube to the retart stand.
2. Fix a delivery tube to the rubber stopper and immerse the second end of the delivery tube into the soap water.
3. Add about 10 m/ of dilute hydrochloric acid to Zn granules and fix rubber stopper to the test tube.
4. Evolved gas forms bubbles in soap water.
5. Bring a lightened candle near to the gas bubbles. We can observe the burning of gas bubble with pop sound.

Result: We can conform that the evolved gas is hydrogen.
Chemical reaction:
Acid + Metal → Salt + Hydrogen
2Hcl_(aq) + Zn_(s) → Zncl_2(aq) + H_2(g)

Additional Experiment :

• Repeat the above experiment with H₂SO₄ and HNO₃.
• We observe the same observation of the HCL experiment.

Conclusion : From the above activities we can conclude that when acid reacts with metal, H₂ gas is evolved.

Activity – 3 Reaction of Bases with metals

Question 4.
Write an activity to show the reaction of bases with metals.
(OR)
Write an activity which proves certain bases produce hydrogen gas when they react with metals.
Answer:
Aim : To show the reaction of bases with metals.

Required Materials :

1. Test tube,
2. Delivery tube
3. Glass trough
4. Candle
5. Soap water
6. Sodium hydroxide (NaOH) Solution
7. Zinc granules
8. One holed rubber stopper

Procedure :

1. Set the apparatus as shown in figure.
2. Take about 10 ml of dilute Sodium hydroxide (NaOH) solution in a test tube.
3. Add a few granules of zinc metal to it.
4. We will observe formation of gas bubbles on the surface of granules.
5. The gas will pass through delivery tube evolved from soap solution as bubbles.
6. Bring burning candle near the gas filled bubble.
7. The gas in the bubble puts off the candle with pop sound.

Result: The evolved gas is hydrogen.

Chemical reaction :
Base + Metal → Salt + Hydrogen

Note : It is better to use cone. NaOH solution for this reaction.

Activity – 4 Reaction of carbonates and metal hydrogen carbonates with Acids

Question 5.
Write an activity to show that all metal carbonates and hydrogen carbonates react with acids to give a corresponding salt. (AS3)
Answer:
Aim : To show that all metal carbonates and hydrogen carbonates react with acids to give a corresponding salt.

Required Materials :

1. Two test tubes
2. Sodium Carbonate (Na₂CO₃)
3. Sodium Hydrogen Carbonate (NaHCO₃)
4. Two holed rubber stopper
5. Thistle funnel
6. Stand
7. Dilute hydrochloric acid
8. Delivery tube
9. Calcium Carbonate (in a test tube)

Procedure :

1. Take a test tube A with 0.5 gm of sodium carbonate.
2. Close the test tube A with two holed rubber cork.
3. Insert a thistle funnel through one hole and insert a delivery tube through the other hole.
4. Pour 2 ml of dilute HC/ to the test tube A.
5. Do the same as above with test tube B with sodium hydrogen carbonate.

Observation :
Carbon dioxide is released from test tube A and B. Passing CO₂ gas through Ca(OH)₂ solution

Chemical Reaction :
Na₂CO_3(s) + 2 HCl_(aq) → 2 NaCl_(aq) + CO_2(g) + H₂O_(l)
Metal Carbonate + Acid → Salt + Carbon dioxide + Water

NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + CO_2(g) + H₂O_(l)
Metal Hydrogen Carbonate + Acid → Salt + Carbon dioxide + Water

Result : All metal carbonates and hydrogen carbonates react with acids to give a corresponding salt.

Activity – 5 Neutralization reaction

Question 6.
Write an activity to find the change of colour in the reaction of an acid with a base (Neutralization) reaction. (AS3)
(OR)
Explain neutralization reaction with an activity.
Answer:
Aim : To test the change of colour in the reaction of an acid with a base.

Required Materials :

1. 2 ml of dilute NaOH (Sodium Hydroxide) solution.
2. Phenolphthalein indicator solution.
3. dilute HCl (Hydrochloric) solution.

Procedure :

1. Take about 2 ml of dilute NaOH solution in a test tube.
2. Add two drops of phenolphthalein indicator solution.

Observation (i) :

1. It turns to red or pink colour.
2. It shows that NaOH is a base.

Experiment (1) : Add dilute HCl solution to the above solution drop by drop.
Observation (ii) : Pink colour disappears due to the reaction of NaOH (base) with HCl (acid).

Experiment (2) : Now add one or two drops of NaOH to the above mixture.
Observation (iii) : Pink colour reappears on adding NaOH.
NaOH + HCl → NaCl + H₂O
base + acid → salt + water
Result: This reaction is called a neutralization reaction.

Activity – 6 Reaction of metallic oxides with acids

Question 7.
Write an activity to show that metal oxide reacts with acid is a neutralization. (AS3)
(OR)
How can you prove metallic acids are basic in nature?
Answer:
1) Take a small amount of copper oxide (CuO) in a beaker.
2) Add dilute HCl slowly while stirring.
3) Copper oxide dissolves in dilute HCl and the solution becomes blueish green colour due to the formation of copper (II) chloride.

Equation : Metal oxide + Acid → Salt + Water
Result: This reaction is same as the reaction of base with acid, (neutralization)

Question 8.
Write an activity to show that non-metallic oxide reacts with base is a neutralization.
Answer:
1) Take a small amount of calcium hydroxide (base).
2) Add CO₂ into it.
3) Salt and water are produced.
Equation : Non-metallic oxide + Base → Salt + Water
Result: It is a neutralization reaction.

Question 9.
Repeat the activity – 7 using alkalis such as sodium hydroxide, calcium hydroxide solutions, etc. instead of acid solutions.
i) Does the bulb glow?
Answer:
Yes, the bulb will glow.

ii) What do you conclude from the results of this activity?
Answer:
Basic solutions are also good conductors of electricity due to released OH^– ions.

iii) What happens to an acid or a base in aqueous solution?
Answer:
Acids produce H⁺ ions and bases produce OH^– ions in aqueous solutions.

iv) Do acids produce ions only in aqueous solution?
Answer:
Yes.

Activity – 8

Question 10.
Do acids produce ions only in aqueous solution? Prove it. (AS3)
(OR)
Acids produce ions only in aqueous solution. Justify your answer with an activity.
Answer:
Procedure :

1. Take about 1.0 g of solid NaCl in a clean and dry test tube.
2. Add some concentrated sulphuric acid to the test tube. .

Observation :

1. A gas comes out of the delivery tube.
2. If we test the gas with dry and wet blue litmus paper, there is no change in colour.

Chemical equation : 2 NaCl_(s) + H₂SO_4(l) > 2 HCl + Na₂SO_4(s)

Conclusion :

1. We can conclude that dry HCl gas (hydrogen chloride) is not an acid.
2. Because we have noticed that there is no change in colour of dry litmus paper.
3. But HCl aqueous solution is an acid because wet blue litmus paper turned into red.

Activity – 9 Reaction of water with acids or bases

Question 11.
Write an activity to show that dissolving of an acid in water is an exothermic process (or) endothermic process. (AS3)
(OR)
What do you observe when water is mixed with acid or base?
Answer:
Experiment :

1. Take 10 ml water in a beaker.
2. Add a few drops of concentrated H₂SO₄ to it and swirl the beaker slowly.
3. Touch the base of the beaker.
4. The base is hot.
5. Do this experiment with other concentrated acids like HCl, HNO₃ Result: This is an exothermic process called dilution.

Activity -10 Strength of acid or base

Question 12.
Write an activity to know whether the acid is strong or weak. (AS3)
Answer:

1. Take dilute HCl in a beaker.
2. Close it with a cardboard and introduce two different colour electrical wires through the holes made on it.
3. Connect a bulb and make the connection as shown in the figure.
4. Do the same replacing dilute HCl with dilute CH₃COOH (acetic acid).

Observation :
The bulb glows brightly in HCl solution, while the bulb’s intensity is low in acetic acid solution.

Result:
More ions are present in HCl solution which is a strong acid than in CH₃COOH solution which is a weak acid.

Activity – 11

Question 13.
Test the pH value of solution given in table. Record your observations. What is the nature of each substance on the basis of your observations?
Answer:

Activity – 12

Question 14.
Write an activity to check the colour change in dilute HCl and antacid solution in addition of methyl orange. (AS3)
Answer:
Procedure :

1. Take dilute HCl in a beaker.
2. Add two to three drops of methyl orange indicator to it.
3. The solution colour turns to red.
4. Now take the same solution and mix antacid tablet powder.

Observation :
Check the colour change.

Result:
The colour of the solution turns to light yellow.

Chemical equation:
2 HCl + Mg(OH)₂ → MgCl₂ + 2H₂O

Activity – 13

Question 15.
How can we test the pH value of the soil? (AS3)
Answer:

1. Take about 2g of soil in a test tube.
2. Add 5 ml water to it.
3. Shake it well.
4. Filter the content.
5. Collect the filtrate in a test tube.
6. Add 2 drops of universal solution to it.
7. Observe the colour.
8. Compare the colour with strip colour on the bottle and find the pH value.
9. In this way we can test the pH of the soil.

Activity – 14

Question 16.
Write the formulae of the following salts and classify them as families based on radicals.
Potassium Sulphate, Sodium Sulphate, Calcium Sulphate, Magnesium Sulphate, Copper Sulphate, Sodium Chloride, Sodium Nitrate, Sodium Carbonate and Ammonium Chloride. (AS4)
Answer:

Sodium family : Na₂SO₄, NaCl, NaNO₃, Na₂CO₃, etc.
Family of chloride salts : NaCl, NH₄Cl, etc.
Family of sulphate salts : K₂SO₄, Na₂SO₄, CaSO₄, MgSO₄, CuSO₄, etc.
Family of carbonate salts : Na₂CO₃ MgCO₃ CaCO₃, etc.

Question 17.
Identify the acids and bases from which they are obtained. (AS4)
Answer:

Activity – 15 pH of Salts

Question 18.
Collect the salt samples like sodium chloride, aluminium chloride, copper sulphate, sodium acetate, ammonium chloride, sodium hydrogen carbonate and sodium carbonate. Dissolve them in distilled water. Check the action of these solutions with litmus papers. Find the pH using pH paper (universal indicator. Classify them into acidic, basic or neutral salts. Identify the acid and base used to form the above salts. Record your observations in table. (AS4)
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 7 Human Eye and Colourful World.

AP State Syllabus SSC 10th Class Physics Important Questions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World 1 Mark Important Questions and Answers

Question 1.
Define the power of lens. (AP June 2016)
Answer:
The reciprocal of focal length is called “power of lens”.

Question 2.
What physical quantity can be found in an experiment done with prism? (AP June 2017)
Answer:

1. Angle of deviation.
2. Refractive index of a prism.

Question 3.
What is the relation between Power and Focal length of the lens? (AP June 2018)
Answer:
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 4.
Draw the diagram of a lens which will be recommended by an eye doctor to a long sighted patient. (TS June 2015)
Answer:

The lens is convex lens.

Question 5.
What is the cause of Presbyopia? (TS March 2015)
Answer:
Presbyopia is vision defect when the ability of accommodation of the eye decreases with ageing.

Question 6.
Draw a ray diagram to show the angle of deviation when a ray of light passes through a glass prism. (TS March 2015)
Answer:

d = angle of deviation

Question 7.
Suggest reasons for the phenomenon associated with the following. The sky appearing blue. (TS March 2015)
Answer:
The reason for blue sky is due to the scattering of light by the molecules of N₂ and O₂ whose size is comparable to the wavelength of blue light.

Question 8.
+50 cm focal length bi-convex lens is recommended to correct the defect of vision of a mart. Find the power of the lens. (TS June 2016)
Answer:
f = +50 cm
Power (P) = \(\frac{100}{\mathrm{f}}\) D (in cm); P = \(\frac{100}{50}\) = 2 D
Power of the bi-convex lens is 2D.

Question 9.
What happens if the eye lens of a person cannot accommodate its focal length more than 2.4 cm? (TS March 2017)
Answer:
The person can able to see certain distance only he cannot see distance objects. For correction, he should use concave lens.

Question 10.
Write the reason for Sun appears red during the Sun-rise and Sun-set. (TS June 2018)
Answer:
Due to the high velocity (wave length) of red right, it reaches our eye without under go scattering. So, sun appears red during sunrise and sunset.

Question 11.
A person is unable to see distant objects. Show the defect of vision of the person with the help of ray diagram. (TS March 2018)
Answer:
1) His vision defect is myopia.
2) Ray diagram

Question 12.
Which molecules of atmosphere act as scattering centres are responsible for the blue sky? (TS June 2019)
Answer:
Oxygen and Nitrogen molecules.

Question 13.
Write any one application of a prism. (AP SA-I; 2019-20)
Answer:

1. Prism is used in scopes like binoculars, telescopes and light houses.
2. Prism is used to create artificial rainbow.

Question 14.
Mention the function of retina in a human eye.
Answer:
It acts as a screen, (which the image is formed) for image formed.

Question 15.
State the role of ciliary muscles in accommodation.
Answer:
It can adjust the focal length of the eye lens.

Question 16.
Why is normal eye not able to see clearly the objects kept closer than 25 cm?
Answer:
The maximum accommodation of a normal human eye is reached when the object is at a distance of 25 cm from the eyes. The focal length of the eye lens cannot be decreased this minimum limit.

Question 17.
What is “Power of accommodation of the eye” (or) “Least distance of clear vision”?
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

Question 18.
What is “Iris” and “Pupil”?
Iris :
The muscular diaphragm between the aqueous humour and the lens is called ‘iris’. Iris is the coloured part that we see in an eye.

Pupil:
The small hole in iris is called ‘pupil’.

Question 19.
What are three common defects of vision?
Answer:
The common defects of vision are i) Myopia ii) Hypermetropia iii) Presbyopia.

Question 20.
What is “Far point”?
Answer:
The point of maximum distance at which the eye lens can form an image on retina is called ‘far point’.

Question 21.
What is power of lens?
Answer:
The reciprocal of focal length is called power of lens. Power of lens focal length = \(\frac{1}{\text { focal length }}\)

Question 22.
What is the function of pupil in human eye?
Answer:
It allows the light falling on iris.

Question 23.
What is the purpose of human eye?
Answer:
The purpose of eye is to see and perceive the objects around us.

Question 24.
What is the principle of the working of human eye?
Answer:
It acts as camera having a lens system forming an invented real image on the light sensitive screen, retina.

Question 25.
What is the nature of the image formed on the retina?
Answer:
Real, inverted and same sized.

Question 26.
What is meant by dispersion of light?
The splitting of white light into its component colours when it passes through the prism is called dispersion of light.

Question 27.
On which factors does the colour of the scattered white light depend?
Answer:

1. Angle of scattering
2. Distance travelled by light
3. Size of the molecules.

Question 28.
Why is normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The focal length of eye lens cannot decrease below 25 cm.

Question 29.
A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from?
Answer:
Myopia (or) short sightedness.

Question 30.
A person suffering from an eye-defect uses lenses of power – 1D. Name the defect he is suffering from and the nature of lens is used.
Answer:
Defect:
Myopia, Nature of lens, Concave/divergence.

Question 31.
Name the two phenomena involved in the formation of rainbow.
Answer:
Dispersion of light and internal reflection.

Question 32.
Identify the following part of the human eye.
1. Where is image of an object formed?
2. Which controls size of pupil?
Answer:

1. Image of an object is formed on retina.
2. Iris controls the size of pupil.

Question 33.
What is the relation between power of lens and focal length (f)?
Answer:
Power of lens (concave/convex)
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 34.
Explain about “Bending of light”.
Answer:

1. When a light travels from rarer to denser medium, it bends towards the normal.
2. And from denser to rarer medium, it moves away from the normal.

Question 35.
What is sclerotic?
Answer:
It is the outermost covering of the eye. It protects the vital internal parts of the eye.

Question 36.
What is “Retina”?
Answer:
Retina is the internal part and the light sensitive surface of the eye. It is equivalent to the photographic film in a camera.

Question 37.
What is “Atmospheric refraction”?
Answer:
When the light rays pass through the atmosphere having layers of different densities and refractive indices, refraction of light takes place. This refraction of light by the earth’s atmosphere is called atmospheric refraction.

Question 38.
What is a “Telescope”?
Answer:
The instrument which is used to see the distant objects such as a star, planet (moon, sun) or distant tree is called telescope.

Question 39.
Have you seen a rainbow in the sky after rain? How is it formed?
Answer:

1. A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on raindrops after the rain.
2. It is formed due to reflection, refraction and dispersion of sunlight by tiny water droplets present in the atmosphere.

Question 40.
“To look at the twinkling of stars is a wonderful experience.” How does it happen?
Answer:
The continuous changing atmosphere (due to varying atmospheric temperature and density) refracts the light from the stars by varying amounts and in different directions from one moment to the next.

Question 41.
Some things appear blue on a misty day. Give two examples.
Answer:

1. The long distance hills covered with thick growth of trees appear blue.
2. The smoke coming from a cigarette or an incense stick (agarbatti) appears blue on a misty day.

Question 42.
Which coloured suits do rescue workers wear?
Answer:
Rescue workers wear orange coloured suits during any rescue operations.

Question 43.
Which colour is best for school buses?
Answer:
Orange or yellow colour is best for school busses.

Question 44.
What is persistence of vision?
Answer:
The time for which the sensation of vision (of an object) continues in the eye is called persistence of vision. It is about 1/16^th part of a second.

Question 45.
Why can’t some people identify some colours?
Answer:
Rods identify the colours in the retina. If some rods are absent, the distinction of colours is not possible. In such cases, persons can’t identify some colours.

Question 46.
Write the reasons for colour blindness.
Answer:

1. Absence of colour responding rod cells in the retina.
2. Due to genetic disorder.

Question 47.
Why does it take some time to see objects in a dim room when we enter the room from bright sunlight outside?
Answer:
In bright light, the size of the pupil is small to control the amount of light entering the eye. When we enter a dim room, it takes some time so that the pupil expands and allows more light to enter and helps to see things clearly.

Question 48.
What is tyndall effect?
Answer:
The phenomenon of scattering of white light by colloidal particles is known as ‘Tyndall effect”.

Question 49.
Give two examples illustrating “Tyndall effect”.
Answer:

• A fine beam of sunlight entering a smoke filled room through a hole. Smoke particles scatter the white light and hence the path of light beam becomes visible.
• Sunlight passing through the trees in forest.
• Tiny water droplets through the trees in forest.

Question 50.
An eye camp was organised by the doctors in a village. What were the benefits to organise such camps in rural areas?
Answer:

1. To make people aware of eye diseases
2. To take proper and balanced diet.

Question 51.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
In the eye, the image distance (distance between eye lens and retina) is fixed and it cannot be changed. So when we increase the distance of an object, there is no change in the image distance.

Question 52.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is at a distance of 25 cm from the eye.
The focal length of the eye lens cannot be decreased below this. Thus an object placed closer than 25 cm cannot be seen clearly by a normal eye.

Question 53.
Write the relation between intensity of scattered light (I) and wavelength (λ).
Answer:
Light of short wavelength is scattered more than the light of long wavelength.
i.e., Intensity of scattered light (I) ∝ \(\frac{1}{\text { wavelength }(\lambda)}\)

Question 54.
Why would the sky look dark if the earth had no atmosphere?
Answer:
If the earth has no atmosphere, no particles present either. Thus no scattering of light. Then, the sky appears dark.

Question 55.
Why do different coloured rays deviate differently in the prism?
Answer:
Because the angle of refraction of different colours is different while passing through the glass prism.

Question 56.
What prevents rainbow from being seen as complete circles?
Answer:
The earth comes in the way of the rainbow and prevents it to form a complete circle.

Question 57.
When a monochromatic light passes through a prism will it show dispersion?
Answer:
No, it will not show any dispersion but show deviation.

Question 58.
Will a star appear to twinkle if seen from free space (say moon)?
Answer:
No, because there is no atmosphere in free space for refraction to take place.

Question 59.
A short-sighted person may read a book without spectacles. Comment.
Answer:
The statement is true, because a short-sighted person has difficulty in observing far off objects.

Question 60.
What is angle of vision?
Answer:
The maximum angle, at which we can see the whole object is called angle of vision.

Question 61.
What is cornea?
Answer:
The front curved portion of eye, which is covered by a transparent protective membrane is called the cornea.

Question 62.
Which lens do you use to correct the eye defect, Myopia?
Answer:
Bi-concave lens are used to correct the eye defect, Myopia.

Question 63.
What happens to power of lens if (i) focal length is increased, (ii) focal length is decreased?
Answer:

1. If focal length is increased, then power of lens decreases.
2. If focal length is decreased, then power of lens increases.

Question 64.
What type of image is formed by magnifying glass?
Answer:
It forms virtual, erect and magnified image.

Question 65.
How is power of lens related to its focal length?
Answer:
Power of lens is inversely proportional to its focal length.

Question 66.
What is “Scattering of light”?
Answer:
The process of re-emission of light in all directions with different intensity is called scattering of light. The re-emitted light is called scattered light.

Question 67.
Ramu is unable to see letters on the blackboard sitting at the last bench in the class¬room. What is the defect from which Ramu is suffering?
Answer:
Ramu is suffering from Myopia.

Question 68.
Vinay is able to read letters in a book beyond certain distance from least distance of distinct vision. What is the eye defect of Vinay?
Answer:
Vinay is suffering from Hypermetropia.

Question 69.
How can an eye lens accommodate its focal length?
Answer:
To see an object comfortably and distinctly, one must hold at a distance about 25 cm from his/her eyes. This distance is called least distance of distinct vision.

Question 70.
Write the uses of “rods” and “cones”.
Answer:
Retina contains about 125 million receptors called rods and cones. Rods identify the colour and cones identify the intensity of light.

Question 71.
Frame some questions, that you are going to ask your friend who is suffering from eye sight.
Answer:

1. Are you not able to see near objects or far objects?
2. Are you not able to see both near and far objects?

Question 72.
Write some more examples to find dispersion of light as VIBGYOR.
Answer:

1. Formation of rainbow.
2. When white light passes through water drop.

Question 73.
Why do stars twinkle?
Answer:
Due to change in atmospheric conditions, density changes so position keeps on changing.

Question 74.
“Sky appears dark to passengers flying at very high altitudes.” Why?
Answer:
At very high altitude, there is no atmosphere. So there is no scattering of light at such heights. So sky appears dark to passengers.

Question 75.
Why are danger signals red? (OR) Why are danger signals shown in red colour?
Answer:
Among the colours of visible light, red has more wavelength and least scattered. Thus, red colour can easily go through fog or mist or smoke without getting scattered. It can be seen from long distance. So red colour is used in universal danger signal.

Question 76.
How can you identify regarding the type of defect a person is suffering from by physically touching his spectacles?
Answer:
By touching the spectacles we can find out whether the lens is concave or convex lens and hence the defect from which he suffers.

Question 77.
A person cannot see objects beyond 1-2 m distinctly. What should be the nature of the defect and what type of lens should be used to correct the defect?
Answer:
The person is suffering from Myopia. It can be corrected by using concave lens.

Question 78.
How do you appreciate the working of “Retina”?
Answer:

1. It is the innermost delicate membrane having a large number of receptors called ‘rods’ and ‘cones’.
2. The rods identify the colour and the cones identify the intensity of light.
3. The retina is a part on which the image of an object is formed.

Question 79.
How do you appreciate the working of “Optic nerve”?
Answer:

1. Optic nerve consists of a large number of fibres.
2. These optic nerve fibres are connected to the rods and cones.
3. Optic nerve fibers transmit the light signals to the brain.

Question 80.
We see advertisements for eye donation on television or in newspaper. Write the importance of such advertisements.
Answer:
Eye donation advertisements are important as :

1. The people aw;are about donation of organs after their death.
2. Sympathetic nature towards others.

Question 81.
An eye donation camp is being organised by social workers in your locality. How and why would you help in this cause?
Answer:

1. We can intimate other people to participate in the camp.
2. As a human being, we should also register our eyes for donation after death.

Question 82.
A short sighted person cannot see clearly beyond 2m. Calculate the power of lens required to correct his vision.
Answer:

Question 83.
How does the power of a lens change if its focal length is doubled?
Answer:
The power get halved.

Question 84.
A person cannot see distinctly objects kept beyond 2m. Which among these lens is useful to correct the defect
a) – 0.2D
b) – 0.5 D
c) + 0.2D
d) +0.5D?
Answer:
The person is suffering from Myopia.
The lens is concave and its focal length f = – 2m.
\(P=\frac{1}{5}=\frac{1}{-2}=-0.5 D\)
So to correct the defect concave lens of – 0.5 D power should be used (b).

Question 85.
Express the power of concave lens of focal length 20 cm with its sign.
Answer:

10th Class Physics 7th Lesson Human Eye and Colourful World 2 Marks Important Questions and Answers

Question 1.
How do you appreciate the working of iris in the eye? (AP June 2018)
Answer:
Iris helps in controlling the amount of light entering the eye through pupil.

Question 2.
What is the reason for the blue colour of the sky? (AP March 2018)
How do you appreciate the role of molecules in the atmosphere in this regard?
Answer:
Blue colour of sky :

1. Atmosphere contains more O₂ and N₂ molacules and they are caused to blue colour of the sky.
2. The size of molecules of O₂ and N₂ are comparable with the blue colour and scatter blue colour only.

Role of molecules in the atomosphere in scattering :

1. Light of certain frequency falls on that atom or molecule.
2. This molecule responds to the light whenever the size of the molecule is comparable to the wavelength of light and vibrates.
3. Due to these vibration, molecule reemits a certain fraction of absorbed energy in all directions. The emitted light is called scattered light.
4. The atoms or molecules are called scattering centres.
5. I appreciate the role of the molecules in scattering.

Question 3.
In which conditions does a rainbow form? Why? (TS June 2015)
Answer:
1) Rainbow forms and appears when,
i) tiny water droplets present in the atmosphere (after rain shower),
ii) sunlight falls on the droplets,
iii) observer watches the rainbow in a specific direction.
2) Rainbow forms due to dispersion of sunlight by tiny droplets, present in the atmosphere, which acts as small prisms.

Question 4.
Draw the ray diagram, showing the correction of defect of vision hyper metropia by using a convex lens. (TS June 2016)
Answer:

Question 5.
Least distance of distinct vision of a person is observed as 35 cm. What lens is useful for him to see his surroundings clearly? Why? (TS March 2016)
Answer:
The least distance of distinct vision is 35 cm. This is more than the least distance of distinct vision of an ordinary person. Hence the person is suffering from Hypermetropia. He has to use double convex lens to see his surroundings clearly.

Question 6.
What happens, if Ciliary muscles do not perform contraction and expansion? Guess and write. (TS June 2018)
Answer:

1. If Ciliary muscles do not perform contraction and expansion, foal length of eye lens do not change.
2. Human eye can see the objects at specific distance only, eye cannot see the object either nearer or far distance.

Question 7.
Write any two situations to observe dispersion of light in your daily life. (TS June 2018)
Answer:
We can observe the dispersion of light in the following situations in our daily life.

1. In the formation of rainbow.
2. When observing sun light through the triangular transparent material like prism, scale edge.
3. At the time of curing of walls of new houses with water.
4. Dispersion of light by inclained plane mirror which is in water.

Question 8.
Write the material that you use to find out the value of refractive index of a prism. What is the necessity of the graph in this experiment? (AP March 2019)
Answer:
The material used to find out the value of refractive index of a prism:
Prism, Piece of white chart, Pencil, Pins, Scale and Protractor.

Necessity of the graph :
To find the angle of minimum deviation graph is required.

Question 9.
Draw a ray diagram showing the correction of myopia eye defect. (TS March 2019)
Answer:
Diagram of Myopia correction :
Note : Draw the diagram using Bi Concave Lens and show the far point (M). Image should form on Retina.

Question 10.
What happen if dispersion and scattering of light do not occur? (TS March 2019)
Answer:
If dispersion does not occur in nature, then there is no rainbow formation and splitting of light into seven different colors. If there is no scattering then the oceans and sky appears to be black. The sun appears white all the time (Including Sunrise and Sunset).

Question 11.
When Mohan viewed white light through a transparent scale, he observed some colours. Predict and write the phenomenon involved in his observation. (AP SCERT: 2019-20)
Answer:

1. The phenomenon involved in his observation is dispersion of light.
2. Splitting of white light into different colours (VIBGYOR) is called dispersion.

Question 12.
A boy who is suffering from eye defect has been given a prescription as -2D. Based on the information given, answer the following questions.
a) Identify the eye defect he is suffering.
b) Write the nature and focal length of the lens. (AP SCERT: 2019-20)
Answer:
a) The boy is suffering from myopia.
b) Nature of the lens :
The lens is biconcave lens.
It is thin at the middle and thicker at the edges.
Focal length of the lens :
Given that power is – 2D

Question 13.
How does eye lens change its focal length? (AP SA-I:2019-20)
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 14.
Kishore wore spectacles. When you saw through his spects the size of his eyes seemed bigger than their original size.
a) Which lens did he use?
b) Explain that defect of vision.
Answer:
a) When we saw through Kishore’s spects the size of his eyes seemed bigger than their original size. This is possible with convex lens only because magnification of the lens is greater than T.

b) The defect he suffers is hypermetropia. This is also called farsightedness.

A person who suffers with this type of defect, he can’t see the objects clearly which are placed near distance because the image is formed beyond the retina. So by using convex lens the rays can be converged on retina.

Question 15.
“God has given the gift for us to see the sunrise and sunset.” Explain the feeling of it.
Answer:

• The sun is visible two minutes before the actual sunrise and remains visible two minutes after the actual sunset.
• The actual sunrise takes place when the sun is just above the horizon.
• The actual sunset takes place when the sun is just below the horizon.

Question 16.
“Smoke coming out of coal fired chimney appears blue on a misty day.” Why?
Answer:

• On a misty day, the air has large amount of tiny particles of water droplets, dust and smoke.
• These tiny particles present in the air scatter blue colour of the white light passing through it.
• When this scattered blue light reaches our eyes the smoke appears blue.

Question 17.
“Motorists use orange light on a foggy day rather than normal white light.” Why?
Answer:

• On a foggy day, the air has large amount of water droplets.
• If a motorist uses white light, the water droplets present in the air scatter large amount of the blue light.
• This on reaching our eyes decreases visibility and hence driving becomes extremely difficult.
• Whereas orange light has longer wavelength and hence it is least scattered.

Question 18.
A rainbow viewed from an airplane may form a complete circle. Where will the shadow of the airplane appear? Explain.
Answer:

• A rainbow viewed from an airplane form a complete circle because the earth does not come along the way of the airplane and rainbow.
• A rainbow is a three dimensional cone of dispersed light it appear as a complete circle.
• The shadow of the airplane appears within the circle of the rainbow.

Question 19.
How do we see colours?
Answer:

• The retina of human eye has a large number of receptors.
• These receptors are of two types i.e., rods and cones.
• The rod cells recognise the colour of light rays, while the cones identify the intensity of light.
• It is these cone cells, which make it possible for a men to see different colours and distinguish between them.

Question 20.
Why do we use lenses in spectacles to correct defects of vision?
Answer:
The process of adjusting focal length is called “accommodation”. This process has to be done by eye itself. Sometimes the eye may gradually lose its power of accommodation. In such condition, the person cannot see the object clearly and comfortably. In this situation, we have to use lenses in spectacles to correct defects of vision.

Question 21.
What is a) far point of the eye and b) near point of the eye?
Answer:
a) The farthest point up to which the eye can see objects clearly without strain (in the eye) is called the far point of the eye. For a normal eye, the far point is at infinity.

b) The minimum distance at which objects can be seen most clearly without strain (in the eye) is called the least distance of distinct vision or the near point of the eye.

Question22.
Write the difference between “Myopia” and “Hypermetropia”.
(OR)
Distinguish between Myopia and Hypermetopia.
Answer:
The eye defect in which people cannot see at long distances but can see nearby objects clearly is called Myopia. The eye defect in which people cannot see near distant objects but can see distant objects is called hypermetropia.

Question 23.
Define the following words.
a) Prism
b) Dispersion of light
c) Scattering of light
Answer:
a) Prism :
A prism is a transparent medium separated from the surrounding medium by at least two plane surfaces which are inclined at a certain angle.

b) Dispersion of light :
The splitting of white light into different colours is called dispersion of light.

c) Scattering of light:
The process of reemission of absorbed light in all directions with different intensities by atoms or molecules is called scattering of light.

Question 24.
Define the words associated with prism with the help of figure.

1) Angle of incidence :
The angle between incident ray and normal is called angle of incidence.

2) Angle of emergence :
The angle between normal and emergent ray is called angle of emergence.

3) Normal:
Perpendicular drawn to the surface of prism.

4) Angle of deviation:
The angle between extended incident ray and emergent ray is called angle of deviation.

Question 25.
State the cause of dispersion, when white light enters a glass prism. Explain with a diagram.
Answer:

• Light is made up of different colours. Each colour travels at its own speed inside a prism.
• Due to this different colours of light bends through different angles with respect to the incident ray, as it passes through a prism.
• The red light bends the least while the violet most.
• Thus, the rays of each colour emerge along different paths and become distinct.
• It is the bond of distinct colours that we see in a spectrum.

Question 26.
What happens to the lens and the ciliary muscles when you are looking at distant objects and near objects?
Answer:
a) The ciliary muscles become relaxed and the lens becomes thin, i.e. its radius of curvature increases. So focal length of eye lens increases for distant object.

b) The ciliary muscles contract and the lens becomes thick, i.e. its radius of curvature decreases. So focal length of eye lens decreases for near objects.

Question 27.
Why does it take some time to see objects in cinema hall when we just enter the hall from bright sunlight? Explain in brief.
Answer:

• The pupil regulates and controls the amount of light entering eye.
• In bright sunlight, the size of pupil is small and when we enter the cinema hall it takes some time for the pupil to expand in size due to dim light.

Question 28.
What are the factors which influence the total angle of deviation?
Answer:

• The angle of incidence at the first surface (i).
• The angle of prism (A).
• Refractive index of the material.

Question 29.
How does eye change its focal length take place in the eyeball?
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 30.
Stars twinkle while planets do not. Why?
Answer:
1) Continuously changing atmosphere refracts light from the stars by different amounts from one moment to the other, when atmosphere refracts more starlight towards us and the stars appear to be bright and when the atmosphere refracts less star-light then the stars appear to be dim.

2) However the planets are nearer to us than the stars, they appear to be comparatively bigger to us so they cannot be considered as a point source, hence no twinkling is seen.

Question 31.
How do earth and stars appear for a person who is on the moon?
Answer:

• For the person who is on the moon, the earth appears blue due to blue colour of sunlight scattered by the earth’s atmosphere reaching him.
• Stars and other heavenly bodies are seen as usual, but without twinkling.

Question 32.
Why does the sky appear dark and black to an astronaut instead of blue?
(OR)
Why does the sky appear dark to the passenger flying at high altitudes?
Answer:

• This is because there is no atmosphere containing air in the outer space to scatter light.
• Since there is no scattered light which can reach our eyes in outer space, the sky looks dark and black there.
• This is why the astronauts who go to outer space find the sky to be dark and black instead of blue.

Question 33.
A person is able to see objects clearly only when these are lying at distance between 60 cm and 250 cm from his eye. What kind of defect of vision is he suffering from?
Answer:
For a normal eye, the near point is at 25 cm and the far point is at infinity. The given person cannot see object clearly either close to the eye or far away from the eye. So he is suffering from presbyopia.

Question 34.
Write the material required in finding the refractive index of a prism.
Answer:
Materials required:
Prism, piece of white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Question 35.
Draw the graph between angle of incidence and angle of deviation.
Answer:

Question 35.
Draw the diagram of scattering of sunlight.
Answer:

Question 35.
How do you appreciate the “Eye Donor”?
Answer:
The human eye is one of the most important sense organs. Without eye we are unable to see the beautiful world. So we have to appreciate “eye donor” for his kindness to give sight to blind people.

Question 36.
How can we appreciate the working of “Iris”?
Answer:

• Iris consists of muscles which help in controlling the amount of light entering the eye through pupil.
• In case of low light the iris makes the pupil to expand and allow more light to enter the eye.
• In case of bright or excess light the iris makes the pupil to contract in order to decrease the amount of light entering the eye. The iris consists of muscles that expand and contract the pupil.

Question 37.
The power of lens is 2.0 D. Find its focal length and state what kind of lens it is.
Answer:
P = \(\frac{1}{f}\) (f in metres)
Given P = 2D
∴ f= \(\frac{1}{P}\) = \(\frac{1}{2}\) = 0. 5 m = 50 cm.
The focal length positive indicates it is a convex lens.

Question 38.
Two convex lenses of powers 1D and 2D are combined together to form a new lens. Then what is the resultant power and focal length of lens?
Answer:
P₁ = 1 D ; P₂ = 2D
Resultant power P = P₁ + P₂ = 1 + 2 = 3D
P = \(\frac{1}{f}\) (f in metres)
∴ f= \(\frac{1}{P}\) = \(\frac{1}{3}\) = 0.3333 m = 33.33 cm.

Question 40.
Figure shows the refraction of light through an equilateral prism. Incident at an angle of 30°. The ray suffers a deviation of 37°. What are the angles marked at A, e and f respectively?
Answer:

Since the prism is an equilateral prism,
A = 60°, also D = 37° and i = 30° (i.e., i₁), e = i₂ = ?
We know that i₁ + i₂ = A + D
30° + e = 60 + 37
e = 97 – 30 = 67°
Also A + f = 180°
f = 180°-60° = 120°

10th Class Physics 7th Lesson Human Eye and Colourful World 4 Marks Important Questions and Answers

Question 1.
Kavya can see distant objects clearly but cannot see objects at near distance. With what eye defect is she suffering? Draw the diagrams showing the defected eye and its correction. (AP June 2016)
Answer:
Kavya is suffering from hypermetropia.
The following diagram shows the defective eye and its correction.

Question 2.
Revathi is a front bench student. She is unable to draw the picture drawn on the blackboard. She got permission from the teacher and sat in the back row. What could be the defect that Revathi is suffering from? Draw the diagram, which shows the correction of the above defect? (AP Mareh 2017)
Answer:
Hypermetropia

Question 3.
An eye specialist suggested a + 2D lens to the person with defect in vision. Which kind of defect in vision does he have? Draw the diagrams to show the defect of vision and its correction with a suitable lens. (TS June 2017)
Answer:
Eye specialist suggested +2D lens, that is convex lens is used to correct Hypermetropia. So the person has Hypermetropia.
Deffect of Vision:

Correction :

Question 4.
How will you calculate the focal length of a biconvex lens that is used to correct the defect of Hypermetropia? Explain it mathematically. (TS March 2017)
Answer:
The person who has hypermetropia cannot see near objects. He can see the objects those are beyond near point (H). For correction of this eye defect the image of the object placed at “least distance of distinct vision (L)” should be at near point (H).
u = -25 cm; v = -d cm

here d > 25, so ‘f gets positive value.
Hence, convex lens should be used.

Question 5.
Mention the required material and chemicals for the experiment of “scattering of light.” Write the experiment procedure. (TS March 2018)
(OR)
Write the required apparatus and chemicals to show the scttering of light experimentally and write the experimental process.
(OR)
How can you demonstrate scatteing of light by an experiment?
Answer:
Aim :
To show the scattering of light.

Material required :
Beaker, sodium thiosulphate, sulphuric acid.

Procedure:

1. Take a solution of sodium-thio-sulphate (hypo) and sulphuric acid in a glass beaker.
2. Place the beaker in an open place where abundant sun light is available.
3. Watch the formation of grains of sulphur and observe the changes in the beaker.
4. We will notice that sulphur participates as the reaction is in progress. At the beginning, the grains of sulphur are smaller in size and as the reaction progress, their size increases due to precipitation.
5. Sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as their size increases.
6. The reason for this is scattering of light.
7. At the beginning, the size of grains is small and almost comparable to the wavelength of blue light. Hence they appear blue in colour.
8. As the size of grains increases, their size becomes comparable to wavelength of other colours.
9. As a result, they acts as scattering centres of all colours.

Question 6.
A boy has been playing games in mobile phone and is suffering from eye defect. The doctor prescribed him to use spectacles of power – 5D. What eye defect is he suffering from? (AP SA-I:2018-19)
Draw a neat diagram which shows the correction of above eye defect.
Answer:
Doctor suggested the boy – 5D lens, that is concave lens. Concave lens is used to correct myopia. So the boy has myopia.
Defect of Vision :

Correction:

Question 7.

Answer the following questions from the above information. (TS June 2019)
i) What is the defect of vision in ‘D’ suffering from? Why does it happen?
ii) Whose defect of vision can be corrected by using Biconcave lens?
iii) Who is suffering with similar defect of vision as of ‘B’?
iv) Who among the above do not have any defect of vision?
Answer:
i) 1) The vision defect of person ‘D’ is presbyopia.
2) Presbyopia happens due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens. This effect can be seen in aged people.
ii) Person A’ defect of vision can be corrected by using Biconcave lens.
iii) Person ’C’ is suffering with similar defect of vision as of ’B’.
iv) Person ‘E’ has no defect of vision.

Question 8.
A prism causes dispersion of white light while a rectangular glass block does not. Explain.
Answer:

• In a prism the refraction of light occurs at two plane surfaces.
• The dispersion of white light occurs at the first surface of prism where its constituent colours are deviated through different angles.
• At the second surface, these split colours suffer only refraction and they get further separated.
• But in a rectangular glass block, the refraction of light takes place at the two parallel surfaces.
• At the first surface, although the white light splits into its constituent colours on refractions, but they split colours on suffering refraction at the second surface emerge out in the form of a parallel beam, which gives an impression of white light.

Question 9.
A convex lens of power 4D is placed at a distance of 40 cm from a wall. At what distance from the lens should a candle be placed so that its image is formed on the wall?
Answer:

So candle should be placed 66.66 cm from the lens.

Question 10.
Explain briefly the reason for the blue of the sky.
Answer:

• Our atmosphere contains different types of molecules and atoms.
• The reason for blue sky is the molecules N₂ and O₂.
• The sizes of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.

Question 11.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:

• For a normal eye, image distance in the eye is fixed.
• This is equal to the distance of retina from the eye lens.
• When we increase the distance of the object from the eye, focal length of eye lens is changed on account of power of accommodation of the eye so as to keep the image distance constant.

Question 12.
A person is able to see objects clearly only when they are lying at distance between 60 cm and 250 cm from his eye. What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain.
Answer:
A bi-focal lens consists of concave and convex lens of suitable focal lengths will be required to correct the defect and to increase his range of vision from 25 cm to infinity. In bi-focal lens, the upper half of the lens is concave lens which corrects distant vision and the lower half is convex which corrects near vision.

Question 13.
Discuss why sun is visible before actual sunrise and after actual sunset?
Answer:

• The sun is visible to us about 2 minutes before the actual sunrise and 2 minutes after the actual sunset because of atmospheric refraction.
• By actual sunrise we mean the actual crossing of the horizon by the sun.

• Figure shows the actual and apparent positions of the sun with respect to the horizon.
• The time difference between actual sunset and the apparent sunset is about 2 minutes.
• The apparent flattering of the sun’s disc at sunrise and sunset is also due to the same phenomenon.

Question 14.
When does the colour of sky appear black for an observer?
Answer:

• In the absence of atmosphere, there will not be any scattering of light and so light will reach our eye, i.e. the sky will appear black instead of blue at night in the absence of light.
• On the moon, since there is no atmosphere, there is no scattering of sunlight reaching the moon surface. Hence to an observer on the surface of moon, no light reaches except the light directly from sun. Thus sky will have no colour and will appear black to an observer on the moon surface. This is applicable for any planet which does not have atmosphere.
• When an astronaut goes above the atmosphere of the earth in rocket he sees the sky black.

Question 15.
Why does the colour of clouds appear black?
Answer:

• The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light.
• Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light does not reach our eye, the clouds seem black.

Question 16.
Give daily life examples of scattering of light by earth’s atmosphere.
Answer:
Some daily life effects of scattering of sunlight by earth’s atmosphere are

1. Red colour of sun at sunrise and sunset.
2. White colour of sky at noon.
3. Blue colour of sky is due to molecules N₂ and O₂.
4. Black colour of sky is due to the absence of atmosphere.
5. Use of red light for the danger signal because it is least scattered by particles due to its greater wavelength.
6. White colour of clouds is due to rise in temperature.

Question 17.
What is the relation between scattering and wavelength of light? Explain.
Answer:

• Scattering is the process of absorption and then re-emission of light energy.
• The air molecules of size smaller than the wavelength of incident light absorb the energy of incident light and then re-emit it without change in its wavelength.
• The intensity of scattered light is found to be inversely proportional to fourth power of wavelength of light.
• The wavelength of violet is least and red light is most, therefore from the incident white light, violet light is scattered the most and red light is scattered the least.

Question 18.
How can we get this (in human eye) same image distance for various positions of objects?
Answer:

• The ciliary muscle attached with eye lens helps to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• F_(max) – 2.5 cm,u = oo, v = 2.5 cm. The parallel rays coming from a distant object are focussed on the retina with 2.5 cm image distance.
• When the eye is focussed on a nearer object the muscles are strained. So the focal length of the eye lens decreases its minimum value.
• F_(min)= 2.27 cm, u = 25 cm, v = 2.5 cm. The rays from an object (u = 25 cm) at L (point of least distance of distinct vision) are focussed on the retina with 2.5 cm image distance.

Question 19.
Does eye lens form a real image or virtual image? Explain it.
Answer:

• Eye lens forms a real image.
• The light that enters the eye forms an image on the retina.

• The image is obtained on a screen (retina) and it is an inverted image.
• So, we can say eye lens forms a real image.

Question 20.
How does the image formed on retina help us to perceive the object without change its shape, size and colour?
Answer:

• The eye-lens forms a real and inverted image of an object on the retina.
• The retina is a delicate membrane which contains about 125 million receptors called ‘rods’ and ‘cones’ which receive the light signals.
• Rods identify the colour.
• Cones identify the intensity of light.
• These signals are transmitted to the brain through about 1 million optic nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 21.
How do you prove that a prism does not produce colours itself?
Answer:

• A white light from a slit ‘S’ is made to pass through prism P which forms spectrum on a white screen AB.
• A narrow slit H is made on the screen AB, parallel to slit S to allow the light of particular colour to pass through it.
• The light of a particular colour is made to fall on the second prism Q placed with its base in opposite direction to that of the prism P.
• The light after passing through the second prism Q is received on another white screen M.
• It is observed that the colour of light obtained on the screen M is same as that of the light incident on the second prism Q through the slit H.

Question 22.
Draw the structure of human eye and explain its parts.
Answer:

Question 23.
What is the part indicated by an arrow mark? What is its working function?
Answer:

• The part indicated by the arrow mark is ciliary muscle.
• Ciliary muscles help to change the focal length of the eye lens.
• When it relaxes the focal length of the eye lens is maximum.
• When it strains, the focal length of the eye lens is minimum.
• In this way the ciliary muscles to which eye lens is attached help to give us clear vision.

Question 24.
Two observers standing apart from each other do not see the “same” rainbow. Explain.
Answer:

• All the rain drops that disburse the light to form rainbow lie within a cone of semi vertical angle 40° to 42°.
• If two observers are standing at a distance apart, they will observe rainbow at different parts on the surface of the cone.
• So the portion of the rainbow observed by an observer depends on the position of the observer.
• Two different observers will form two different cones with the observer standing at the vertex of the cone, therefore rainbow seen by them will be different.

Question 25.
A prism with an angle A = 60° produces an angle of minimum deviation of 30°. Find the refractive index of material of the prism.
Answer:
Given, A = 60° and D = 30°

Question 26.
A person cannot see the objects distinctly, when placed at a distance less than 50 cm. Calculate the power and nature of the lens he should be using to see clearly the object placed at a distance of 25 cm from his eyes.
Answer:

Question 27.
A person cannot see the objects distinctly, when placed beyond 2 m.
Calculate the power and nature of the lens he should be using to see the distant objects clearly.
Answer:
For myopia the focal length = – far point distance = – 2m
Power = \(\frac{1}{\mathrm{f}}=\frac{1}{-2}\) = 0.5 D.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 6 Refraction of Light at Curved Surfaces.

AP State Syllabus SSC 10th Class Physics Important Questions 6th Lesson Refraction of Light at Curved Surfaces

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 1 Mark Important Questions and Answers

Question 1.
Suppose you are inside the water in a swimming pool. Your friend is standing on the edge. Do you find your friend taller or shorter than his actual height? Why? (AP June 2018)
Answer:
Friend is seemed to be taller. Because of refraction of light.

Question 2.
What happens to the image, if a convex lens is made up of two different transparent materials as shown in figure? (TS March 2016)

Answer:
The convex lens is made up of two different materials. So the refractive i these two materials will be different. Hence two images will be formed.

Question 3.
Write the list of materials required for the experiment to find the focal length of a convex lens. (TS June 2017)
Answer:
Convex Lens, Scale, Piece of paper, Sunrays.
(OR)
Convex Lens, V-Stand, Candle, Match box, Screen, Scale.

Question 4.
Complete the following ray diagram. (TS March 2019)

Answer:

The parallel rays coming with some angle to principal axis meet on focal plane.

Question 5.
If the object is placed between the focal point and the optical centre of a convex lens, what will be the characteristics of the image formed? (AP SA-1:2019-20)
Answer:
Object is placed between F₂ and optic centre P :

Nature :
Virtual, erect and magnified.

Position :
Same side of the lens where object is placed.

Question 6.
For a concave lens, what type of image will be formed if the object is placed at the centre of curvature? (AP SA-1:2019-20)
Answer:
Same size of object, inverted and real image will be formed.

Question 7.
Write lens formula. (AP SA-I:2019-20)
Answer:
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Question 8.
What is a lens? (or) Define lens.
Answer:
A lens is formed when a transparent material is bounded by two spherical surfaces.

Question 9.
What is a double convex lens?
The lens having two spherical surfaces bulging outwards is called double convex lens.

Question 10.
What about the thickness of double convex lens?
Answer:
It is thick at the middle as compared to edges.

Question 11.
What is a double concave lens?
Answer:
The lens having two spherical surfaces curved inward is called a double concave lens.

Question 12.
Write about the thickness of concave lens.
Answer:
It is thin at the middle and thicker at the edges.

Question 13.
What is centre of curvature?
Answer:
The centre of sphere which contains the part of curved surface is called centre of curvature.

Question 14.
What is radius of curvature?
Answer:
The distance between the centre of curvature and curved surface is called radius of curvature.

Question 15.
What is the mid point of lens called?
Answer:
The mid point of lens is called pole (or) optical centre.

Question 16.
What is a focus?
Answer:
The point where rays converge or the point from which rays seem emanate is called focal point (or) focus.

Question 17.
What is the distance between pole and focal point called?
Answer:
Focal length.

Question 18.
What happens if the ray passes through principal axis?
Answer:
It will be undeviated.

Question 19.
Where do light rays travelling to principal axis converge?
Answer:
They converge at focus.

Question 20.
What happens to light rays passing through focus?
Answer:
The path of the rays is parallel to principal axis after refraction.

Question 21.
What is a focal plane?
Answer:
A plane which is perpendicular to principal axis at the focus is called focal plane.

Question 22.
What is lens formula?
Answer:
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Question 23.
On what factor does focal length of a lens depend?
Answer:
It depends on refractive index of the medium, object distance and image distance.

Question 24.
What is lens maker formula?
Answer:
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)

Question 25.
What happens to be image formed by a convex lens if its lower part is blackened?
Answer:
Every part of a lens forms a complete image. If the lower part of the lens is blackened the complete image will be formed but its intensity will be decreased.

Question 26.
From which point of lens are all the distances are measured?
Answer:
The optical centre of lens.

Question 27.
Is it possible for a lens to Act as a convergent lens in are medium and a divergent lens in another?
Answer:
Yes. A convergent lens is placed in a higher refractive index of medium the nature of the lens changes i.e., it acts as divergent lens.

Question 28.
What are paraxial rays?
Answer:
The rays which move very close to the principal axis which can be treated as parallel are called paraxial rays.

Question 29.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in air to speed of light in any medium.

Question 30.
Give mathematic expression for power lens and explain the terms in the formula.
Answer:
Power (P) = \(\frac{1}{f}\)
where f is focal length of lens.

Question 31.
If the size of image is same as object through a convex lens, then where is the object placed?
Answer:
The object is placed at centre of curvature.

Question 32.
How will you identify a concave lens by touching it?
Answer:
A concave lens is thinner at centre and thicker at edges.

Question 33.
How will you identify a convex lens by touching it?
Answer:
A convex lens is thicker at centre and thinner at edges.

Question 34.
Give the sign conventions for lenses with regard to the object and image distance.
Answer:
The distance measured in the direction of incident ray is taken as positive.
The distance measured against the direction of incident ray is taken as negative.

Question 35.
Give the sign conventions for lenses with regard to the height of objects and images.
Answer:
All the heights of objects and images above principal axis are positive and below the axis are negative.

Question 36.
When light of two colours A and B passes through a plane boundary, A is bent more than B. Which colour travels more slowly in the second medium?
Answer:
Colour A travels slowly.

Question 37.
What type of lens behaviour will an air bubble inside water show?
Answer:
It will act as a concave lens.

Question 38.
Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?
Answer:
Yes. A lens is placed in a medium of a high refractive index than that of the lens then nature of lens changes (M_L > M_g).

Question 39.
The image formed by a lens is always erect and diminished. What is the nature of lens?
Answer:
Given that the lens is forming an image which is always erect and diminished. So it is virtual also. Such type of image is formed by concave lens.

Question 40.
If a student observed an image of same size with a convex lens of focal length 20 cm, then where should he keep the object in front of lens?
Answer:
Because the student got image of same size the object should be placed at a distance of twice the focal length, i.e. 40 cm.

Question 41.
For an object placed at a distance of 20 cm in front of convex lens, the image formed is at a distance of 20 cm behind the lens. Find the focal length of lens.
Answer:
The object distance and image distance are same. So the object is kept at twice the focal length. So the focal length of the convex lens is 10 cm.

Question 42.
A doctor suggested spectacles for a student which has negative focal length. Which type of lens is that?
Answer:
Focal length negative indicates that it is a concave lens.

Question 43.
What happens to a light ray which passes through optical centre?
Answer:
The light ray which passes through optical centre does not deviate.

Question 44.
When do you get image at infinity with a convex lens?
Answer:
When the object is at the focal point.

Question 45.
When do you get a virtual image with a convex lens?
Answer:
When the object is placed between focus and pole.

Question 46.
Is focal length of a lens zero? If not, why?
Answer:
No, focal length of lens never equals to zero because it is the distance between focal point and optical centre.

Question 47.
A thin lens has a focal length of 12 cm. Is it a convex leps or a concave lens?
Answer:
It is a convex lens, because f is positive.

Question 48.
Name the different apparatus where we are using the convex and concave lenses.
Answer:
The magnifying lense, telescope, microscope.

Question 49.
Draw the given diagram in your answer book and complete it for the path of ray of light beyond the lens.

Answer:

Question 50.
The diagram below shows two incident rays P and Q which emerge as parallel rays R and S respectively. The appropriate device used in the box is ……..

Answer:
The rays are diverging and they produced the parallel from the device after refraction. So the device is concave lens.

Question 51.
The following figure shows the incident and refracted rays pass through a lens kept in the box. Draw the lens and complete the path of rays.

Answer:
The incident rays 1 and 2 have converged after refraction. So the lens is convex.

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 2 Marks Important Questions and Answers

Question 1.
A convex lens is made of five different materials as shown in the figure. How many images does it form? Why? (AP March 2017)

Answer:
The given convex lens is made up of five different materials.
So they have different refractive indices / different focal lengths.
Hence they form five different images.

Question 2.
The focal length of a converging lens is 20cm. An object is 60cm from the lens. Where will be image be formed and write characteristics of the image. (AP March 2018)
Answer:
Focal length = f = 20 cm (+ 20cm)
Object distance = u = 60 cm (- 60cm)
Image distance = v = ?

Here, (-) indicates inverted images.
m = \(\frac{1}{2}\) < 1 indicates diminished image.
Image forms between F₁ and 2F₁.
Characteristics of the image :

1. real
2. inverted
3. diminished.

Question 3.
When a light rays enters a medium with refractive index n2 from a medium with refractive index n, at curved interface with radius of curvature R is given by
\(\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}\)
Now assume that the surface is plane and rewrite the formula with suitable changes.
Answer:
Assume that the interface is plane surface
Then R becomes infinity
R = ∞ (or) R = 1/0
Substitute the above value in the given equation

Question 4.
Two convex lenses of same focal length are fixed in a PVC pipe at a distance double to their focal length. What happens if a boy sees the moon with that arrangement? (TS March 2017)
Answer:

• The rays coming from moon are parallel. The first lens converges the rays at focus.
• The converging point is the focus of second lens. So the second lens convert the diverging rays into parallel.
• Hence, in the rays of moon, there will be no change when we see moon with this arrangement or without this arrangement.

(OR)

This arrangement does not make any difference in the rays coming from moon.
The moon appears same if we see directly or with this arrangement.

Question 5.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

• When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
• Lens do not diverge or converge the light.
• Light do not get refracted when it passes through that lens.

Question 6.
Complete the ray diagram given below (TS March 2018)

Answer:

Question 7.
The refractive index of convex lens material is 1.46. The refractive index of Benzene and water is 1.5 and 1.0 respectively. How does the lens behaves when it is kept in Benzene and water? Given and write. (TS March 2018)
Answer:

• When the convex lens with refractive index 1.46 is kept in Benzene with refractive index 1.5, then the lens acts as a diverging lens.
• If the same lens is kept in water whose refractive index is 1, then it acts as a converging lens.

Question 8.
Write the applications of lenses in day to day life. (AP SCERT: 2019-20)
Answer:
Uses of lenses in day to day life :

• Lenses are used for correcting eye defects.
• They are used as magnifying lenses.
• They are used in microscopes, telescopes, binoculars.
• They are used in cinema projectors and cameras.

Question 9.
Water lens is made of double convex lens of radius of curvature “R”. Write lens makers formula for water lens. (AP SA-I: 2019-20)
Answer:
1) Radius of curvatures of water lens are R₁ = R₂ = R and n = 1.5.
2) Sign conversion R₁ = + R₁, R₂ = -R₂.
3) Lens makers formula

Question 10.
Find the focal length of plane convex dens if its radius of curvature is R and its refractive index is n.
Answer:
Given lens is plano-convex lens; radius of curvature = R
Refractive index = n

Question 11.
In a classroom, four friends found out the focal length of a lens by conducting an experiment. The value came out to be 12.1cm, 12.2cm, 12.05 cm, 12.3 cm. The friends discussed the reasons for the differences or defects. Mention those reasons.
Answer:
Students got different focal lengths.

1. By observing the values they got all positive values. This indicates they are given by convex lens.
2. All the students got exact interger but different decimal value.

Reasons :
The difference in values is due to least count errors, parallax errors, random errors and systematic errors, etc.

Question 12.
How will you decide whether a given piece of glass is a convex lens, concave lens or a plane plate?
Answer:
Hold the given piece of glass over some printed matter.

1. If the letters appear magnified, then the given piece of glass is convex lens.
2. If the letters appear diminished, then the given piece of glass is concave lens.
3. If the letters appear to be same size, then it is a plane glass piece.

Question 13.
State the type of lens used as a magnifying glass. Draw a labelled ray diagram to show that the image of the object is magnified.
Answer:

A single convex lens is used as a magnifying glass, i.e. for seeing small object magnified. When the object to be seen in between the focus and optical centre of the lens, a vertical, erect magnified image of the object is formed as shown and convex lens is and to act as magnifying glass AB’ is the magnified image of AB.

Question 14.
Give conventions used in lenses.
(OR)
Write the signs of convex and concave lens using in drawing ray diagrams.
Answer:

• All distances are measured from optical centre of the lens.
• Distances measured along the direction of the incident light are taken as positive.
• The distances against the incident light are taken as negative.
• The heights measured vertically above the axis, are taken as positive.
• The heights measured vertically down the axis, are taken as negative.

Question 15.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Answer:
Yes, the statement is correct.
For magnified virtual image :
The object should be placed between optic centre (C) and focus (F) (< 20)

For magnified real image:
Placed between focus (F) and centre of curvature (2F) (20 – 40)

Question 16.
Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building ? What is the approximate focal length of this lens?
Answer:
As the image is real, therefore the lens use is convex lens. The distance of the real image formed by a convex lens from the lens decreases as the object distance from the lens increases. Hence, the screen has to be moved towards the lens to obtain the sharp image of the building. Approximate focal length of the lens =15 cm as the rays of light coming the window pane are considered to come from infinity. These rays of light are focussed by the convex lens at its focus, (i.e. on the screen).

Question 17.
What do you see when your friend brings a sheet of paper on which arrow was drawn behind the empty cylindrical shaped transparent vessel? Why do you see a diminished image?
Answer:
We will see a diminished image of the arrow.

When the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass and enters air, then it again undergoes refraction curved surface of the vessel and comes out into the air. In this way, light travels in two media, comes out of the vessel and forms a diminished image.

Question 18.
Using the formula of refraction at curved surfaces, write the formula for plane surfaces.
Answer:
For curved surfaces the formula for refraction is \(\frac{\mathrm{n}_{2}}{\mathrm{v}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)}{\mathrm{R}}\)

For plane surface, the radius of curvature (R) approaches infinity. Hence 1/R becomes zero.
\(\frac{\mathrm{n}_{2}}{\mathrm{v}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=0 \Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{v}}=\frac{\mathrm{n}_{1}}{\mathrm{u}}\)

Question 19.
Explain how a convex lens behaves on converging lens and diverging lens.
Answer:
The convex lens behaves as a converging lens, if it is kept in a medium with refractive index less than the refractive index of the lens. It behaves like a diverging lens when it is kept in transparent medium with greater refractive index than that of lens.
e.g. : Air bubble in water behaves like a diverging lens.

Question 20.
When does Snell’s law fail?
Answer:

• Snell’s law fails when light is incident normally on the surface of refracting medium.
• Both media have same refractive index.

Question 21.
If on applying sign convention for lens the image distance obtained is negative, state the significance of negative sign.
Answer:

• Negative sign of image distance means the image is virtual and erect.
• It is formed on the same side of the object with respect to lens.

Question 22.
Magnification of lens is found to be +2. What type of lens is that?
Answer:
Magnification +ve indicates the image is erect and virtual.
Magnification 2 indicates it is magnified.
Magnified virtual image is formed by only convex lens.

Question 23.
For same angle of incidence in media A, B and C the angle of refractions are 30°, 25° and 20° respectively. In which medium will the velocity of light be minimum?
Answer:
In medium R the velocity of light is minimum because it has greater refractive index. Refractive index and velocity of light in a medium are inversely proportional. So in medium R the velocity is minimum.

Question 24.
The radius of curvature (twice the focal length) of a convex lens is 40 cm. A student wants to get various images of following types (a) enlarged virtual image, (b) enlarged real image, (c) image of same size, (d) diminished image.
In order to get these images where should the object should be kept in front of convex lens?
Answer:
a) The object should be kept in less than 20 cm (i.e. less than focal length).
b) In order to get enlarged real image, object should be kept between 20 cm to 40 cm in front of lens.
c) In order to get image of same size object should be kept at a distance of 40 cm from the lens.
d) In order to get diminished image the object should be kept beyond 40 cm from the lens.

Question 25.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images.
Answer:

• Yes, the statement is correct.
• A convex lens of focal length 20 cm will produce a magnified virtual image if object is placed at a distance less than 20 cm from the lens.
• A convex lens of focal length 20 cm will produce magnified real image if the object is placed at a distance more than 20 cm and less than 40 cm.

Question 26.
A concave mirror and a convex lensare held in water. What changes, if any, do you expect in their focal length?
Answer:
The focal length of a concave mirror independent of the medium and A convex lens depends on medium when they are placed in water.
The focal length of the mirror – Does not change.
The focal length of the convex lens – Changes (means increases 4 times).

Question 27.
When does a convex lens behave like a diverging lens? Given example.
Answer:
A Convex lens behaves like a diverging lens when it is kept in a tranparent medium with greater refractive index than that of the lens.
Eg : An air bubble in water behaves like a diverging lens.

Question 28.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

• Suppose two rays are originated from the bottom of the pond.
• As these rays get refracted into air, they bend away from the normal.
• When these two refracted rays produced backwards they seem to meet at a point higher than the bottom of the pond.
• This point gives the apparent position of the bottom of the pond.
• Thus the pond appears to be shallower.

Question 29.
What happens to the image formed by a convex lens if its lower part is blackened?
Answer:

1. Every part of lens forms complete image.
2. If lower part of the lens is blackened, the complete image will be formed.
3. But its intensity will decrease.

Question 30.
Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?
Answer:

• Yes, the type of lens changes, if it is placed in medium having higher refractive index that of lens.
• For example, convex lens acts as converging lens when it is placed in a medium of lower refractive index otherwise it behaves like a diverging lens.

Question 31.
Draw the different types of convex and concave lens.
Answer:

Question 32.
Complete the ray diagram and give reason.

Answer:

The light ray which passes through optical centre does not undergo refraction. So it goes straight.

Question 33.
How do you appreciate the refraction at plane surfaces and at curved surfaces?
Answer:
The refraction of curved surfaces are used in various aspects such as

1. In microscope to enlarge microscopic objects.
2. In telescopes to see celestial objects.
3. To correct eye defects like myopia, hypermetropia and presbyopia.

So refraction at curved surfaces is thoroughly appreciated.

Question 34.
An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.
Answer:

Image distance is negative that indicates it is a virtual and erect image.

Question 35.
A bird is flying at the height 3m above the river surface while a fish is 4 m below the surface. At what depth would the fish appear to the bird ? At what height the bird would appear to the fish? (given a/w = 4/3)
Answer:
Given that refractive index of air / water = \(\frac{4}{3}\)
The height the bird would appear to fish = 4 + \(\frac{4}{3}\) × 3 = 4 + 4 = 8m

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 4 Marks Important Questions and Answers

Question 1.
Complete the ray diagram when an object is placed between F₂ and 2F₂. (AP June 2017)

Answer:

Question 2.
An object is placed at the following distances from a convex lens of focal length 10 cm.
(a) 8 cm.
(b) 15 cm.
(c) 20 cm.
(d) 25 cm.
Which position of the object will produce……. (TS March 2015)
(i) a diminished, real and inverted image?
(ii) a magnified, real and inverted image?
(iii) a magnified, virtual and erect image?
(iv) an image of same size as the object?
Justify your answer in each case.
Answer:
i) d (or) 25 cm
Reason : Object placed between centre of curvature and focal point.

ii) b (or) 15 cm
Reason : Object placed beyond centre of curvature.

iii) a (or) 8cm
Reason : Object placed between focal point and optic centre.

iv) c (or) 20 cm
Reason :Object placed at centre of curvature.

Question 3.
The ray diagrams showing the image formed by a convex lens are given in the following table. From these diagrams complete the table. (TS June 2016)

Answer:

Question 4.
Explain the behaviour of light rays in any four situations of their incidence on a convex lens. (TS March 2016)
Answer:
1) A ray passing along the principal axis is undeviated.

2) Any ray passing through optic centre is also undeviated.

3) The rays passing parallel to principal axis converge at focus.

4) The rays passing through the focus will take a path parallel to principal axis.

Question 5.
Draw the ray diagrams to find the images when an object is placed in front of the lens (i) at a distance of 8 cm, and (ii) at a distance of 10 cm on the principal axis of a convex lens whose focal length is 4 cm. Write the characteristics of images in both the cases. (TS June 2017)
Answer:
(i) Ray diagram :

Characteristics of Image :
i) Size of the image equal to the size of the object,
ii) Inverted image,
iii) Real image,
iv) Image formed at C.

(ii) Ray diagram

Characteristics of Image :
i) Image size is less than that of object size,
ii) Inverted image,
iii) Real image,
iv) Image is formed between F & C.

Question 6.
A double concave lens with the refractive index (n) = 1.5 is kept in the air. Its two spherical surfaces have radii R₁ = 20 cm and R₂ = 60 cm. Find the focal length of the lens. Write the characteristics of the lens. (TS March 2017)
Answer:

Hence f = – 30 cm (Here minus indicates that the lense is divergent)
Characteristics of the biconcave lens :

1. It is a diverging lens.
2. It is thin at the middle and thicker at the edges.

Question 7.
Draw ray diagrams for a double concave lens of focal length 4 cm, when objects are placed at 3 cm and 5 cm on principal axis. Write characteristics of images. (TS June 2018)
Answer:
i)

ii)

Characteristics of images :

1. Image formed between P and F
2. Diminished image
3. Errected image
4. Virtual image

Question 8.
Write the characteristics of the images which are formed when objects are placed at 50cm and 75cm on the principle axis of a convex lens with focal length of 25 cm. (TS March 2018)
Answer:
i) Object is placed at 50cm.

Characteristics of the image :

1. Image forms at 2Fp
2. Image is real,
3. Image is inverted,
4. Image is same size

ii) Object is placed at 75cm
f = +25cm; u = -75cm; v = ?

Characteristics of the image :

1. Image forms between F₁ and 2F₁
2. Image is real
3. Image is inverted
4. Image is diminished.

Question 9.
Write the role of lenses in our daily life. (AP March 2019)
Answer:
The role of lenses in our daily life :

1. Used for correcting eye defects.
2. Used as magnifying lens.
3. Used in Microscopes.
4. Used in Telescopes.
5. Used in Binoculars.
6. Used in Cinema Projectors.
7. Used in Cameras.

Question 10.
Draw the ray diagrams for the following positions of objects in front of a convex lens mention the characteristics of the image. (AP SCERT: 2019-20)
a) Object is placed beyond 2F₂.
b) Object is placed between focal point and opint center.
Answer:
a)

b)

a) Characteristics of the Image :

1. real
2. inverted
3. diminished.

b) Characteristics of the Image :
If we place an object between focus and optic centre, we will get an image which is virtual, erect and magnified.

Question 11.
The focal length of a convex lens is 2 cm. Draw the ray diagram of an object placed on principal axis at the ‘C’ of lens and at a distance of 3 cm from its optic centre.
Answer:
1) Object is placed on principal axis of a convex lens at ‘C’.

2) Object is placedon principal axis at a distance of 3 cm from its optic center.

Question 12.
Using biconvex lens, a point image is made on its principal axis S. Let us assume that we know optical centre P and its focus F. We also know PF > PS. Draw the ray diagram to identify the point source and give reasons.
Answer:

Given lens is biconvex lens and given condition is PF > PS’ means image is formed between optic centre (P) and Focus (F).

According to Snell’s law this condition is possible when the object is also placed between P and F. Because reflected rays are divergent.

Question 13.
Write about the behaviour of light rays when they incident on a lens.
Answer:
1) Situation I:
Ray passing through the principal axis.
⇒ It is not deviated.

2) Situation II:
Ray passing through the pole.
⇒ It is also undeviated.

3) Situation III:
Rays travelling parallel to the principal axis.
⇒ They converge at focus or diverge from the focus.

4) Situation IV :
Ray passing through focus.
⇒ It will take a path parallel to principal axis after refraction.

5) Situation V :
Parallel rays fall on a lens making some angle with principal axis.
⇒ They converge at a point or diverge from a point lying on a focal plane.

Question 14.
Write characteristics of image formed due to convex lens at various distances.
Answer:

Question 15.
Write characteristics of image formed by a concave lens at various distances.
Answer:

Question 16.
You are given a convex lens of focal length 10 cm. Where would you place an object to get a real inverted and highly enlarged image of the object? Draw a ray diagram.
Answer:
If an object is placed at the focus of the lens it forms real, inverted and highly enlarged image. Thus, the distance of the object from the optical centre of the lens is equal to the focal length of the lens =10 cm.
The ray diagram is as shown

Question 17.
Derive the formula of image formation in refraction at curved surfaces.
Answer:
1) Object at infinity :
The rays coming from the object at infinity are parallel to principal axis and converge to the focal point after refraction. So, a point-sized image is formed at the focal point.

2) Object placed beyond the centre of curvature on the principal axis :
When an object is placed beyond the centre of curvature 2F₂, a real, inverted and diminished image is formed on the principal axis between F₁ and 2F₁

3) Object placed at the centre of curvature :
When an object is placed at the centre of curvature (2F₂) on the principal axis, a real, inverted image is formed at 2F₁ which is same size as that of the object.

4) Object placed between the centre of curvature and focal point:
When an object is placed between centre of curvature (2F₂) and focus (F2), we will get an image which is real, inverted and magnified. This image will form beyond 2F₁.

5) Object located at focal point:
When an object is placed at focus (F₂), the image will be at infinity.

6) Object placed between focal point and optic centre :
If we place an object between focus and optic centre, we will get an image which in virtual, erect and magnified.

Question 18.
Distinguish between convex lens and concave lens.
Answer:

Question 19.
The ray diagram given below illustrates the experimental set up for the determination of the focal length of a converging lens using a plane mirror.

1) State the magnification of the image formed.
2) Write the characteristics of the ipiage formed.
3) What is the name given to the distance between the object and optical centre of the lens in the following diagram?
Answer:

1. The magnification of the image formed is unity (or 1).
2. The image is a) real and b) inverted is at the same position of the object.
3. The distance between the object and the optical centre of the lens is called object distance.

Question 20.
A concave lens made of a material of refractive index n₁ is kept in medium of refractive index n₂. A parallel beam of light incident on the lens. Complete the path of rays of light emerging from the concave lens if
i) n₁ > n₂
ii) n₁ = n₂
iii) n₁ < n₂.
Answer:
i) When n₁ > n₂, light goes from rarer to denser medium. Therefore, in passing through a concave lens it diverges.

ii) When n₁ = n₂, there is no change in medium. Therefore no bending or refraction occurs.

iii) When n₁ < n2, light goes from a denser to rarer medium. Therefore, in passing through a concave lens it converges.

Question 21.
One half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:

• Every part of a lens forms an image.
• For formation of image we require only two light rays to converge.
• Therefore, if the lower half of the lens is covered, it will still form a complete image.
• However the intensity of the image will be reduced.
• This can be verified experimentally by observing the image of distant object like tree on a screen, when lower half of the lens covered with a black paper.

Question 22.
Complete the following table if the object is placed at various positions in front of a convex lens.

Answer:

Question 23.
A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations.

a) From the above table, find the focal length of lens without using lens formula.
b) Which set of observations is incorrect and why?
c) In which case the size of the object and image will be same? Give reason for your answer.
Answer:
a) From the observations, it is clear that for u = 30, v = 30 cm. This means this value must be equal to twice the focal length of the convex lens.
∴ Focal length of convex lens = 30/2 = 15 cm

b) The observation (5) is not correct because if u = 15 cm i.e., the object is kept at focus so the image should be at infinity and not at 70 cm.

c) For twice the focal length we know size of object = size of image. So when object is kept at 30 cm the size of object and image are same.

Question 24.
Draw the ray diagrams when incident ray striking a convex surface or a concave surface moving from one medium to another medium.
Answer:

Question 25.
Draw ray diagrams of image formed by a convex lens at various distances.
Answer:
1) Object at infinity.

2) Object placed beyond the centre of curvature (2F₂).

3) Object placed at the centre of curvature.

4) Object placed between 2F₂ and F₂.

5) Object at the focal point.

6) Object placed between F and P

Question 26.
Write about the focal length of the lens with diagram.
Answer:

• A parallel beam of light incident on a lens converges to a point as shown in figure (a) or seems to emanate from a point on the principal axis as shown in figure (b).
• The point of convergence (or) the point from which rays seem to emanate is called focal point or focus (F).
• Every lens has two focal points.
• The distance between the focal point and optic centre is called the focal length of lens denoted by ‘f’
• To draw ray diagram for lenses, we need two more points in addition to focal points F₁ and F₂.
• These points are equidistant from centre of the lens and also equal to double the focal length. So we call them 2F₁ and 2F₂.
• For drawing ray diagrams related to lenses, we represent convex lens with a symbol £ and concave lens with J as shown in the figure c and d.

Question 27.
The diagram shows an object OA and its image IB formed by a lens. The image is same size as the object.
a) Complete the ray diagram and locate the focus of lens by labelling it as F.
b) State whether the lens is convex or concave.
Show it in the diagram.

Answer:

a)

1. Optical centre goes undeviated, therefore to find the optical centre, join A to B to meet the line 01 at the point P which gives the position of optical centre of the lens.
2. Draw a line CP through P perpendicular to the line 01 to represent lens.
3. Draw another ray AC from the point A parallel to the principal axis 01 to meet the lens line CP at a point C.
4. This ray AC will reach the image point B while passing through the focus, therefore join C to B to meet line 01 at a point F which is the focus of the lens. The completed ray diagram is shown above.

b) Since the image is real and inverted, the lens is convex.

c) Since the size of object and image of equal, the object must be at a distance of twice the focal length, i.e., at 2F₂.

Question 28.
The diagram shows an object AB placed on the principal axis of the lens L. The two foci of lens are F₁ and F₂. The image formed by the lens is erect, virtual, and diminished.

i) Draw the outline of the lens L used and name it.
ii) Draw a ray of light starting from B and passing through ‘O’. Show the same ray after refraction by the lens.
iii) Draw another ray from B which is incident and parallel to the principal axis. Show how it emerges after refraction from the lens.
iv) Locate the final image formed.
Answer:
i) Since the image formed by the lens is erect, virtual, and diminished, the lens is concave.
ii) The light ray BO incident at the optical centre ‘O’ of the lens, passes undeviated as OC after refraction by the lens.
iii) The light ray BP is incident and parallel to the principal axis. It emerges as PQ after refraction which appears to diverge from the second focus F₂ of the lens.
iv) The refracted rays OC and PQ do not actually meet, but they appear to diverge from a point B’ (i.e. they meet at a point B’ when they are produced backwards).
v) Thus, B’ is the complete image A’B’ is obtained by drawing perpendicular from B’ on the line F₂OF₁. The image is formed between the optical centre O and focus F₂ of the lens.

Question 29.
Figure below shows the refracted ray BC through a concave lens and its foci marked as F₁ and F₂. Complete the diagram by drawing the corresponding incident ray and also give reason.

Answer:

1. Figure shows the refracted ray parallel to the principal axis. Therefore, the incident ray must be travelling towards the focus F₂.
2. Thus, to find incident ray, F₂ is joined to the starting point B of the refracted ray and produced backward as BA.
3. Then AB is the required incident ray.
4. The completed diagram is shown below.

Question 30.
State the type of lens used as a simple magnifying glass. Draw a labelled diagram to show that the image of the object is magnified.
Answer:

• A single convex lens is used as a magnifying glass, i.e. for seeing small object magnified.
• When the object to be seen between the focus and optical centre of the lens, a virtual, erect and enlarged image is formed.
• So a convex lens acts as magnifying glass for object AB as shown in the figure.

Question 31.
Radii of biconvex lens are equal. Let us keep an object at one of the centres of curvature. Refractive index of lens is ‘n’. Assume lens is in the air. Let us take R as the radius of the curvature.
a) How much is the focal length of the lens?
b) What is the image distance ?
c) Discuss the nature of the image.
Answer:
Radii of curvatures (R) of biconvex lens are equal, so R₁ = R₂ = R


c) The nature of the image is inverted and v < u.

Question 32.
Refractive index of a lens is 1.5. When an object is placed at 30 cm, image is formed at 20 cm. Find its focal length. Which lens is it? If the radii of curvature are equal, then what is its value?
Answer:
Object distance = u = – 30 cm (Infornt of the lens)
Image distance = v = 20 cm

Question 33.
A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall?
Answer:

Question 34.
A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. If image distance is thrice the focal length, find object distance, image distance and nature of image.
Answer:
Given that
Focal length of lens = + 20 cm
Object distance = -u
Image distance (v) = + 3u

So the object is between F₂ and 2F₂. So the image beyond 2F₁ it is real, inverted, and magnified.

Question 35.
What are the various applications of lens?
Answer:

• The objective lens of a telescope, camera, slide projector, etc. is a convex lens which forms real and inverted image of object.
• Our eye lens is also a convex lens. The eye lens forms the inverted image of the object on the retina.
• The eye defects are corrected by lenses.
• A magnifying glass is nothing but a convex lens of short focal length fitted in a steel (or plastic) flame provided with a handle.
• In spectroscope, convex lenses are used for obtaining a pure spectrum.
• A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 4 Acids, Bases and Salts.

AP State Syllabus SSC 10th Class Chemistry Important Questions 4th Acids, Bases and Salts

10th Class Chemistry 4th Lesson Acids, Bases and Salts 1 Mark Important Questions and Answers

Question 1.
Take some water in a test tube and add concentrated H₂SO₄ to it. Shake the test tube well. If you touch the bottom of the test tube, you feel it as hot. Now, instead of H₂SO₄, if you add NaOH pellets to water in another test tube and touch the bottom, what do you observe? (TS June 2015)
Answer:
The bottom of test tube is also hot because reactions of acids, bases with water are exothermic reactions.

Question 2.
What happens if the copper sulphate crystals taken into dry test tube are heated? (TS June 2016)
Answer:

• When copper sulphate crystals are heated, water present in crystals is evaporated and the salt turns white.
• Evaporated water appears as droplets on the walls of the test tube.
• Blue coloured copper sulphate (CuSO₄ 5H₂O) is turned into white colour because 5H20 molecules are evaporated from crystals.

Question 3.
Why does the soil of agricultural lands get tested for pH?
Answer:
Plants require a specific pH range for their healthy growth. So, finding pH of a soil suggested the farmers to treat the fields with acidic or basic substances to maintain the required pH range.

Question 4.
Write the molecular formulae of common salt and baking soda which are widely used at home. (TS June 2017)
Answer:
Common Salt: NaCl; Backing soda : NaHCO₃

Question 5.
Mention the precautions to take while conducting an experiment to prove acids produce ions only in aqueous solutions. (TS June 2018)
Answer:

• Testing of the evolved gas by using dry litmus paper first. Then with wet litmus paper.
• Use gaurd tube containing calcium chloride.

Question 6.
What are antacids?
Antacids are mild alkalies. These are used for getting relief from acidity and indigestion and sometimes even headache. When taken orally, it reacts with hydrochloric acid present in the stomach and reduces its strength by consuming some of it.
Ex: Milk of magnesia.

Question 7.
Tap water conducts electricity whereas distilled water does not. Why?
Answer:
Tap water contains some impurities in the form of salts. Due to presence of salts, it conducts electricity. Distilled water is free from all kinds of salts and hence does not conduct electricity.

Question 8.
What do you mean by dilution of an acid or base? Why is it done?
Answer:
Dilution of an acid or base means mixing an acid or base with water. This is done to decrease the concentration of ions per unit volume. In this way the acid or the base is said to be diluted.

Question 9.
What is a universal indicator?
Answer:
An indicator which passes through a series of colour changes over a wide range of H₃O⁺ ions concentration is called universal indicator.

Question 10
What is tooth decay?
Answer:
Tooth enamel is chemically calcium phosphate Ca₃(PO₄)₂. It starts corroding when pH falls below 5.5. Food particles left in the mouth degrade to produce acid which lower the pH of the mouth. This is called tooth decay.

Question 11.
Define Alkalis and give some examples.
Answer:
Alkalis : An alkali is a base that dissolves in water.

Examples :
i) Sodium Hydroxide (NaOH),
ii) Potassium Hydroxide (KOH),
iii) Magnesium Hydroxide (Mg(OH)₂)

Question 12.
Why should we not taste or touch alkalis?
Answer:
We should not taste or touch alkali. Because they are corrosive.

Question 13.
Salts conduct electricity. Why?
Answer:
Salts contains ions. So they conduct electricity.

Question 14.
Why are calcium sulphate hemihydrates called Plaster of Paris?
Answer:
Calcium sulphate hemihydrates are used as plaster for supporting fractured bones in the right position. So, it is called Plaster of Paris.

Question 15.
Why does an aqueous solution of acid conduct electricity?
Answer:
An aqueous solution of acid liberates H⁺ ions. This makes the aqueous solution of acid to conduct electricity.

Question 16.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer:
When a solution of an acid is diluted, concentration of H₃O⁺ ions decreases.

Question 17.
What is pH?
A. pH is a scale for measuring hydrogen ion concentration in a solution. It is the negative logarithm of H⁺ concentration.
pH = – log [H⁺].

Question 18.
How is pH of a solution related to the [H₃O⁺] of that solution?
Answer:
The presence of H₃O⁺ ions indicate us whether it is a strong acid or weak acid.

Question 19.
There are two solutions of pH values 6 and 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:

• The solution whose pH value 6 is acid and has more hydrogen ion concentration.
• The solution of pH value 8 is basic and has less hydrogen ion concentration.

Question 20.
Can you give example for use of olfactory indicators in daily life?
Answer:
Examples of olfactory indicators : Onion, vanilla extract.

Question 21.
How do acids neutralize bases?
(OR)
How do acids and bases react with each other?
Answer:
According to Arrhenius theory acids produce H⁺ ions and bases produce OH^– ions in aqueous media.
The combination of H⁺ and OH^– ions is called ‘neutralization’.
Thus acids neutralize bases.

Question 22.
How strong are acids and base solutions?
Answer:
The acids of pH value as much less as possible have more concentration [pH < 7], Basic nature increases as pH value increases.

Question 23.
What do you say about salts of both weak acid and weak base?
Answer:
The pH of aqueous solutions of salt obtained from both weak acid and weak base is nearly 7.

Question 24.
Which base is used for removing permanent hardness of water?
Answer:
Sodium carbonate is used for removing permanent hardness of water.

Question 25.
Name two antacids used to get rid of our indigestion problem.
Answer:
Magnesium hydroxide and a mild base (baking soda).

Question 26.
Under what soil conditions would a farmer would treat the soil of his fields with quicklime (calcium hydroxide) or calcium carbonate?
Answer:
When the field has acidic nature, the farmer uses quicklime or calcium carbonate to neutralize it.

Question 27.
Write the formulas of Gypsum and Plaster of Paris.
Answer:
The formulae of Gypsum is CaSO₄ . 2H₂O and Plaster of Paris is CaSO₄ . ½H₂O.

Question 28.
Write any two Acid Base indicators.
Answer:

1. Methyl orange
2. Phenolphthalein.

Question 29.
Which salt is used in the manufacture of borax?
Answer:
Washing soda (Na₂CO₃.10H₂O)

Question 30.
What is family of salts? Give examples.
Answer:
Salt having the same positive or negative radical is called family of salts.
Eg : Family of sodium salts : NaCl, Na₂SO₄
Family of chloride salts : NaCl, KCl.

Question 31.
‘A’ is a substance which is acidic and it is added in solution to preserve pickles. What is A and what is the name given to its dilute solution?
Answer:
‘A’ is acetic acid and its dilute solution is called vinegar.

Question 32.
What are the chemical names of the following?
1) Baking soda
2) Gypsum
Answer:

• The chemical name of baking soda is sodium hydrogen carbonate (NaHCO₃).
• The chemical name of gypsum is calcium sulphate dihydrate (CaSO₄ • 2 H₂O).

Question 33.
Write the water of crystallisation of following compound.
a) Hydrous copper sulphate
b) Washing soda
c) Gypsum
d) Plaster of Paris
Answer:

Question 34.
Why don’t we use a strong base like NaOH as antacid?
Answer:
Strong bases like potassium hydroxide (KOH), sodium hydroxide (NaOH) are corrosive in nature. So they can harm the internal organs. Therefore we should not use them as antacid.

Question 35.
What do you mean by HsO⁺ ion?
Answer:
Hydrogen ions cannot exist as base ions. They associate with water molecules and exist as hydrated ions with each H⁺ attached by 4 to 6 water molecules. For this we represent H⁺ as hydronium ion, H³O⁺.
H⁺ + H₂O → H₃O⁺

Question 36.
Which indicator is useful at all pH? Why?
Answer:
Universal indicator is useful to test solutions of all pH because it gives different colours at different pH range.

Question 37.
Which substance is useful in removing permanent hardness of water?
Answer:
The permanent hardness of water is due to chloride and sulphate salts of magnesium and calcium, which can be removed by adding washing soda.

Question 38.
Given two examples for strongest bases.
Answer:
Sodium hydroxide – NaOH
Potassium hydroxide – KOH

Question 39.
What is the confirmation test for hydrous and anhydrous salt?
Answer:

• On heating hydrous salt in a test tube it will form water droplets on the sides of test tube.
• On heating anhydrous salt in a test tube it will not form water droplets on the sides of test tube.

Question 40.
P.O.P, cement, calcium chloride should be stored in moisture proof containers. Why?
Answer:

• P.O.P, cement and calcium chloride react with moisture (H₂O) in the atmosphere and set into hard solid masses.
• To avoid availability of moisture they should be stored in moisture proof containers.

Question 41.
Give some examples for hydrous and anhydrous salts.
Eg : For hydrous salts :

1. CuSO₄ . 5H₂O
2. Na₂CO₃ . 10H₂O
3. CaSO₄ . 2H₂O

Eg : for anhydrous salts :

1. NaCl
2. MgCl₂
3. Na₂CO₃

Question 42.
How are bitter and sour taste substances tested without testing?
Answer:

• Sour taste substances turn blue litmus to red.
• Bitter taste substances turn red litmus to blue. By these tests we can test them as acids and bases.

Question 43.
Do the metallic oxides react with acids?
Answer:
Yes, metallic oxides are basic in nature. They react with acids and form salt and water.

Question 44.
Does non-metallic oxide react with base?
Answer:
Yes, non-metallic oxide is acidic in nature. It reacts with base and forms salt and water.

Question 45.
Why does dry HCl gas not change the colour of the dry litmus paper?
(OR)
Prove that dry HCl gas is not an acid but HCl aqueous solution is an acid using an activity.
Answer:

• Dry hydrogen chloride gas is not an acid. Hence it can’t turn blue litmus into red.
• Hydrochloric acid is an aqueous solution. Hence it can turn blue litmus into red.

Question 46.
Do you know that the atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid? Do you think life can exist on this planet?
Answer:

1. No, it is not possible.
2. When pH value decreases, the survival of living organisms becomes difficult.
3. Hence there is not any possibility of life on Venus.

Question 47.
Why do acids not show acidic behaviour in the absence of water?
Answer:
Acids don’t show acidic behaviour in the absence of water as H⁺ ions are absent in them.

Question 48.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When a base like NaOH (Sodium hydroxide) is dissolved in water, it liberates (OH^–) ions.
Equation :

OH^– ion concentration increases.

Question 49.
How does the nature of the solution change with change in concentration of H⁺_(aq) ions?
Answer:
The concentration of H⁺ ions is responsible for the acidic nature of a substance.
If [H⁺] > 1 × 10^-7 mol/lit the solution is acidic.
If [H⁺] < 1.0 × 10^-7 mol/lit the solution is basic.

Question 50.
Do basic solutions also have H⁺_(aq) ions? If yes, then why are these basic?
Answer:
Yes. Basic solutions have OH^–_(aq) ions and bases have less number of H₃O⁺ ions. H⁺_(aq) ions are less in base. In basic solution [OH^–] > [H⁺].

Question 51.
What do acids have in common?
Answer:
(Acids have sour taste and conduct electricity. They release H₂ (Hydrogen) gas on reacting with metals). All acids have H⁺_(aq) ions.

Question 52.
What do bases have in common?
Answer:
Bases are slippery to touch and bitter to taste. All bases have OH^–_(aq) ions.

Question 53.
Why does pure acetic acid not turn blue litmus to red?
Answer:
Pure acetic acid is a weak acid so it does not have sufficient H⁺_(aq) ion to change the colour of blue litmus to red.

Question 54.
What will happen if the pH value in your body increases?
Answer:
It affects our digestion system.

Question 55.
A student checked pH of a salt solution and found that its pH is more than 7. How is that type of salt formed?
Answer:

• When a strong base reacts with weak acid then the solution is basic in nature. So its pH is more than 7.
• For example when acetic acid reacts with sodium hydroxide the salt formed has basic nature.

Question 56.
Explain the procedure that you follow to reduce water from a given salt.
Answer:
Procedure to reduce water from a given salt :

1. Take a boiling test tube.
2. Drop given salt in the test tube.
3. Heat the test tube gently.
4. Water from salt evaporates.
5. In this way we can reduce the water from salt.

Question 57.
Write the observations, when the hydrated salt or unhydrated salt is heated.
Answer:

• When hydrated salt is heated water droplets form inside the walls of test tube and sometimes blue or green colour salt turns into white colour.
• When unhydrated salt is heated it does not form water droplets inside the test tube walls and colour also does not change.

Question 58.
On heating the hydrated salt it loses water molecules present in it. To show this what are the equipment required?
Answer:
1) On heating the hydrated salt it loses water molecules present in it.
2) To show this the given equipment are required

1. Boiling tube
2. Test tube holder
3. Burner

Question 59.
Try to collect the information to reasons for calling calcium sulphate hemihydrates as Plaster of Paris (POP).
Answer:

• Gypsum plaster (or) Plaster of Paris (POP) is produced by heating gypsum to about 300°F.
• A large gypsum deposit is found at Montmartre in Paris (France).
• This gave the name Plaster of Paris to calcium sulphate hemihydrates.
• The term plaster can refer to gypsum.

Question 60.
Is the substance present in antacid tablet acidic or basis?
Answer:
The substance present antacid is weakly basic.

Question 61.
Give pH of neutral, acid and base.
Answer:

Question 62.
Which nature of Plaster of Paris makes its importance? Appreciate it.
Answer:
Plaster of Paris is a white powder. It is very soft and can be used to make toys, materials for decoration and to make surfaces smooth.

But on mixing with water, it changes to a hard solid mass (Gypsum). This is the important character of Plaster of Paris (POP).

Question 63.
What is acid rain? How does it affect our aquatic life?
Answer:
When the pH of rain water is less than 5.6 it is called acid rain. When acid rain flows into the rivers, it lowers the pH of the river water. Since our body works within a narrow pH range close to 7 the survival of aquatic life in river water mixed with rain water becomes difficult.

Question 64.
Why are pickles and sour substances not kept in brass and copper vessels?
Answer:
Pickles and sour substances contain acidic nature which may react with brass and copper vessels to produce toxic substances.
So, we don’t keep them in brass arid copper vessels.

Question 65.
Can you suggest some examples of use of pH in everyday life?
Answer:
Uses of pH in everyday life :

1. pH value helps us to identify acids, bases and neutrals.
2. If pH value is less in our mouth, it leads to tooth decay. We can find it as the reason for our tooth decay.
3. pH value helps us to know about acid rain.

Question 66.
Write any two uses of Bleaching powder.
Answer:

• It is used for disinfecting drinking water to make it free of germs.
• It is used as a reagent in the preparation of chloroform.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 2 Marks Important Questions and Answers

Question 1.
What value of pH in the mouth leads to tooth decay? Why? (TS June 2015)
Answer:

• Tooth decay starts when the pH of the mouth is lower than 5.5.
• Tooth enamel, made of calcium phosphate is the hardest substance in the body.
• It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5.
• Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth.

Preventions :

1. Clean the mouth after eating food.
2. Using tooth pastes, which are generally basic neutralize the excess acid and pre¬vent tooth decay.

Question 2.
Equal lengths of Magnesium ribbons are taken in two test-tubes X and Y. Hydro¬chloric acid is added to test-tube X and Acetic acid is added to test-tube Y. In which test-tube, the reaction will be more vigorous? Why? (TS March 2015)
Answer:
The speed of the reactions is higher in X test tube than Y test tube.

Reason :
Due to strong acidic nature, Hydrochloric acid reacts very fast with magnesium ribbon.

Question 3.
Name the four chemicals that are obtained from common salt and write their molecular formulae. (TS March 2015)
Answer:
Chemicals that can be obtained from common salt are

1. Sodium Hydroxide – NaOH
2. Baking soda / Cooking soda / Caustic soda / Sodium bicarbonate / Sodium Hydrogen carbonate. – NaHCOv
3. Washing soda / Sodium carbonate – Na₂CO₃ 10H₂O
4. Bleaching powder / Calcium Oxy Chloride – CaOCl₂

Question 4.
Observe the information given in the table and answer the questions given below the table. (TS March 2017)

i) Which one of them may be the neutral salt among A, B, C?
ii) What may happen when some drops of phenolphthalein is added to the substance B?
Answer:
i) C
ii) Pink Colour

Question 5.
Why do we use antacids? Write it’s nature. (TS March 2018)
Answer:
Pain and irritation will be caused in stomach during the acidity problem/indigestion problem. Antacids used to neutralize the excess acid in the stomach and gives relief from acidity Antacids are basic in nature.

Question 6.
Which product will form when CaO is dissolved in water? How do you find the nature of product? (TS March 2018)
Answer:
CaO reacts with water and gives calcium hydroxide [Ca(OH₂)]. The nature of the calcium hydroxide will be tested with red litmus paper or pH paper.

Calcium Hydroxide turns red litmus into blue. Thus we can say that ca(OH)₂ is basic in nature.

Ca(OH)₂ shows pH value more than 7. Thus we can say that Ca(OH)₂ is basic in nature.

Question 7.
How do you test the nature of the solution formed by dissolving CaO in water? What is the nature of the solution? (TS June 2019)
Answer:

• The solution formed by dissolving CaO in water is tested with red litmus paper, it turns into blue colour, (or) It is tested with methyl orange, it turns into yellow in colour.
• The solution of CaO and water is basic in nature.

Question 8.
Write the experimental procedure to test carbon dioxide gas. (AP SCERT: 2019-20)
Answer:

1. Pass the CO₂ gas through lime water [Ca(OH)₂].
2. The lime water appears as milky white.
3. The reaction is Ca(OH)₂ + CO₂ → CaCO₃ ↓ + H₂O.
4. The milky white is caused by CaCO₃.

Question 9.
Write two reactions of acids with carbonates and metal hydrogen carbonates. (AP SA-I : 2019-20)
Answer:
Reactions :

Question 10.
What do acids have in common?
Answer:
Common characteristics of acids :

1. Similar chemical properties.
2. Acids generate hydrogen gas on reacting with metals.
3. Hydrogen is common to acids.
4. Acids are sour in taste and turn blue to red when react with bases form salt and water.

Question 11.
What do bases have in common?
Answer:
Common characteristics of bases :

1. Bitter in taste.
2. Soapy in nature.
3. Turn red to blue colour.
4. On heating decompose into metal oxides and water.
5. React with acids to form salt and water.
6. Produce OH^– ions in aqueous solution.

Question 12.
How is bleaching powder produced?
Production of bleaching powder :
Bleaching powder is produced by the action of chlorine on dry slaked lime (Ca(OH)₂).
Equation :

Question 13.
What can you conclude about the ideal soil pH for the growth of plants in your region?
Answer:
1) Soil is considered the ‘skin of the earth’. The soil pH plays a vital role in the growth of a plant and it influences plant nutrition.
2) Soil pH strongly affects the nutrients required for the plant growth.
3) The nutrients may be stored on soil colloids, and live or dead organic matter, but may not be accessible to plants due to extremes of pH.

Conclusion :
For optimum plant growth, the generalized content of soil components by volume should be roughly 50% solids (45% mineral & 5% organic matter), and 50% voids of which half is occupied by water and half by gas.

Question 14.
How can you prepare turmeric indicator? What is the use of it?
Answer:
i) Turmeric indicator is prepared from turmeric.
ii) It has red colour in basic solution.

Question 15.
Name two salts and write their formulae which possess water of crystallization.
Answer:

1. Hydrous copper sulphate. Its formula is CuSO₄ . 5H₂O
2. Gypsum. Its formula is CaSO₄ . 2H₂O

Question 16.
What is neutralization reaction? Give two examples.
Answer:
When an acid reacts with base it forms salt and water. This reaction is called neutralisation reaction.
e.g.: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
H₂SO_4(aq) + 2Na0H_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

Question 17.
All alkalis are bases but all bases are not alkalis. Do you agree with the statement? If yes, why?
Answer:
Yes, I agree with the statement. Because alkalis are those bases which are soluble in water. So all alkalis are bases but all bases are not alkalis.

Question 18.
Why are solutions of acids, bases and salts good conductors of electricity?
Answer:
For passage of electricity through a material or substance charged particles are required. In metals charged particles are electrons whereas in solutions ions are charged particles which carry electrical energy. Solutions of acids, bases and salts undergo ionisation and produce ions. So they are good conductors of electricity.

Question 19.
What is strength of acid ? What are the factors that influence strength of acid?
Answer:
The extent which an acid undergoes ionisation is called strength of acid.
Factors influence strength of acid :

1. Degree of ionisation.
2. Concentration of hydronium ions produced by acid.

Question 20.
Why are organic acids weak acids when compared with mineral acids?
Answer:

• Strength of acid depends on extent of ionisation.
• Organic acids do not undergo 100% ionisation. Their ionisation is less than 30%. There is equilibrium between ionised and unionised molecules whereas mineral acids undergo complete ionisation.
• So mineral acids behave like strong acids when compared with organic acids.

Question 21.
“Acids do not contain OH^– ions”. Do you agree with this statement? If not, why?
Answer:
No. I do not agree with the statement because all acids also contain OH^– ions but in acid solutions, H⁺ ions are more than OH^– whereas in bases OH^– ions are more than H⁺.

Question 22.

If B is calcium chloride, what are A, C, D and E?
Answer:
A is calcium carbonate or calcium hydrogen carbonate.
C is water and D is carbon dioxide.
E is calcium carbonate and F is water.

Question 23.
Some salts are given below. Classify them into hydrous and anhydrous salts. Sodium carbonate, Sodium chloride, Sodium hydrogen carbonate, Copper sulphate, Hypo, Magnesium Sulphate (epsum salt)
Answer:
Hydrous salts :

1. Hypo (Na₂S₂O₃ • 2H₂O)
2. Epsum (MgSO₄ • 7H₂O)
3. Copper sulphate (CuSO₄ • 5H₂O)

Anhydrous salts :

1. Sodium chloride (NaCl)
2. Sodium carbonate (Na₂CO₃)
3. Sodium hydrogen Carbonate (NaHCO₃)

Question 24.
If someone in the family is suffering from a problem of acidity, which of the following would you suggest as a remedy : lemon juice, vinegar or baking soda solution? Which property do you think of while suggesting the remedy?
Answer:

• I suggest baking soda solution. As acidity can be neutralized by baking soda solution, we can use it.
• Neutralizing property of baking soda solution.

Question 25.
Why are curd and sour substances not kept in copper vessels?
Answer:
Curd and sour substances contain acids which react with copper vessels and form poisonous substances. So curd and sour substances should not be kept in copper vessels.

Question 26.
Which gas is liberated when acids react with metals? Give one example.
Answer:
When acids react with active metals they release hydrogen gas.
Zn_(s) + 2HCl_(aq) → Zncl_2(aq) + H₂ ↑_(g)

Question 27.
Why should pickles not be stored in metallic containers?
Answer:
Pickles contain acids which react with metallic containers and form poisonous substance. So they are kept in plastic containers.

Question 28.
Solution x turned blue litmus red and Solution y turned red litmus blue.
a) What products could be formed when x and y are mixed?
b) Which gas is released when we put magnesium pieces in solution x?
c) Will any chemical reaction take place when zinc pieces are put in solution y?
d) Which of the above solutions contain more hydrogen ions?
Answer:
Given solution Y turned blue litmus into red so, Y is an acid.
Given solution ‘y’ turned red litmus into blue so, ‘y’ is a base.
a) The reaction of an acid (x) with a base to give a salt and water.
b) When we put magnesium pieces in solution releases hydrogen gas.
c) When zinc pieces are put in solution y, a chemical reaction will take place there.
d) Acids contain more H+ ions in the given solutions, Y has more H+ ions because it is an acid.

Question 29.
Acid should be added to water but not water to the acid. Why?
Answer:

• The dissolving of an acid or base in water is an highly exothermic process. Care must be taken while mixing concentrated HNO₃ or concentrated H₂SO₄ with water.
• The acid must always be added slowly to water with constant stirring.
• If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause bums.
• The glass container may also break due to excessive local heating.

Question 30.
Explain the procedure to confirm the given salt is a hydrous or anhydrous.
Answer:

1. Take given salt in a test tube
2. Observe the colour of salt
3. Heat the test tube gently
4. Observe the colour of salt and also moisture (droplets) inside of the test tube walls.
5. If its colour changes or forms water droplets, it is hydrous salt.
6. Otherwise, it is anhydrous salt.

Question 31.
Categorize the following as acids, bases, and salts :
Lemon juice, salt water, soap water, tamarind juice, surf water, lime water.
Answer:
Acids :

1. Lemon juice
2. Tamarind juice

Bases :

1. Soap water
2. Surf water
3. Lime water

Salts :

1. Salt water

Question 32.
Classify the following salts as family of salts having same cation or anion and prepare a table.
Potassium sulphate, Sodium sulphate, Calcium sulphate, Magnesium sulphate, Copper sulphate, Sodium chloride, Sodium nitrate, Sodium carbonate and Ammonium chloride.
Answer:

Question 33.
Observe the table and answer the following questions.

1) Which of the solutions among these is a strongest base?
2) Which body fluid has slightly basic nature?
3) What is the nature of pure water?
4) Which body fluid is strongest acid?
Answer:

1. Sodium hydroxide because its pH is 13.
2. Blood because its pH is 7.3.
3. Pure water is neutral in nature because its pH is 7.
4. Gastric juice because its pH is 1.2.

Question 34.
The diagram given below shows the removal of water crystallisation. Find error in the diagram.

Answer:
Error in the diagram

1. Test tube is placed towards observer. It causes bums on his hands.
2. So, test tube should be placed away from the observer.

Question 35.
What are the uses of Plaster of Paris?
Answer:

• The substance which doctors use as plaster for supporting fractured bones in the right position.
• Plaster of Paris is used for making toys, materials for decoration and for making surfaces smooth.

Question 36.
What are the applications of pH in daily life?
Answer:
1) In medical science :
The pH values of urine and blood are taken for diagnosis of various diseases.

2) In dairies :
Milk has pH of 6.6. A change in the pH of milk indicates that milk has turned sour.

3) In agriculture :
For better growth of crops the pH of the soil is regularly tested.
For examples :

1. Citrus fruits require slightly alkaline soil.
2. Rice requires acidic medium.
3. Sugarcane requires neutral soil.

4) In technology:
Organic and biochemical reactions are carried out under control pH.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 4 Marks Important Questions and Answers

Question 1.
Draw a neat diagram showing a base solution in water conducts electricity. Why the solution of sugar/glucose in water do not conduct electricity? (AP March 2017)
Answer:

The solution of sugar/glucose in water do not conduct electricity because there is no H⁺ ions in the solution.

Question 2.
Explain an activity to show the water of crystallisation in CuSO₄ • 5H₂O. (AP June 2018)
Answer:

• Take a few crystals of copper sulphate in a dry test tube and heat the test tube.
• We observe water droplets on the walls of the test tube and salt turns white.
• Add 2 – 3 drops of water on the sample of copper sulphate obtained after heating.
• We observe the blue colour of copper sulphate crystals is restored.

Question 3.
Read the information given in the table and answer the following questions. (TS March 2016)

a) List out the acids in the above table.
Answer:
The acids are HCl and lemon juice.

b) What is the nature of the solution which gives pink colour with Phenolphthalene solution?
Answer:
The nature of the solution which turns pink colour with phenolphthalene solution is basic.

c) List out the neutral solutions in the above table.
Answer:
The neutral solutions are distilled water and NaCl.

d) Name the strongest acid and the strongest base among the given solutions.
Answer:
The strongest acid is HCl and the strongest base is NaOH.

Question 4.
Observe the following table and answer the questions given below. (TS June 2o17)
The table contains the aqueous solutions of different substances with the same concentrations and their respective pH values.

i) Which one of the above acid solutions is the weakest acid? Give a reason.
Answer:
Weakest acid is ‘C’. Because its pH value is less than 7 and it is nearer to 7.

ii) Which one of the above solutions is the strongest base? Give a reason.
Answer:
Strongest base is ‘D’. Because it’s pH value is near to 14.

iii) Which of the above two produce maximum heat when they react ? What does that heat energy called?
Answer:
B, D produce maximum heat when they react. This heat energy is known as neutralization energy.

iv) Which one of the above solutions has the pH equal to that of the distilled water? What is the name given to solutions of that pH value?
Answer:
‘G’ has the pH equal to that of the distilled water. These type of solutions are known as neutral solution.

Question 5.
List out the materials required to test whether the solutions of given acids and bases contain ions or not. Explain the procedure of the experiment. (TS March 2017)
Answer:
Required Materials :
Beaker, Bulb, Graphite rods, connecting wires, 230 V AC current, water, different acids, bases.

Experimental Procedure :

1. Connect the two connecting wires to the graphite rods.
2. Keep the graphite rods into the beaker, take care that two graphite rods do not touch each other.
3. Arrange a bulb in the circuit.
4. Pour dilute acid into the beaker.
5. Connect the ends of the connectors to 230 V AC.
6. In this way, change the acid / base and do the experiment.

The bulb glows in the experiment when the beaker contains acid or base. Hence, when the bulb glows we can say that acid or base contain ions.

Question 6.
List out the material for the experiment “when Hydrochloric acid reacts with NaHCO₃ and evolves CO₂“. Write the experimental procedure. (TS March 2018)
Answer:
Required material : Stand, test tubes, delivery tube, thistle funnel, two hole rubber corks, Ca(OH)₂, NaHCO₃, HCl.

Experimental procedure :

1. Take NaHCO₃ in a test tube and fix two hole cork to the test tube.
2. Fix thistle funnel in one hole of cork and insert delivery tube in the second hole of the cork. Insert the second end of the delivery tube in the other test tube which is containing Ca(OH)₂/lime water.
3. Pour dil. HCl into the test tube using thistle funnel.
4. Due to chemical reaction, gas is evolved and pass into the Ca(OH)₂ through delivery tube. It turns in to milky. We can conclude that it is CO₂ gas.

Question 7.
Prepare a table based on the colour responses of acid, base and salt with indicators such as indicators. (AP SA-I:2018-19)
Answer:

Question 8.
Draw universal pH value indicator and identify different substances. (AP SA-I : 2019-20)
(OR)
Draw a neat diagram showing variation of pH with the chage in concentration of H⁺_(aq) ion and OH^–_(aq) ions.
Answer:

Variation of pH with the change in concentration of H⁺_(aq) ions and OH^–_(aq) ions.

Question 9.

Answer the following questions by using above information. (TS June 2019)
1) Which of the above is neutral solution?
2) Which of the above is used to neutralize the acidity in stomach?
3) Which is the strong acid among the above solutions?
4) What is the colour of Phenolphthalein indicator in NaOH solution?
Answer:

1. Distilled water
2. Milk of Magnesia
3. Gastric juice
4. Pink

Question 10.
If the pH values of solutions X, Y and Z are 13, 6 and 2 respectively, then
a) Which solution is a strong acid? Why?
b) Which solution contains ions along with molecules of solution?
c) Which solution is a strong base? Why?
d) Does the pH value of a solution increase or decrease when a base is added to it? Why?
Answer:
The strength of an acid (or) an alkali can be tested by using pH value of a solution. If the value of a pH of a solution is less, then that solution exhibits acidic nature.

If the value of a pH of a solution is more, then that solution exhibits basic nature.
pH value of a solution “X” is 13
pH value of a solution “Y” is 6
pH value of a solution “Z” is 2

a) Solution ‘Z’ is strong acid because its pH is 2.
b) Among given solutions, solution X is weakest acid. Weak solution contains ions along with molecules of solution. So X exhibits like this character.
c) Solution X is strong base. Because its pH is 13.
d) If base is added to solution ‘Z’, then its pH will increase.

Question 11.
Distinguish between acids and bases.
Answer:

Question 12.
Explain chlor-alkali process.
Answer:
When electricity is passed through an aqueous solution of sodium chloride, it decomposes to form sodium hydroxide. The process is called chloralkali process because of the products formed chlor for chlorine and alkali for sodium hydroxide.
2 NaCl (aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

Chlorine gas is given off at the anode and hydrogen gas at the cathode and sodium hydroxide is formed near the cathode.

Question 13.
Define the following. Give one example for each.
a) Strong acid
b) Strong base
c) Weak acid
d) Weak base.
Answer:
a) Strong acid :
The acid which undergoes 100% ionisation is called strong acid.
e.g.: HCl, H₂SO₄

b) Strong base :
The base which undergoes 100% ionisation is called strong base.
e.g.: NaOH, KOH

c) Weak acid:
The acid which undergoes less than 100% ionisation is called weak acid.
e.g.: CH₃COOH, H₂CO₃

d) Weak base:
The base which undergoes less than 100% ionisation is called weak base.
e.g.: NH₄OH, Mg(OH)₂

Question 14.
Write any four chemical properties of acids.
Answer:
Chemical properties of acids :
1) Active metals react with acids and liberate hydrogen gas.
Zn + HCl → ZnCl₂ + H₂ ↑

2) Acids react with bases to form salt and water.
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

3) Acids react with metallic oxides to form salt and water.
MgO_(s)+ 2HCl_(aq) → MgCl_2(aq) + H₂O_(l)

4) Acids react with carbonates and hydrogen carbonates and release carbon dioxide gas.
CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq)+ H₂O_(l) + CO_2(g) ↑
Ca(HCO₃)_2(l) + 2HCl_(aq) → CaCl_2(aq) + 2H20((| + 2CO_2(g) ↑

Question 15.
Write the formulae of the following salts.
a) Sodium sulphate
b) Ammonium chloride
Identify the acids and bases for which the above salts are obtained. Also write chemical equations for the reactions between such acids and bases. Which type of chemical reactions are they?
Answer:
a) Formula of sodium sulphate is Na2S04. When sulphuric acid reacts with sodium hydroxide it forms sodium sulphate.
H₂SO_4(aq)+ 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

b) Formula of Ammonium chloride is NH4C/. When Ammonium hydroxide reacts with hydrochloric acid it forms Ammonium chloride.
NH₄OH_(aq) + HCl_(aq) → NH₄Cl_(aq) + H₂O_(l)

Question 16.
Write balanced equations to satisfy each statement,
a) Acid + Active metal → Salt + Hydrogen
Answer:
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

b) Acid + Base → Salt + Water
Answer:
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

c) Acid + Carbonate / Hydrogen carbonate → Salt + Water + Carbon dioxide
Answer:
CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + H₂O_(l) + CO_2(g) ↑
NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(g) ↑

d) Metal oxide + Acid → Salt + Water
Answer:
CaO_(s) + 2 HCl_(aq) → CaCl_2(aq) + H₍₂₎O_(l)

e) Non metal oxide + base → Salt + Water
Answer:
CO_2(g) + 2 NaOH_(aq) → Na₂CO_3(aq) + H₂O_(l)

Question 17.
Give important products obtained from chloralkali process.

Answer:
Result:
Five water molecules are present in one formula unit of copper sulphate. Water of crystallization proves that the crystals contain a fixed quantity of water in them.

Question 18.
Give the equations for the preparation of each of the following.
i) Copper sulphate from copper (II) oxide.
Answer:
CuO + H₂SO₄ → CuSO₄ + H₂O

ii) Potassium sulphate from potassium hydroxide solution.
Answer:
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

iii) Lead chloride from lead carbonate.
Answer:
PbCO₃ + 2 HCl → PbCl₂ + H₂O + CO₂

Question 19.
How are the following salts prepared?
1) Calcium sulphate from calcium carbonate
Answer:
When calcium carbonate is treated with sulphuric acid it forms calcium sulphate.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

2) Lead carbonate from lead nitrate
Answer:
When lead nitrate is treated with carbonic acid we will get lead carbonate.
Pb(NO₃)₂ + H2CO₃ → PbCO₃ + 2 HNO₃

3) Sodium nitrate from sodium hydroxide
Answer:
When sodium hydroxide is reacted with nitric acid it will form sodium nitrate.
NaOH + HNO₃ → NaNO₃ + H₂O

4) Magnesium carbonate from magnesium chloride
Answer:
When magnesium carbonate is reacted with hydrochloric acid it forms magnesium chloride.
MgCO₃ + 2 HCl → MgCl₂ + H₂O + CO₂

Question 20.
Which of the following reactions are considered as neutralization reactions? Why?
1) NaOH + HCl → NaCl + H₂O
2) CaO + 2 HCl → CaCl₂ + H₂O
3) CO₂ + 2 NaOH → Na₂CO₃ + H₂O
4) SiO₂ + CaO → CaSiO₃
Answer:
All of them are considered as neutralization reactions.

1. An acid (HCl) reacts with base (NaOH) and forms salt and water. So it is a neutralization reaction.
2. Here metallic oxide which is basic in nature reacts with acid and forms salt and water. So it is also a neutralization reaction.
3. In third case non-metallic oxide (acidic oxide) reacts with base (NaOH) and forms salt and water. So it is also a neutralization reaction.
4. In fourth case a metallic oxide (CaO) reacts with non-metallic oxide (SiO₂) and forms salt. So it is also a neutralization reaction in the absence 6f water.

Question 21.
Which metals produce hydrogen gas when they are reacted with bases like NaOH and KOH? Write the chemical equations for the reactions.
Answer:
Zinc, aluminium and lead react with bases like NaOH and KOH and produce hydrogen gas.

Question 22.
i) A solution has a pH of 7. How would you increase its pH and decrease its pH? Explain.
Answer;
We can increase the pH of a solution by adding base because we know that bases have pH > 7. We can decrease the pH of a solution by adding an acid because acidic solution have pH < 7.

ii) If a solution changes the colour of litmus from red to blue, then what can you say about its pH?
Answer:
Bases can change red litmus into blue. So the pH of the solution is greater than 7.

iii) What can you say about pH of a solution that liberates carbon dioxide from sodium carbonate?
Answer:
Acids react with carbonates and liberate hydrogen gas. So the pH of the solution is less than 7.

Question 23.
Write the pH values of some solutions.
Answer:

Question 24.
Fill the following table of results of reactions between some substances (acids, bases, neutral substances) and indicators.

Answer:

Question 25.
The pH values of six solutions A, B, C, D, E, F are given as 5,2,1,3,7 and 9 respectively. Which solution is
a) Neutral
b) Strongly alkaline
c) Strongly acidic
d) Weakly acidic?
Arrange the pH in increasing order of Hydrogen ion concentration.
Answer:
a) Solution E is neutral.
b) Solution F is Alkaline.
c) Solution C is strongly acidic.
d) Solution A is weakly acidic.
e) Solution B is strongly acidic.
f) Solution D is strongly acidic.
g) Ascending order of increase of Hydrogen ion concentration is F, E, A, D, B, C.

Question 26.
Collect information about various organic acids different occurring naturally and prepare a table.
Answer:

Question 27.
Complete the table.

Answer:

Question 28.
Fill the table.

Answer:

Question 29.
Draw a diagram to show the reaction of acids with metals.
Answer:

Reaction of Zinc granules with dil. HCl and testing hydrogen gas by a burning candle

Question 30.
Draw a diagram to show that all metal carbonates and react hydrogen carbonates
Answer:

Question 31.
What are the uses of Bleaching powder?
Answer:

• It is used for bleaching cotton and linen in the textile industry for bleaching wood pulp in paper industry and for bleaching washed clothes in laundry.
• Used as an oxidizing agent in many chemical industries.
• Used for disinfecting drinking water to make it free of germs.
• Used as a reagent in the preparation of chloroform.

Question 32.
What are the uses of Baking soda?
Answer:
1) Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place.

Carbon dioxide produced during the reaction causes bread or cake to rise making them soft and spongy.

2) Sodium hydrogen carbonate is also an ingredient in antacids. Being alkaline, it neutralizes excess acid in the stomach and provides relief.

3) It is also used in soda-acid, fire extinguishers.

4) It acts as mild antiseptic.

Question 33.
What are the uses of Washing soda?
Answer:

• Sodium carbonate (washing soda) is used in glass, soaps and paper industries.
• It is used in the manufacture of sodium compounds such as borax.
• Sodium carbonate can be used” as a cleaning agent for domestic purposes.
• It is used for removing permanent hardness of water.

Question 34.
Write the chemical formulae of the following :
i) Bleaching powder
ii) Sodium Chloride
iii) Slaked lime
iv) Baking Soda
v) Washing Soda
vi) Gypsum
vii) Plaster of Paris
viii) Acetic acid
ix) Sodium Hydroxide
x) Limestone
Answer:
i) Bleaching powder = CaOCl₂
ii) Sodium Chloride = NaCl (Common Salt)
iii) Slaked lime (or) lime water = Ca(OH)₂
iv) Baking Soda = NaHCO₃
v) Washing Soda = Na₂CO₃, 10H₂O
vi) Gypsum = CaSO₄ . 2H₂O
vii) Plaster of Paris = CaSO₄ . ½H₂O
viii) Acetic acid = CH₃COOH
ix) Sodium Hydroxide = NaOH
x) Limestone = CaCO₃

Question 35.
What are the various applications of neutralization?
Answer:

• The acidity of soil is reduced by adding slaked lime.
• The sting of yellow wasps contains alkalis. If acetic acid is rubbed on affected area, they are neutralized.
• Ants and bees have formic acid in their stings which can be neutralised by applying soap and some other alkali.
• Antacids tablets contain magnesium hydroxide, persons suffering from acidity are administered these tablets.
• The affect of nettle plant leaves is neutralized by leaves of dock plant.

Question 36.
Mention two situations where you use hydrated and unhydrated salts in your daily life.
Answer:

• NaCl is unhydrated salt. It flows freely when filled in a container.
• NaHCl (Baking soda) is unhydrated salt. It flows freely when filled in a container.
• NaCO₃ . 10H₂O (washing soda) is hydrated salt. It leaves wetness inside the container.
• CaSO₄ . 2H₂O (Gypsum) is also hydrated salt. On careful heating of gypsum it loses water molecules partially to become (CaSO₄ . ½H₂O) P.O.P. It is used in hospitals as plaster for supporting fractured bones in right position.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 12 Electromagnetism.

AP State Syllabus SSC 10th Class Physics Important Questions 12th Electromagnetism

10th Class Physics 12th Lesson Electromagnetism 1 Mark Important Questions and Answers

Question 1.
Draw the diagram showing the magnetic field lines of bar mannet. (TS June 2016)
Answer:

Question 2.
Correct the diagram according to Lenz law and draw it again. (TS March 2017)
Answer:

Question 3.
What is the use of slip – ring in AC motor? (TS June 2018)
Answer:
Uses of slip rings :
Slip rings are used to change the direction of current in the coil continuously.

Question 4.
Draw the direction of magnetic lines force, assuming that the current is flowing into the page. (AP SCERT: 2019-20 )
Answer:

Question 5.
What happens when a current carrying coil is placed in a uniform magnetic field? (TS June2019)
Answer:
The rectangular coil comes into rotation in clock – wise direction because of equal and opposite pair of forces acting on the two sides of the coil.

1. If the direction of the current in the coil unchanged it rotates half clock – wise and comes to half and rotates in anticlock – wise direction.
2. If the direction of the current in the coil changed after the first half rotation, the coil continuously rotates in a same direction.

Question 6.
Write the name of the device that converts mechanical energy into electrical energy. (AP June 2019)
Answer:
Generator.

Question 7.
Name some sources of direct current.
Answer:
Dry cell, lead-acid battery.

Question 8.
Which sources produce alternating current?
Answer:
A.C generator, thermal power station, hydroelectric stations.

Question 9.
What is the role of split ring in an electric motor?
Answer:
The split rings are used to change the direction of current flowing through the coil.

Question 10.
Write one method of inducing current in the coil.
ANswer:
By pushing or pulling a bar magnet into or away from the coil we can induce current. Name two safety measures commonly used in electric circuit, i) Fuse ii) Earthing

Question 11.
On what factors does the magnetic induction at the centre of the coil depend?
Answer:
It depends on current, number of turns and radius of the coil.

Question 12.
Which is more dangerous AC or DC?
Answer:
AC is more dangerous.

Question 13.
State two serious hazards of electricity.
Answer:

• If a person touches the live wire, he gets severe shock which may prove fatal.
• Short-circuiting can cause a spark which may lead to fire in a building.

Question 14.
Why is earthing of electrical appliances recommended?
Answer:
To protect the user from any accidental electrical shock caused due to leakage of current.

Question 15.
Why is a spark produced at the place of short circuit? Why is the spark in white colour?
The resistance of circuit decreases, and a sudden flow of large current heats up the live wire and vapourises the metal. This causes spark. The metal of wire becomes very hot and naturally emits white light.

Question 16.
What is electromagnetic induction?
Answer:
Mechanical energy can be converted into electrical energy by moving a magnet inside a coil.

Question 17.
What is Maxwell’s right hand screw rule?
Answer:
The direction of current is the direction in which the tip of the screw advances and direction of ration of the screw gives the direction of magnetic lines of force.

Question 18.
What type of energy transformation takes place in electric generator?
Answer:
Electrical energy from mechanical energy.

Question 19.
Where are the electromagnets used?
Answer:
In electric generators and televisions.

Question 20.
What is electromagnet?
Answer:
When current carrying conductor is wound over a magnetic material like soft iron it gets magnetized.

Question 21.
What are the different types of power stations?
Answer:
Electrical energy is produced in different power stations from mechanical energy of water, meat energy, and nuclear energy.

Question 22.
If the current in the coil is in anti-clockwise, then what would be the face of the coil?
Answer:
It behayes as north pole.

Question 23.
If the current in the ceil is in clockwise, then what would be the face of the coil?
Answer:
It behaves as south pole.

Question 24.
What is the frequency of the A.C. supplied in your house?
Answer:
It is approximately 50 Hz.

Question 25.
What type of current is generated in electric power station?
Answer:
Alternating current.

Question 26.
What is the shape of magnetic lines due to straight current carrying conductor?
Answer:
They are concentric circles.

Question 27.
What is a transformer?
Answer:
It is a device which increases or decreases the voltage.

Question 28.
State two ways by which speed or rotation of electric motor can be increased.
Answer:

1. By increasing strength of the current.
2. By increasing number of turns in the coil.

Question 29.
What happens if an iron piece is dropped between two poles of strong magnet?
Answer:
Eddy current is produced in it. These eddy currents oppose the motion of the piece of iron. So it falls as it is moving through a viscous liquid.

Question 30.
If a copper rod carries a direct current, then where will be the magnetic field in the conductor?
Answer:
It will be both inside and outside the rod.

Question 31.
In what form is the energy in a current carrying coil stored?
Answer:
It is stored in the form of magnetic field.

Question 32.
What is solenoid?
Answer:
A solenoid is a long wire wound in a close packed helix.

Question 33.
What is the pattern of field lines inside a solenoid around when current carrying solenoid?
Answer:
Parallel to each other.

Question 34.
List any two properties of magnetic field lines.
Answer:

• Inside the magnet they start from south pole and end at north pole whereas outside the magnet they start at north pole and end at south pole.
• Two magnetic lines of force never intersect each other.

Question 35.
Why does the picture appear distorted when the bar magnet is brought close to the screen of a television?
Answer:
Picture on a television screen is due to motion of the electrons reaching the screen. These electrons are affected by magnetic field of bar magnet.

Question 36.
What is meant by electromagnetic induction?
Answer:
Whenever there is continuous change of magnetic flux linked with a closed coil, current generated in the coil is called electromagnetic induction.

Question 37.
State the Lenz’s principle.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 38.
What is induced emf?
Answer:
Change in magnetic flux produces emf in the circuit called induced emf.

Question 39.
What do you mean by magnetic effect of current?
Answer:
Current carrying conductor produces a magnetic field around it. This is called magnetic effect of current.

Question 40.
What is the direction of magnetic field at the centre of the coil carrying current in (i) clockwise, (ii) anti-clockwise direction?
Answer:
i) Along the axis of coil inwards.
ii) Along the axis of coil outwards.

Question 41.
Why does a current carrying freely suspended solenoid rest along a particular direction?
Answer:
A current carrying solenoid behaves like a bar magnet.

Question 42.
What effect will there be on a magnetic compass when it is brought near a current carrying solenoid?
Answer:
The needle of the compass will rest in the direction of magnetic field due to solenoid at that point.

Question 43.
How is magnetic field due to solenoid carrying current affected, if a soft iron bar is introduced inside the solenoid?
Answer:
The magnetic field increases when iron bar is introduced inside the solenoid.

Question 44.
What happens to magnetic field if we reverse the current direction?
Answer:
The magnetic field also gets reversed.

Question 45.
How do magnetic field lines inside a current carrying solenoid appear?
Answer:
They form along the axis and parallel to each other.

Question 46.
In which of the following cases does the electromagnetic induction occur?
i) A current is started in a wire held near a loop of wire.
ii) The current is stopped in a wire held near a loop of wire.
iii) A magnet is moved through a loop of wire.
iv) A loop of wire is held near a magnet.
Answer:
In first three cases there is a change in magnetic flux. So electromagnetic induction occurs in first three cases.

Question 47.
Why must an induced current flow in such a direction so as to oppose the change producing it?
Answer:
So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Question 48.
What is the maximum force acting on the current (i) carrying conductor of length (l) in the presence of magnetic field (B)?
Answer:
F = Bil

Question 49.
A charged particle q is moving with a speed v perpendicular to the magnetic field of induction B?
Find the equation of time period of the particle.
Answer:
\(\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{Bq}}\)

Question 50.
Name two safety measures commonly used in electric circuit.
Answer:
a) Fuse
b) Earthing

Question 51.
On what factors the magnetic induction at the centre of coil depends?
Answer:
a) Number of turns
b) Radius of the coil.

Question 52.
Write the formula for magnetic flux passing through an Area A with an angle θ.
Answer:
Flux ΦV= BA cos θ

Question 53.
Write the Lenz’s law.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 54.
What is the difference between AC and DC generator?
Answer:
In AC generator, ends of coil are connected to two slip rings.
In DC generator ends of coil are connected to two half split rings.

Question 55.
What are the uses of electromagnet?
Answer:
It is used in electric bells, electric motors, telephone diaphragms, etc.

Question 56.
What is the principle of Electric motor?
Answer:
When a rectangular coil is placed in magnetic field and current is passed through it, two equal and opposite forces act on the coil which rotates it continuously.

Question 57.
What factors are influence the speed of rotation of the motor?
Answer:

1. Strength of current
2. Number of turns
3. Area of the coil
4. Strength of magnetic field

Question 58.
Which two physical quantities are interrelated in Oerstead experiment?
Answer:
Electricity and magnetism are interrelated.

Question 59.
Which property of proton can change while it moves freely in a magnetic field?
Answer:
When a proton (the charge) moves in a magnetic field, then magnetic force is acting on proton. So its momentum changes.

Question 60.
Which type of conductors are producing magnetic field?
Answer:

1. Long straight current carrying conductor.
2. Circular loop.
3. Solenoid.

Question 61.
How much force acting on a neutron particle is moving with velocity V in a mag¬netic field with induction B?
Answer:
Zero, because neutron is charge less particle.

Question 62.
What are the instruments used in A.C Generator?
Answer:
Rectangular coil, brushes, slip rings, and magnetic poles.

Question 63.
What are the ways to produce the induced current in a coil?
Answer:
When a magnet is moved towards or away from coil or there is a relative motion between coil and magnet a current is induced in the coil.

Question 64.
At the time of short circuit, what happens to the current?
Answer:
At the time of short circuit, the current in the circuit increases heavily becomes the resistance of the conductor becomes almost zero.

Question 65.
A wire with green insulation is usually the line wire of an electric supply. Is it true?
Answer:
It is false, the wire with green insulation is the earth wire, not the line wire.

Question 66.
Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
Yes. current will be induced in coil ‘B’, because flux linked the coil ‘B’ changes with respect to time.

Question 67.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Electric fuse and Miniature Circuit Breaker (MCB).

Question 68.
An attemating current has frequency of 50Hz. How many times does it change its direction in one second?
Answer:
Frequency of AC = 50 Hz ⇒ 50 cycles in one sec. So it reverses its direction 100 times in one second.

Question 69.
Under what orientation, the induced current produced in moving conductor in a magnetic field can be maximum?
Answer:
The current induced in a conductor is maximum when direction of motion of conductor is at right angle to the magnetic field.

Question 70.
How could you make the coil rotate continuously in motor?
Answer:
The direction of current through the coil is reversed every half rotation, the coil will rotate continuously and the same direction.

Question 71.
What is the formula for induced cmf when change the magnetic flux?
Answer:
Induced EMF = \(-\frac{\Delta \phi}{\Delta t}\)

Question 72.
The magnetic flux is varying with time. Which cases E.M.F is induced?

Answer:
In OA and BC cases, E.M.F is induced.

Question 73.
In above problem, how much EMF is induced in BC curve?
Answer:

Question 74.
What is the equation of motional E.M.F?
Answer:
Motional E.M.F = Blv;
where B = magnetic induction,
l = length of rod,
v = velocity of rod.

Question 75.
When a magnet and a coil are moving same direction with same speed. Then induced E.M.F in coil is zero. Why?
Answer:
E.M.F = \(\frac{-\Delta \phi}{\Delta \mathrm{t}}\), but both moving same direction, so change in flux linked with coil is zero i.e., ∆Φ = 0

Question 76.
What is shape of conductor is drawn when current is passing through conductor?
Answer:

Question 77.
Draw the symbols of North and South pole when depends on current.
Answer:

Question 78.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:

Question 79.
Draw the diagram of AC and DC current varying with time.
Answer:

Question 80.
Where are electric motors used?
Answer:
Electric fans, water-pumps, coolers.

Question 81.
Mention two uses of solenoid.
Answer:
It is used in electric bells, fans, and motors.

Question 82.
Mention applications of electromagnetic induction.
Answer:
It is used in devices which convert mechanical energy into electrical energy.

Question 83.
What are the advantages of an electromagnet over permanent magnet?
Answer:

1. An electromagnet can produce a strong magnetic field.
2. The strength of electromagnet can be changed.
3. The polarity of electromagnet can be changed.

10th Class Physics 12th Lesson Electromagnetism 2 Marks Important Questions and Answers

Question 1.
Anand appreciated the law behind the making of ‘generator’. Name the law and state it. (AP June 2017)
Answer:
1) The law behind the making of ‘generator’ is Faraday’s law.

2) Faraday’s Law :
“Whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Question 2.
Explain Oersted experiment to show that Electricity and Magnetism were related phenomena. (AP June 2018)
Answer:

• Place a compass needle underneath a wire and then turn on electric current.
• Immediately the needle of compass shows the deflection. By this we can conclude that electricity and magnetism are related phenomena.

Question 3.
With the help of the given figure, the teacher explained that magnetic field lines are closed lines and not open lines. Write the questions which you will ask to rest whether the given statement is right or wrong. (TS June 2015)

Answer:

• Are there any magnetic field lines inside the magnet?
• If magnetic field lines are there inside the magnet, what is the direction of field lines inside the magnet?
• What is the direction of field lines outside the magnet?

Question 4.
State Right-hand rule with a labelled diagram. (TS March 2015)
Answer:

(OR)

• Right hand rule indicates the direction of magnetic force acting on a moving charge.
• It is used when velocity and field are perpendicular to each other. If the fore finger points towards the direction of velocity of charge or current (I), middle finger points to the direction of field (B), then thumb gives the direction of force (F).

Question 5.
A coil of insulated Copper wire is connected to a Galvanometer. (TS March 2015)
What happens, if a bar magnet is ………….
1) pushed into the coil?
2) withdrawn from inside the coil?
3) held stationary inside the coil?
Answer:

1. If a bar magnet is pushed into the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
2. If a bar magnet is withdrawn from inside the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
3. If a bar magnet is held stationary inside the coil, then the needle in Galvanometer does not get deflected. Because current is not generated in the coil.

Question 6.
Compare the magnetic field lines of force formed around a current carrying solenoid with the magnetic field lines of force of a bar magnet.
Answer:

Magnetic field lines of a bar magnet	Magnetic field lines of a solenoid
2) The direction of the field lines of the outside the magnet is from north to south pole.	2) The direction of the field lines of the outside the solenoid is from north to south pole.
3) The direction of the field lines of the inside of the magnet field looks like south to north pole.	3) The direction of the field lines of the inside of the solenoid is from south to north pole.
4) These lines are closed loops.	4) These lines are also closed loops.
5) We cannot find the field lines inside the bar magnet.	5) We can find the field lines inside the solenoid.
6) These field lines are same as field lines formed by a solenoid.	6) These field lines are also same as field lines formed by a bar magnet.
7) More field lines are found at poles.	7) More field lines are found at poles.

Question 7.
Which energy we get from an electric motor? Write two daily life applications of the electric motor. (TS June 2017)
Answer:

• We get mechanical energy from electric motor.
• In our daily life we use motor in
  i) Mixies
  ii) Grinders
  iii) Water Pumps
  iv) Fans / Coolers, etc.

Question 8.
List out the material required for Oersted experiment and mention the precautions to be taken in the experiment. (TS March 2019)
Answer:
Materials required for Oersted experiment are :

1. Thermocol sheet.
2. Wooden sticks.
3. Copper wire of 24 gauge.
4. Battery.
5. Key.
6. Magnetic compass.

Precautions to be taken are :

1. Copper Wire is made through the slits of the wooden sticks tightly.
2. Arrange/complete the circuit correctly.

Question 9.
What happens when magnetic flux passing through a coil changes continuously? Where does this process is used? (TS June 2019)
Answer:
1) When there is a continuous change of magnetic flux passing through a coil a current is generated in the coil.

2) This process is used in

1. Induction stoves
2. Security checking entrance/exit doors.
3. ATM cards scanners.

Question 10.
Distinguish between AC and DC.
Answer:

Question 11.
Distinguish between AC motor and DC motor.
Answer:

Question 12.
Distinguish between AC generator and DC generator.
Answer:

Question 13.
What is Faraday’s law of electromagnetic induction? Write its expression.
(OR)
State the Faraday’s law of electromagnetic induction. Write the equation of this law.
Answer:
The induced EMF generated in a closed loop is equal to the rate of change of magnetic flux passing through it.
Induced EMF = Change in flux / time
ε = ∆Φ / ∆t

Let Φ₀ be flux linked with single turn. If there are N turns of the coil, the flux linked with the coil is NΦ₀.
∴ ε = NAΦ₀ / ∆t

Question 14.
State the right-hand thumb rule. How does the rule help us?
Answer:
When you curl your right hand fingers in the direction of current thumb gives the direction of magnetic field.

It is useful to find the magnetic field direction as well as current direction.

Question 15.
A flat rectangular coil is rotated between the pole pieces of a horseshoe magnet. In which position of coil with respect to magnetic field, will the emf (i) be maximum, (ii) be zero, (iii) change its direction.
Answer:
i) The emf is maximum when the plane of coil is parallel to the magnetic field.
ii) The emf is zero when the plane of coil is normal to the magnetic field.
iii) The emf will change its direction when the plane of coil passes from the position normal to the magnetic field.

Question 16.
State two factors on which the magnitude of induced emf depend?
Answer:
The magnitude of induced emf depends on the following two factors.

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes i.e., the rate of change of magnetic flux.

More the change in magnetic flux, more is the emf induced. Further if more rapid the magnetic flux changes, more is the emf induced.

Question 17.
How do you increase the magnetic field of solenoid?
Answer:
The magnetic field of solenoid can be increased by the following two ways.

1. by increasing the number of turns of winding in the solenoid.
2. by increasing the current through the solenoid.

Question 18.
State the function of split ring in a DC motor.
Answer:

• The split ring acts as a commutator in a DC motor.
• With the split ring, the direction of current through the coil is reversed after every half rotation of coil.
• Thus the direction of couple rotating the coil remains unchanged and the coil continues to rotate in the same direction.

Question 19.
A DC motor is rotating in a clockwise direction. How can the direction of rotation be reversed?
Answer:
The direction of rotation of motor can be reversed by interchanging the connections at the terminals of the battery joined to the brushes of the motor.

Question 20.
State the effect of inserting a soft iron core within the coil of DC motor.
Answer:

• On inserting a soft iron core within the coil of a DC motor, the speed of rotation of coil increases.
• The reason is that the strength of magnetic field between the pole pieces of magnet increases due to which the deflecting couple on coil increases.

Question 21.
State condition when magnitude of force on a current-carrying conductor placed in a magnetic field is (a) zero, (b) maximum.
Answer:
a) When current in conductor is in direction of magnetic field,
b) When current in conductor is normal to the magnetic field.

Question 22.
A flat coil ABCD Is freely suspended between the pole pieces of a U – shaped permanent magnet with the plane of coil parallel to the magnetic field.
a) What happens when current is passed in the coil?
b) When will the coil come to rest?
c) Name the instrument which makes use of the principle stated above.
Answer:
a) The coil will experience a torque due to which it will rotate.
b) The coil will come to rest when its plane becomes normal to the magnetic field.
c) DC motor.

Question 23.
Why is it more difficult to move a magnet towards a coil which has a large number of turns?
Answer:

1. Emf induced in the coil becomes more when the number of turns in the coil are made large.
2. So it is difficult to move a magnet towards a coil which has a large number of turns.

Question 24.
A coil ABCD mounted on an axle is placed between the poles N and S of permanent magnet as shown in figure.

a) In which direction will the coil begin to rotate when the current is passed through the coil in direction ABCD by connecting a battery at the ends A and D of the coil.
b) Why is commutator necessary for continuous rotation of the coil?
Answer:
a) Anti-clockwise direction.
b) Commutator is needed to change the direction of current in the coil after each half rotation of coil.

Question 25.
Draw the magnetic field lines in uniform magnetic field.
Answer:

Question 26.
Draw a labelled diagram to make an electromagnet from soft iron bar AB. Mark the polarity at its ends. What precaution would you observe?
Answer:

The labelled diagram is shown in figure. The polarity at the end A where the current is clockwise, is south (S), while at the end B where the current is anti-clockwise is north (N).
Precaution :
The source of current must be the DC source.

Question 27.
State the principle of an electric motor. Name some appliances in which the Electric motor is used.
Answer:
Current carrying coil rotates when it is kept in a uniform magnetic field. It is the working principle of electric motor.
Appliances containing electric motor are :

1. Fans,
2. Mixies,
3. Grinders,
4. Machines, etc.

10th Class Physics 12th Lesson Electromagnetism 4 Marks Important Questions and Answers

Question 1.
How can you verify with experiment “The magnetic field lines are closed loops”? (AP March 2017)
Answer:

• Place a retort stand on the plank as shown in the figure.
• Pass a copper wire through hole of the plank and rubber knob of the retort stand in such a way that the wire be arranged in a vertical position and not touch the stand.
• Connect the two ends of the wire to a battery via switch. Allow the current flows through wire.
• By sprinkling the iron fillings around the wire, we can observe the magnetic field lines are in circular shape.

Conclusion :
Hence it is proved that “Magnetic field lines are closed loops”.

Question 2.
Name the device that converts electrical energy into mechanical energy. Draw its diagram and label the parts. (AP March 2018)
Answer:
1) The device that converts electrical energy into mechanical energy is motor.
2) Diagram of motor

Question 3.
List out the materials required for the Oersted experiment of electromagnetism. Write the procedure of the experiment. What do you understand by this experiment? (TS March 2016)
Answer:
List of material required for Oersted experiment:
A thermocol sheet, two small sticks, insulated copper wires, 9 V battery, switch, magnetic compass.

Procedure :

1. Take a thermocol sheet and fix two thin wooden sticks of height 1 cm which have small slit at the top of their ends.
2. Arrange a copper wire so that it passes through these slits and makes a circuit.
3. The circuit consists of a 9 V batten’, key, and copper wires connected in series.
4. Now keep a magnetic compass below the wire.
5. Now switch on the circuit and observe the compass needle.
6. Change the directions of current and observe the compass needle.

Observation :

1. When current is passed through circuit, we observe deflection of compass needle in one direction.
2. When the direction of current is changed, the compass needle deflects in another direction.
3. This shows that a current carrying conductor possesses magnetic field around it.

Question 4.
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid. (TS June 2017)
Answer:

• Fix a white paper on a wooden plank.
• Make two holes to the plank at appropriate distance.
• Make some more holes parallel to these two holes.
• Insert a copper wire through these holes. It looks as a coil.
• Connect the two ends of the coil to a battery and a switch in series.
• When swich is on current flows through the wire.
• Sprinkle some iron filings around the coil and tap the plank.

Observation :
An orderly pattern of iron filing is seen on the paper. These are magnetic field lines. The magnetic field lines set up by solenoid resemble those that of a bar magnet.

Question 5.
Why the current-carrying straight wire which is kept in a uniform magnetic field, perpendicularly to the direction of the field bends aside? Explain this process with a diagram showing the direction of forces acting on the wire. (TS March 2017)
Answer:

Uniform magnetic field (due to horse shoe magnet)

Magnetic field due to current carrying straight wire

Net field formed due to the above two fields

Explanation :
The net field in upper part is strong and in lower part it is weak. Hence a non-uniform field is created around wire. Therefore the wire tries to move to the weaker field region.

Question 6.
List out the apparatus and experimental procedure for the experiment to observe a current-carrying wire experiences a magnetic force when it is kept in uniform magnetic field. (TS June 2018)
Answer:
Required apparatus :
i) Horseshoe magnet,
ii) Conducting wire,
iii) Battery, switch

Experimental procedure :
1) Arrange the circuit :
Take a wooden plank and arrange two wooden sticks with slits and arrange a conductor through the sticks and make a circuits with switch, battery.

2) Put horse shoe magnet:
Arrange the horse shoe magnet on the conductor in such a way that the conductor should be in between the two poles of the magnet.

3) Deflections in conductor :
Allow the current to pass through the circuit. We can find that conductor deflects up wards or down wards.

4)

Question 7.

Answer the following questions by observing above diagram. (TS March 2018)
1. Which device function of working does the above figure gives?
Answer:
Motor

2. What is the angle made by AB and CD with magnetic field?
Answer:
90°

3. What are the directions of magnetic forces on sides AB and CD?
Answer:
By applying right hand rule to get the directions of magnetic force. At ‘AB’, the magnetic force acts inward perpendicular to field of the magnet and on ‘CD’, it acts outwards.

4. What is the net force acting on the rectangular coil?
Answer:
The net force on ‘AB’ is equal and opposite to the force on ‘CD’ due to external magnetic field because they carry equal currents in the opposite direction. Sum of these forces is zero. Similarly, the sum of the forces on sides ‘BC’ and ‘DA’ is also zero.

So, net force on the rectangular coil is zero.

Question 8.
Explain the working process of induction stove. (TS March 2019)
Answer:

• An induction stove works on the principle of electromagnetic induction.
• A metal coil is kept just beneath the cooling surface.
• It carries alternating current (AC) so that AC produce an alternating magnetic field.
• When you keep a metal pan with water on it, the varying magnetic field beneath it crosses the bottom surface of the pan, and EMF is induced in it.
• Induced EMF produces induced current in the metal pan.
• The pan has a finite resistance.
• The flow of induced current produces heat in it.
• That heat is conducted to the water. In this way, induction stove works and water will be heated.

Question 9.
Which device is used to convert mechanical energy into electrical energy? Draw a neat diagram and label the parts of this device. (TS March 2019)
Answer:
1) Dynamo is used to convert mechanical energy into electrical energy. They are also called Generators.

2) AC Generator (or) DC Generator

AC Generator

DC Generator

Question 10.
A coil is hung as shown in the figure. A bar magnet with north pole facing the coil is moved perpendicularly

a) How does the magnetic flux passing through the coil change?
b) State the direction of the flow of the current induced in the coil, keeping the direction of bar magnet in view.
c) Draw the diagrams showing the magnetic field formed due to bar magnet at the surface of the coil and the magnetic field formed due to induced current.
d) Explain the reason for induced current.
Answer:
a) A bar magnet with north pole facing the coil is moved perpendicularly, the magnetic flex increases when passing through the coil.
b) The direction of the flow of the. current induced in the coil, keeping the direction of bar magnet is anti-clockwise due to north pole. KT
c) Φ = 0

Plane of coil is parallel to ‘B’.
d) Electromagnetic induction is the reason for induced current.

Question 11.
Conductor of length T moves perpendicular to its length with the speed V. Length of the conductor is perpendicular to the magnetic field of the conductor. Let us assume that electrons could move freely in the conductor and the charge of an electron is ‘e’.

a) What is the magnetic force acting on electron in the conductor?
b) In which direction does the above force act?
c) What effect does this force have on motion of electrons?
Answer:
a) Magnetic field acting on the electron inside the conductor is = F_m = e (\(\bar{V} \times \bar{B}\)) = BeV
This field acts from P to Q.

b) Consider in the field P and Q are ends of a conductor. ‘Q’ will act as negative end and ‘P’ will act as positive end then flow passes from P to Q means downwards.

c) The force on electrons shows an effect creates a potential difference at the ends of the rods.
∴ BeV = eE ⇒ E = BY

Question 12.
A charged particle q is moving with a speed V perpendicular to the magnetic field induction B. Find the radius of the path and time period of the particle.
Answer:

1. Let us assume that the field is directed into the page.
2. Then the force experienced by the particle F = qvB.
3. We know that this force is always directed perpendicular to velocity.
4. Hence the particle moves along a circular path and the magnetic force on a charged particle acts like a centripetal force.
5. Let r be the radius of the circular path.

Question 13.
Explain different ways to induced current in a coil.
Answer:

• Moving a north pole of a magnet into a coil.
• Withdrawing north pole from a coil.
• Moving a south pole of magnet into a coil.
• Withdrawing a south pole of a magnet from a coil.
• Moving a coil towards a magnet.

Question 14.
What are the similarities between a current-carrying solenoid and a bar magnet?
Answer:
1) The magnetic field lines due to current carrying solenoid are identical to those of bar magnet. Thus a current-carrying solenoid behaves just like a bar magnet with fixed polarities at the ends. The end at which the direction of current is clockwise behaves like a south pole and the end at which current is anti-clockwise behaves like a north pole.

2) A current-carrying solenoid when suspended freely, will set itself in the north south direction exactly in the same manner as a bar magnet does.

3) A current-carrying solenoid also acquires the- attractive property of magnet. If iron filings are brought near the solenoid, it attracts them when current flows through the solenoid.

Question 15.
What are the dissimilarities between a current-carrying solenoid and a bar magnet?
Answer:

• The magnetic field strength due to solenoid can be altered by altering current in it, while the magnetic field strength of a bar magnet cannot be changed.
• The direction of magnetic field due to solenoid can be reversed by reversing the direction of current in it, but the direction of magnetic field of the bar magnet cannot be changed.

Question 16.
Compare electromagnet with a permanent magnet.
Answer:

Question 17.
What are the characteristics of magnetic field lines due to current in a loop (or circular coil)?
Answer:

• The magnetic lines are nearly circular in the vicinity of coil.
• Within the space enclosed by the wire the magnetic field lines are in the same direction.
• Near the centre of loop, the magnetic field lines are nearly parallel and the magnetic field may be assumed to be uniform in a small space near the centre.
• At the centre, the magnetic lines are along the axis of the loop and at right angles to the plane of the loop.
• The magnetic field lines become denser if the strength of current in the loop is increased and there are more number of turns in the loop.

Question 18.
A straight conductor passes vertically through a cardboard having some iron filings sprinkled on it.
a) Show the setting of iron filings when current is passed in the downward direction and then the cardboard is gently tapped. Draw arrows to represent the direction of magnetic field lines.
b) What changes occur if
i) current is increased ?
ii) the single conductor is replaced by several parallel conductors each carrying same current flowing in the same direction?
c) Name the law used by you to find the direction of magnetic field lines.
Answer:

a) The figure shoy/s, the pattern in which iron filings will set themselves. The arrows show the direction of magnetic field lines.
b) i) The arrangement of iron filings remains unchanged, but they become denser and Cardboar get arranged up to a larger distance from the conductor when the strength of current is increased and it is effective up to a larger distance from the conductor.
ii) The magnetic field at a point due to each conductor will be in same direction, so they will be added up. Thus the magnetic field strength is increased and it is effective up to a large distance so the magnetic field lines come closer and iron filings get arranged up to a larger distance.
c) Right hand thumb rule.

Question 19.
In figure A and B represent two straight wires carrying equal currents in a direction normal to the plane of paper inwards.
a) Sketch separately the magnetic field lines produced by each current.
b) Give a reason why the magnetic field at K (mid point of the line joining A and B) will be zero.
c) What will be the effect on the magnetic field at the point K if the current in wire B is reversed?

Answer:
a) Figure shows the sketch of magnetic field A and B.

b) The point K is equidistant from the wires A and B, and the wires A and B carry equal currents. So the magnetic fields at K due to wires A and B are equal in magnitude, but opposite in direction. Due to the wire A, it is downwards in the plane of paper, while due to the wire B, it is upwards in the plane of paper. So the net magnetic field at the point K is zero as the two fields cancel each other.
c) On reversing the direction of current in the wire B, the direction of magnetic field due to current is reversed at the point K, i.e. it becomes downwards in the plane of paper.

Question 20.
The diagram given below shows two coils X and Y. The coil X is connected to a battery S and a key K. The coil Y is connected to a galvanometer G.

When the key K is closed state the polarity.
i) At the end B of the coil X.
ii) At the end C of the coil Y.
iii) At the end C of the coil Y if the coil Y is (a) moved towards the coil X, (b) moved away from the coil X.
Answer:
i) On closing the key K, the current at the end B of the coil X is anti-clockwise, therefore at this end there is a north pole.

ii) While closing the key, polarity at the end C of the coil Y will be north. But there will be no polarity at the end C of the coil Y when the current becomes steady in the coil X

iii) a) With the key K closed, while the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
b) With the key K closed, while the coil Y is moved away from the coil X, the polarity at the end C of the coil Y is south.

Question 21.
How do you increase the speed of rotation of coil in a DC motor?
Answer:
The speed of rotation of coil can be increased by

1. Increasing the strength of current.
2. Increasing the number of turns in coil.
3. Increasing the strength of magnetic field. To increase the strength of magnetic field a soft iron core can be inserted within the coil.

Question 22.
Explain Lenz’s law with an activity.
Answer:
Lenz law :
The induced current will appear in such a direction that it opposes the change in the flux in the coil.

Explanation :

1. We know that when a bar magnet is pushed towards a coil with its north pole facing the coil an induced current is set up in the coil.
2. Let the direction be clockwise then current carrying loop behaves like a magnet with its south pole facing the north pole of bar magnet.
3. In such a case, the bar magnet attracts the coil. Then it gains kinetic energy. This is contradictory to conservation of energy.
4. Hence our assumption is wrong. So correct induced current direction is anti-clockwise.
5. Let us see a case where the bar magnet is pulled away from the coil with the north pole facing the coil. In such case, the coil opposes the motion of bar magnet to balance the conversion of mechanical energy into electric energy.
6. It happens only when the north pole of the magnet faces the south pole of the coil.
7. So, the direction of induced current in the coil must be in anti-clockwise direction.
8. In simple terms, when flux increases through coil, the coil opposes the increase in the flux and when flux decreases through coil, it opposes the decrease in the flux. This is Lenz law.

Question 23.
What are the uses of electromagnet?
Answer:
Electromagnets are mainly used for the following purposes.

1. For lifting and transporting the large masses of scrap, girders, plates, etc. especially to the places where it is not convenient to take the help of human labour.
2. For loading furnaces with iron.
3. For separating the magnetic substances such as iron from other debris.
4. For removing pieces of iron from wounds.
5. In several electrical devices such as electric bell, telegraph, electric thumb, electric motor, relay, microphone, loudspeaker, etc.
6. In scientific research, to study the magnetic properties of a substance in a magnetic field.

Question 24.
Write the differences between AC generator and DC motor.
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 14 Carbon and its Compounds.

AP State Syllabus SSC 10th Class Chemistry Important Questions 14th Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds 1 Mark Important Questions and Answers

Question 1.
Define Isomerism. (AP March 2016)
Answer:
The phenomenon of possessing same molecular formula but different properties by the compounds is known as “Isomerism”.

Question 2.
Give the names of the functional groups. (AP March 2018)
a) – COOR
b) – OH
Answer:
a) Ester
b) Alcohol

Question 3.
How do you explain the role of Oxygen in combustion process? (TS March 2015)
Answer:
Oxygen helps the combustion (or) No combustion will take place without oxygen.
Ex : C + O₂ → CO₂

Question 4.

Predict and write the products. (TS March 2016)
Answer:

Question 5.
Write two uses of nano tubes. (TS June 2017)
Answer:

1. Nano tubes are used as molecule wires.
2. In intigrated circuits nano tubes are used to connect the components together.
3. Nano tubes are used to incert Bio-molecules into the single cell.

Question 6.
Write two uses of Ethanol in day to day life. (TS March 2018)
Answer:
Ethanol is used in
i) Preparation of Alchoholic drinks
ii) Preparing tincture iodine
iii) Preparing cough syrup and tonics

Question 7.
Write the atomic structure of the following carbon compound. 3, 7-dibromo-4, -6 dichloro – oct-5-ene-l, 2-diol. (TS March 2019)
Answer:

Question 8.
Thanish added acetic acid along with concentrated sulphuric acid to ethanol what would be his observation during the experiment? (AP SCERT: 2019-20)
Answer:

1. He may observe that the resulting mixture is a sweet odoured substance.
2. The substance is ethyl acetate, an ester.

Question 9.
Why do the various micelles present in water do not come together to form a precipitate? Guess the reason. (TS June 2019)
Answer:
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.

Question 10.
Mention any two uses of graphite in day to day life. (TS June 2019)
Answer:
Uses of graphite in day to day life :

1. Pencil lead.
2. Lubricant.

Question 11.
What is “Allotropy”?
Answer:
The property of an element to exist in two or more different forms due to the difference in their atomic arrangement is called “Allotropy” and the different forms are called allotropes.

Question 12.
‘Diamond is a bad conductor of heat.’ Why?
Answer:
Diamond is a bad conductor of heat due to lack of free electrons.

Question 13.
What is ‘cleavage’?
Answer:
Cleavage is a property of splitting of crystals of some minerals in certain directions to produce a flat, even surface.

Question 14.
“Diamond is the hardest natural substance but is brittle.” Why?
Answer:
Diamond is the hardest natural substance but is brittle and can be broken due to the property of cleavage.

Question 15.
Explain about high refractive index of diamond.
Answer:
Diamond has a high refractive index, due to which most of the light that enters the diamond gets reflected back internally. This internally reflected light is responsible for the brilliance of a diamond.

Question 16.
What is catenation?
Answer:
Catenation is the phenomenon in which atoms of same element join together to form long chains.

Question 17.
What is an alkyl group?
Answer:
If one hydrogen is removed from an alkane, it is called alkyl group.
Ex : CH₄ → methane
CH₃ → methyl group

Question 18.
What is polymerization?
Answer:
The reaction in which a large number of identical and simple molecules join together to form a large molecule is called ‘polymerization’.

Question 19.
What do you understand by a ‘Functional group’?
Answer:
A group of atoms in carbon compounds showing characteristic properties is called a functional group.

Question 20.
Name some functional groups.
ANswer:
Alcohol – OH, Aldehyde – CHO, Ketone – > C = O, Carboxylic acid (- GOOH), ester (-COOR), and amine – NH₂ are some important functional groups.

Question 21.
What is pyrolysis?
Answer:
Decomposition of a compound on heating in the absence of air is called pyrolysis.

Question 22.
What is hydrocarbon?
Answer:
Compounds containing only carbon and hydrogen are called ‘hydrocarbons’.
Ex : Alkanes (Saturated hydrocarbons),
Alkenes and Alkynes (Unsaturated hydrocarbons).

Question 23.
What is ‘Saturated hydrocarbon’? (Or) What is an alkane?
Answer:
The valency of carbon is 4, of all the valencies of carbon, are satisfied, the resultant hydrocarbons are referred to as ‘saturated hydrocarbons’ or alkanes. Their general formula is C_nH_2n+2.

Question 24.
What are ‘Unsaturated hydrocarbons’?
Answer:
The hydrocarbons containing one or more double bonds or triple bonds between two carbon atoms are called ‘unsaturated hydrocarbons’.
Ex : C₂H₆ and C₃H₆, etc.

Question 25.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons having at least one (C = C) double bond in their structures, Alkenes are also called olefins. Their general formula is C_nH_2n.
Ex : Ethylene (C₂H4) and propene (C₃H₆), etc.

Question 26.
What are alkynes?
Answer:
Alkynes are unsaturated hydrocarbons having at least one (\(C \equiv C\)) triple bond in their structures. Their general formula is C_nH_2n-2.
Ex: Acetylene (\(\mathrm{HC} \equiv \mathrm{HC}\))

Question 27.
Mention the natural sources of carbon compounds.
Answer:
Plants, wood, natural gas, coal, petroleum, etc. are the natural sources of carbon compounds.

Question 28.
Explain about methanol (or) methyl alcohol.
Answer:
Methanol is the simplest alcohol, It is the first member of the homologous series of alcohol. It is also known as wood alcohol, as it was initially obtained by the destructive distillation of wood.

Question 29.
What is organic chemistry?
Answer:
The chemistry of carbon compounds (excluding the carbonates, bicarbonates, carbides, cyanides, carbon dioxide, and carbon monoxide) is called organic chemistry. The large number of organic compounds necessitated their study in separate branch of chemistry, known as organic chemistry,

Question 30.
What is halogenation?
Answer:
Alkanes react with halogens in the presence of sunlight. For example, when a mixture of methane and chlorine is exposed to sunlight, a hydrogen atom of methane is replaced by a chlorine atom,

Question 31.
How many rings are there in buckminsterfullerene?
Answer:
In buckminsterfullerene, there are 32 rings, of them 12 are pentagonal rings and 20 are hexagonal rings.

Question 32.
Give example for homologous series.
Answer:
CH₄ and C₂H₆ → These differ by a – CH₂ unit.
and C₂H₆ and C₃H₈ → These differ by a – CH₂ unit.

Question 33.
What is hybridisation?
Answer:
The intermixing of orbitals to form equivalent new orbitals is called hybridisation.

Question 34.
What are nanotubes?
Answer:
Nanotubes are allotropic form of carbon.

Question 35.
What are homologous series?
Answer:
The ierles of carbon compound in which successive compounds differ by -CH₂ unit is called homologous series.

Question 36.
Write the molecular formula of the fourth member of the homologous series of alcohols.
Answer:
CH₃ – CH₂ – CH₂ – CH₂ – OH

Question 37.
What is a catalyst?
Answer:
The substance which does not take part in chemical reaction but changes the rate of reaction.

Question 38.
Why are oils liquids at room temperature?
Answer:
Oils are unsaturated compounds so they are in liquid state.

Question 39.
Why are fats solids at room temperature?
Answer:
They are saturated compounds so they are in solid state.

Question 40.
Do you know the police detect whether suspected drivers have consumed alcohol or not? Explain.
Answer:
Orange Cr₂O₇^2- changes bluish green Cr³⁺ during the process of the oxidation of alcohol. The length of die tube that turned into green is the measure of die quantity of alcohol that had been drunk.

Question 41.
What is pka?
Answer:
The negative value of logarithm of dissociation constant of an acid.

Question 42.
What is Saponification?
Answer:
Alkaline hydrolysis of triesters of higher fatty acids producing soaps is called saponification.

Question 43.
What is a soap?
Answer:
Sodium or potassium salt of fatty acid.

Question 44.
What is micelle?
Answer:
A spherical aggregate of soap molecules in water is called micelle.

Question 45.
What change will you observe if you test soap with litmus papers?
Answer:
Red litmus turns into blue.

Question 46.
Write the valency of carbon in CH₃ – CH₃, CH₂ = CH₂ and \(\mathrm{HC} \equiv \mathrm{CH}\)?
Answer:
The valency of carbon in CH₃ – CH₃ is 4.
The valency of carbon in CH₂ = CH₂ is 3.
The valency of carbon in \(\mathrm{HC} \equiv \mathrm{C}\) – H is 2.

Question 47.
Out of butter and groundnut oil which is unsaturated in nature?
Answer:
Groundnut oil is unsaturated in nature.

Question 48.
What are hydrophobic and hydrophilic parts in soap?
Answer:

Question 49.
Name the carboxylic acid used as preservative.
Answer:
Acetic acid is used as preservative.

Question 50.
Why does graphite act as a good conductor of electricity?
Answer:
Graphite is a good conductor of electricity because of delocalized x electron system.

Question 51.
Among objects made of glass and diamond, which one shines more? Why?
Answer:
Diamond shines more because of low conical angle of 24,4° and also high refractive

Question 52.
Write IUPAC names of the following compounds.

Answer:
a) 2, 2, 3, 3 – tetra methyl butane
b) 3-chloro butan-l-oic acid.

Question 53.
What is the difference between combustion and oxidation reaction?
Answer:
Combustion is an oxidation reaction where a compound is burnt in the presence of oxygen, whereas oxidation is addition of oxygen which does not require any burning.

Question 54.
Write the order of priority of functional groups for naming carbon compounds.
Answer:

Question 55.
What is glycerol?
Answer:
The trihydroxy alcohol is called glycerol.

Question 56.
What do you mean by CMC?
Answer:
CMC means Critical Micelle Concentration.

Question 57.
Name the simplest chloride of saturated hydrocarbon.
Answer:
Chloro methane or methyl chloride.

Question 58.
Write the IUPAC name of next homolog of CH₃CH₂CHO.
Answer:
The next homolog of CH₃CH₂CHO is CH₃CH₂CH₂CHO (its IUPAC name is butanol). Since homologs differ by – CH₂.

Question 59.
How do physical properties like boiling point and melting point vary as the number of carbon atoms increases in a homologous series?
Answer:
There is regular gradation in physical properties of homologous series. So the physical properties like boiling point and melting point vary.

Question 60.
What is meant by Hybridisation?
Answer:
Mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals.

Question 61.
Write any two uses of Graphite.
Answer:
i) Conductor
ii) Lubricant

Question 62.
Write any two examples to Amorphous form of carbon.
Answer:
i) Coke
ii) coal
iii) charcoal.

Question 63.
Write any two examples to crystalline forms of carbon.
Answer:
i) Diamond
ii) graphite

Question 64.
What are the applications of Buckminster fullerene?
Answer:
i) Antioxidants
ii) Anti aging and damage agent in cosmetic sector.

Question 65.
What is meant by catenation?
Answer:
Binding of an element to itself through covalent bonds to form chain or ring molecules.

Question 66.
Write any one use of nanotubes.
Answer:
i) Used as molecular wires.
ii) Used in integrated circuits.

Question 67.
On which reason, graphite is used as lubricant and as the lead in pencils?
Answer:
Graphite has free electrons.

Question 68.
How many isotopes are there for C₄H₁₀, what are they?
Answer:
i) n – Butane
ii) Iso – Butane

Question 69.
CH₃ – CH = CH – CH₃, how many sigma bonds are present in the above compound?
Answer:
11

Question 70.
Write the IUPAC name of Ethyle alcohol.
Answer:
Ethanol.

Question 71.
Classify the following into alkanes, alkenes and alkynes.
C₁₂ H₂₂, C₁₀ H₂₂, C₁₁ H₂₂
Answer:
i) C₁₀ H₂₂ – Alkanes
ii) C₁₁ H₂₂ -Alkenes
iii) C₁₂ H₂₂-Alkynes

Question 72.
Hi ……… I am carboxylic acid. I am used in the making vinegar, who am I?
Answer:
Acetic acid.

Question 73.
What does IUPAC represent?
Answer:
International Union of Pure and Applied Chemistry.

Question 74.
Write any one example for esterification reaction.
Answer:
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 75.
A compound with molecules formula C₂H₆O is used in cough syrup. Identify the compound.
Answer:
Ethyl Alcohol.

Question 76.
Which substance is added for the denaturation of ethyl alcohol?
Answer:
Pyridine.

Question 77.
What is the abbreviation of CMC?
Answer:
Critical Micelle Concentration.

Question 78.
Write the names of polar end and non-polar end in a soap.
Answer:
Polar end – COO^– Na⁺, Non-polar end – R.

Question 79.
Write the IUPAC name of the alcohol which one carbon atom.
Answer:
Methanol.

Question 80.
Write the chemical equation which indicates the preparation of ethanol industrially?
Answer:

Question 81.
What is the formula of chloroform? Write its one use.
Answer:
CHCl₃, Anesthetic.

Question 82.
Which type of hydrocarbons are participate in addition reaction?
Answer:
Unsaturate Hydrocarbons.

Question 83.
What are the oxidising agents used in oxidisation of C₂H₅?
Answer:
K₂Cr₂O₇, KMn0₄.

Question 84.
What is meant by catalyst?
Answer:
To change die rate of reaction without itself undergoing any permanent chemical change.

Question 85.
What are the main constituents of LPG?
Answer:
Butane, Methane.

Question 86.
What is the difference between saturated and unsaturated hydrocarbons?
Answer:
Saturated house single bonds, unsaturated have multiple bonds.

Question 87.
Describe a test for carboxylic acid.
Answer:
React with metals liberate hydrogen gas.

Question 88.
What is meant by denatured alcohol?
Answer:
Unfit for human consumption by adding one or more chemicals.

Question 89.
Complete the following equation.
CH₄ + 2O₂ →
Answer:
CH₄ + 2O₂ → CO₂ + 1H₂O

Question 90.

i) In the above substance, what is the hybridisation of 3^rd carbon?
Answer:
sp²

ii) What is the hybridisation of 4^th carbon?
Answer:
sp³

Question 91.
What is the main misuse of Ethanol?
Answer:
Drinking.

Question 92.
What is gasohol?
Answer:
10% Ethyl alcohol with gasoline.

Question 93.
Write any two uses of Ethyle alcohol.
Answer:
i) Good solvent
ii) Additive to automotive gasoline.

Question 94.
Write two IUPAC name

Answer:
3 – Chloro 1 – Butane

Question 95.
Name the following functional groups.
i) – COOR
ii) R – COOH
Answer:
i) – COOR (Ester)
ii) R – COOH (Carboxylic acid)

Question 96.
Name the crystalline allotrope of carbon which conducts electricity.
Answer:
Graphite.

Question 97.
Ravi gets confused while understanding the between R – COOH and R – OH functional groups, ask him one question to classify it.
Answer:
i) What is carboxylic acid?
ii) What is Alcohol?

Question 98.
Formic acid (HCOOH)
Farmaldehyde (HCHO)
Methanol (CH₃OH), then answer the following questions.
i) Which is present in ants?
Answer:
HCOOH (Formic acid).

ii) Which is used to preservation of dead bodies?
Answer:
HCHO (Farmaldehyde).

Question 99.
Write the symbolic representation showing the functional groups.
i) amine
ii) amide
Answer:
i) R – NH₂
ii) R – CONH₂

Question 100.
How many sigma and pi-bonds present in Acetylene?
Answer:
\(\mathrm{HC} \equiv \mathrm{CH}\) ; σ bonds – 3 ; π bonds – 2

Question 101.
Which of the following will give substitution reactions?
CH₄, C₃H₆, C₃H₄, C₅H₁₂, C₄H₈
Answer:
CH₄, C₅ H₁₂

Question 102.
Which of the following will give addition reactions?
CH₄, C₃H₆, C₃H₄, C₅H₁₂, C₄H₁₀
Answer:
C₃H₆, C₃H₄

Question 103.
What is a homologous series?
Answer:
Same functional group, difference between successive members is a simple structural unit – CH₂.

Question 104.
Name the hydrocarbon which is used in the artificial ripening of fruits?
Answer:
C₂H₄

Question 105.
Define fermentation process.
Answer:
Chemical break down of a substance by bacteria, yeast or other microorganisms.

Question 106.
Define functional group.
Answer:
They are specific substituents within molecules that are responsible for die characteristic chemical reactions.

Question 107.
Which hydrocarbons participate in sp² hybridisation?
Answer:
C₂H₄

Question 108.
Name the following compounds,
i) CH₃ – CH₂ – Br
Answer:
1 – Bromo Ethane

ii)

Answer:
Ethanol

iii)

Answer:
2 – Butanone

iv)

Answer:
2, 3 – dichloro Butane

Question 109.
Which constituents are present in tincture Iodine?
Answer:
i) Iodine
ii) Alcohol.

Question 110.
Write the uses of esters in daily life.
Answer:
i) Solvents
ii) Plasticizers

Question 111.
Name the gas evolved when acetic acid reacts with sodium hydrogen carbonate.
Answer:
The gas liberated is carbon dioxide.

Question 112.
Name the organic acid present in vinegar. Write its chemical formula.
Answer:
The acid present in vinegar is acetic acid. Its formula is CH₃COOH.

Question 113.
Why is graphite a good conductors’of electricity?
Answer:
Graphite has free electrons.

Question 114.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Tbtravalency.

Question 115.
Why are alkanes called as paraffins?
Answer:
Low reactivity.

Question 116.
Draw two possible structures with formula C3HgO and what they are called?
Answer:

Question 117.
Draw structure of 3 – methyl pentan-3-ol.
Answer:

Question 118.
Draw the shape of soap molecule.
Answer:

Question 119.
Draw the shape of Micelle.
Answer:

Question 120.
Draw the shape of methane.
Answer:

Question 121.
Draw the structure of pentanoic acid.
Answer:

Question 122.
How do you appreciate the role of diamond in space probes?
Answer:
Since it has the ability to filter out harmful radiations, it is used in making protective windows for space probes.

Question 123.
How do you appreciate the role of acetic acid as a preservative?
Answer:

• Dilute acetic acid is used as a food preservative in the preparation of pickles and sauces,
• As vinegar, it is also used as an appetiser for dressing food dishes.

Question 124.
How do you appreciate the role of diamond in surgery?
Answer:
A sharp edged diamond is used as a tool to remove cataract in eye surgery.

10th Class Chemistry 14th Lesson Carbon and its Compounds 2 Marks Important Questions and Answers

Question 1.
Draw the simple figure of a soap molecule. (AP March 2016)
Answer:
Structure of soap molecule :

Question 2.
Draw the structure of the methane molecule. Write its bond angle. (TS June 2015)
Answer:
The bend angle in methane is 109°2 8′.

Question 3.
a) Why are vegetable oils healthy as compared to vegetable ghee? (TS March 2015)
b) Write the IUPAC name of

Answer:
a) Because vegetable oils contain unsaturated fatty acids or vegetable oils are easily digestible.
b) 3 – Mono chloro butene (or) 3 Chloro butene

Question 4.
What are alkenes? Write the general formula of alkenes. Give an example for alkenes. (TS June 2017)
Answer:

• Unsaturated hydrocarbons those are having carbon * carbon double bond are known as alkenes.
• The general formula of Alkenes is CHH2h.
• Example : Ethelene (C2H4).

Question 5.

Based on the diagram, answer the following.
1) Write the name of the compound.
2) Write the name of functional group in the structure. (AP March 2019)
Answer:

1. The compound is 2, 3-di ethyl-cycle hexan-1-ol.
2. Alcohol (OH) is the functional group in the structure.

Question 6.
Identify the functional groups in the following compounds and write IUPAC names.

Answer:

The IUPAC name of the compound Is 2 – Chloro-Butan 1-ol.

The IUPAC name of the compound is 3 – Methyl-2-Butan-one.

Question 7.
Draw the structure of butanoic acid C₃H₇COOH.
Answer:
Formula of butanoic acid is C₄H₅O₂.

Question 8.
What is ‘Isomerism’?
Answer:
Compounds having same molecular formula but different structures are called isomers, and the phenomenon is called isomerism.
Ex: C₄H₁₀ exists an n-hutane and iso-butane.

Question 9.
How do you detect leakage in the cylinder?
Answer:

• To detect any leakage of gas from die cylinder, a strong-smelling substance like ethyl mercaptan (C₂H₅ SH) is added to die gas.
• Then the leakage can be easily detected by the foul smell of die ethyl mercaptan.

Question 10.
How is LPG gas useful for environment?
Answer:

• Because of its heat producing capacity (calorific value), it is considered to be a good fuel.
• It bums without producing smoke. Hence, it does not cause any pollution.
• It is a dean fuel and can be conveniently handled.

Question 11.
How is ethanol useful in pharmaceutical industry?
Answer:

• Solutions in ethanol are often prepared in pharmaceutical industry, these solutions are known as tinctures.
• For example, a solution of Iodine and potassium iodide in ethanol is called tincture of iodine.
• It is also used as an important raw material for the synthesis of many organic compounds, for example, ethanol, ethanoic acid, ethanoie anhydride, esters, chloroform, etc.

Question 12.
How are synthetic detergents harmful for environment?
Answer:

• Some synthetic detergents resist biodegradation, i.e. they are not decomposed by micro-organisms such as bacteria.
• Hence, they cause water pollution in lakes and rivers.
• They tend to persist for a long time, making the water unfit for aquatic life.

Question 13.
Explain about allotropic forms of carbon.
Answer:

Question 14.
Diamond is considered to be the purest form of carbon. How can we prove it?
Answer:
When diamond is heated in oxygen alone, it bums at about 800° C and forms carbon dioxide leaving no residue. This proves that diamond to be the purest form of carbon.

Question 15.
Why does carbon not form C⁴⁺? Why?
Answer:

• Electronic configuration of carbon is 1s²2s²2p².
• If carbon loses four electrons from the outer shell, it will form C⁴⁺ ions.
• This requires huge amount of energy which is not available normally.
• Therefore C⁴⁺ formation is not possible.

Question 16.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Carbon is unable to form C⁴⁺ ion as well as C^4- ion. So carbon has to satisfy its tetra- valency by sharing electrons with other atoms. So it mainly forms covalent bonding.

Question 17.
Define Allotropy. What are the allotropic forms of carbon?
Answer:
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. The allotropic forms of carbon are graphite, diamond, etc.

Question 18.
Identify the unsaturated compounds of the following.
a) CH₃ – CH₂ – CH₂
b) CH₃ – CH = CH₃
c)

Answer:
a) CH₃ – CH₂ – CH₂ saturated compound.
b) CH₃ – CH = CH₃ unsaturated compound.
c)

Question 19.
Define Isomers. Write structural formula of isomers of butane.
Answer:
Compounds having same molecular formula but different properties are called isomers.
Isomers of butane :
1) CH₃ – CH₂ – CH₂ – CH₃
Butane

Question 20.
What happens when a small piece of sodium is dropped into ethanol?
Answer:
When a small piece of sodium is dropped into ethanol it releases hydrogen gas and forms sodium ethoxide.
2C₂H₅OH + 2 Na → 2C₂H₅ONa + H₂

Question 21.
What type of reaction takes place between ethane and chlorine?
Answer:
Substitution reaction takes place between ethane and chlorine in die presence erf sunlight
CH₄ + Cl₂ → CH₃Cl + HCl
CH₃Cl +Cl₂ → CH₂Cl₂ + HCl
CH₂Cl₂ + Cl₂ → CHCl₃ + HCl
CHCl₃ + Cl₂ → CCl₄ + HCl

Question 22.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

1. Catenation
2. Isomerism.

Question 23.
How could you name the following compounds?
a) CH₃ – CH₂ – CH₂ – Br
b) CH₃ – CH₂ – CH₂ – CH₂
Answer:
a) Bromo propane
b) Hexyne

Question 24.
Give examples for primary, secondary and tertiary amines.
Answer:
Primary amine – CH₃NH₂
Secondary amine – CH₃ – NH – CH₃

Question 25.
Write the following conversions.
1) Ethanol to Ethene
2) Ethene to Ethanol
3) Methane to carbon tetra chloride.
Answer:
1) Ethanol reacts with cone. H₂SO₄ at about 170°C to give ethene.

2) Ethanol is prepared from ethene by the addition of water vapour in the presence of catalyst P₂O₅.

3) Methane reacts with chlorine in the presence of sunlight. Hydrogen atoms of CH₄ are replaced by chlorine at

Question 26.
Name the following compounds and which one is saturated among them.
a) CH₃ – \(\mathrm{C} \equiv \mathrm{H}\) – CH₃
b) CH₃ – CH = CH – CH₃
c) CH₃ – CH₂ – CH₂ – CH₃
Answer:
a) 2-Butyne
b) 2 – Butene
c) Butane

Butane does not show any double or triple bonds. Its valency is completely satisfied with formation of single bond. So it is a saturated compound.

Question 27.
How do you identify the given organic compound contains carboxylic acid functional group?
Answer:

• On adding carbonates and bicarbonates the compound containing carboxylic acid group evolves carbon dioxide gas.
• When warmed with alcohol and cone. H₂SO₄ a pleasant fruity smell is produced due to formation of ester.

Question 28.
Explain briefly about the structure of “Diamond”.
Answer:

• In a diamond, each carbon atom is surrounded by four other carbon atoms.
• In these carbon atoms, each carbon atom undergoes in its excited state sp3 hybridisation.
• These are placed at the four corners of a regular tetrahedron.
• This results in a 3-dimensional network of carbon atoms.
• So diamond is in three dimensional structure.

Question 29.
Explain briefly about the structure of “Graphite”.
Answer:

• In graphite, each ‘C’ is surrounded by three other ‘C’ atoms.
• The ‘C’ atoms are arranged in layers.
• In the layer structure, the carbon atoms are in trigonal planar environment.
• Each layer consists of a 2-dimensional hexagonal network.

Question 30.
Diamond is an extremely bad conductor of electricity.” Why?
Answer:
1) In diamond, each carbon atom is covalently bonded with four other carbon atoms.
2) So, the four outermost electrons of a carbon atom are engaged or trapped in the covalent bonds, having no free electrons making it a bad conductor of electricity.

Question 31.
Why is diamond hard but graphite is smooth and slippery?
Answer:
Diamond has sp³ hybridisation with tetrahedral environment. As C – C bonds are very strong any attempt to distort the diamond structure requires large amount of energy. Hence diamond is one of the hardest material.

Whereas graphite has sp² hybridisation with layer structure with trigonal planar environment. The layers tend to slide on one another. So graphite is smooth and slippery.

Question 32.
An organic compound X with a molecular formula C₂H₆O undergoes oxidation within presence of alkaline KMnO₄ to form a compound Y. X on heating in presence of con. H₂SO₄ at 443 K gives Z. Which on reaction with Br₂ and decolorizes it? Identify X, Y, and Z and write the reactions involved.
Answer:
X is ethanol.

Question 33.
Complete the following reactions.

Answer:

Question 34.

What are A and B?
Answer:
1) Alkynes undergo addition reaction in the presence of nickel catalyst and hydrogen to form Alkene.

Question 35.
Draw the structure for the following compounds.
a) Propanoic acid
b) Chlorobutane
c) Hexanone
d) Pentanal
Answer:
a) CH₃CH₂COOH
b) CH₃CH₂CH₂CH₂Cl
c) CH₃CH₂CH₂CH₂COCH₃
d) CH₃CH₂CH₂CH₂CHO

Question 36.
Give IUPAC names of the following compounds. If more than one compound is possible, name all of them.
i) A chloride derived from butane.
ii) A ketone derived from pentane.
Answer:
i) The following chlorides are possible for butane.

ii) The following ketones are possible for pentane.

Question 37.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆?
b) Give IUPAC names of the above possible compounds and represent them in structure.
c) What is the difference between those
Answer:

b) The IUPAC names of compounds are propene and cycle propane.
c) The main difference Is that the first compound Is alkene-an unsaturated compound and second is cyclo alkane-a saturated compound.

Question 38.
Draw isomeric forms of C₆H₁₄.
Answer:
Isomers of hexane :

Question 39.
How do you appreciate the role of carbon in everyday life?
Answer:

• Major components of our daily food have carbohydrates, proteins, fats, etc. which are all made up of carbon compounds.
• The fibres of cloth are made up of cellulose and other types of materials, which are all carbon compounds.
• Cement and steel form the core of any of the modern buildings. Carbon bestows steel with hardness, while limestone (CaCO₃) a major constituent of cement also contains carbon.

Question 40.
How do you appreciate the role of oxygen in combustion process?
Answer:

• When the oxygen supply is insufficient, the fuels burn incompletely producing mainly a yellow flame.
• When the oxygen supply is sufficient, the fuels burn completely producing a blue flame.

Question 41.
How do you appreciate the role of Ethanol as a fuel?
Answer:

• A material which is burnt to obtain heat is called a fuel. Since ethanol burns with a clear flame giving a lot of heat, it is used as a fuel.
• Some countries add ethanol to petrol to be used as a fuel in cars. Thus ethanol is used as an additive in petrol.
• Ethanol alone can also be used as a fuel for cars.

Question 42.
What are the uses of fullerenes?
Answer:
Fullerenes are under study for potential medical use such as specific antibiotics to target resistant bacteria and even target cancer cells such as melanoma.

Question 43.
Write the HJPAC names of the following compounds.
i) CH₃ – CH₀ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂OH
ii) CH₃ – CH₂ – CH = CH- CH₂ – \(\mathrm{C} \equiv \mathrm{CH}\)
iii) CH₃ – CH₂ – CH₂ – CH₂ – CHO
iv) CH₃ – CH₂ – CH₂ – CH₂ – COOH
Answer:

1. nananol
2. 4- ene – 1 heptyne
3. pentanal
4. pentanoic acid

Question 44.
What are the uses of alcohol?
Answer:

• Alcohols are goods solvent for resin and gums.
• Ethanol is used in the thermometers because of its low freezing point.
• One of the products of ethyl alcohol is chloroform, which is used as an aesthetic.
• 10% ethanol in gasoline is a good motor fuel.
• It is used in medicines such as tincture iodine, cough syrups and many tonics.

Question 45.
What are the uses of acetic acid?
Answer:

• 5 to 8% solution of acetic acid in water is called vinegar and is used widely as a preservative in pickles.
• Used as a laboratory reagent.
• Used in the production of perfumes, dyes, esters, etc.
• Used in medicine.

10th Class Chemistry 14th Lesson Carbon and its Compounds 4 Marks Important Questions and Answers

Question 1.
Write IUPAC names for the following carbon compounds.

Answer:
A) 2 – methyle pentane – 3 – ol
B) 3 – chloro, 4 – Methyle hexanoic acid
C) 2 Bromo – Bute – 2 – ene
D) 2, 5 Dimethyle hexane

Question 2.
(AP June 2017)
Observe the given carbon compound and answer the following questions.
a) Give numbering to the carbons in the given compound according to IUPAC rules.
b) Name the functional group present in the given compound.
c) Name the word root for the given carbon compound.
d) Write the IUPAC name of the given compound.
Answer:
a)

b) The given compound contains functional group – OH. It is an alcohol.
c) Word root: The number of carbon atoms present in the molecules is called word root. Here the word root is (C₅) – pent.
d) IUPAC name of the given compound is pent 4 – ene 2 – ol.

Question 3.
Alkanes are considered as Paraffins. So, they undergo substitution reactions but not addition reactions. Explain with suitable example. (AP March 2017)
Answer:

Question 4.

Observe the structure and answer the following.
a) Write the name of principal functional group present in the compound.
b) Identify the parental chain in the compound.
c) What are the substituents in the above compound?
d) Name the above compound as per IUPAC nomenclature. (AP June 2018)
Answer:
a) Ketone
b)

c) Methyl group ; Hydroxy group
d) 7 – hydroxy – S – methyl heptan – 2 – one

Question 5.
In the table given below, fill the information in the empty boxes and give answers to the following questions. (TS June 2015)
a Write the general formula of alkanes from the table.

b) How many a bonds are there in C₃H₆?
c) What sequential order did you notice in the molecular formulae?
d) There exist single bonds between carbon atoms of alkanes. Do you agree with this statement? Give reasons.
Answer:

a) The general formula of Alkanes is C_nH_2n+2
b) The number of o bonds in C₂H₆ are 7.
c) Two successive alkanes are differed by – CH₂ group.
d) Except Methane all other alkanes have single bonds between carbon atmos because it is a saturated hydro carbon.

Question 6.
Why do we call alkanes as paraffins? Explain the substitution reactions of alkanes. (TS June 2016)
Answer:
a) 1. Alkanes are saturated hydrocarbons with least reactivity.
2. Therefore they are called paraffins.
3. Parum = little and affins = affinity.

b) 1. A reaction in which one atom or a group of atoms in a given compound is replaced by other atom or group of atoms is called a substitution reaction.
2. Alkanes have single bonds and undergo substitution reactions.

3. For example :
Methane (CH₄) reacts with chlorine in the presence of sunlight.

Question 7.
Write the types of Allotropes of Carbon. Give any three examples of each. (TS March 2016)
Answer:
The allotropes of carbon are classified into two types. They are
i) Amorphous forms,
ii) Crystalline forms.
Examples :
Amorphous forms :
Coal, coke, wood charcoal, animal charcoal, lamp black, gas carbon, petroleum coke, sugar charcoal, etc.

Crystalline forms :
Diamond, graphite, and buckminsterfullerene.

Question 8.
Write any 4 characteristic features of homologous series of Organic compounds. (TS March 2016)
Answer:
Homologous series :
The series of carbon compounds in which two successive compounds differ by – CH2 unit is called homologous series.

Characteristic features of homologous series :

1. They have one general formula.
   Ex : Alkane (C₄H_{2n + 2}), Alkene (C₄H_2n), Alkyne (C₄H_2n-2)
2. Successive compounds in their series possess a difference of (- CH₂) unit.
3. They possess similar chemical properties due to the same functional group.
4. They show a regular gradation in their physical properties.

Question 9.
List out the materials required to conduct the experiment to understand the esterification reaction. Explain the procedure of the experiment. How can you identify that an ester is formed in this reaction?(TS March 2017)
Answer:
Required Material :
Test tube, beaker, tripod* burner, water, wire guage, ethanol (absolute alcohol), glacial acetic acid, concentrated sulphuric acid.

Procedure :

1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in the test tube.
2. Warm it in a water bath or in a beaker containing water for atleast five (5) minutes.
3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
   If we smell sweet odour from the beaker, we can confirm that ester is formed.

Question 10.
Explain the Isomerism and Catenation properties of carbon. (TS March 2018)
Answer:
Catenation properties of carbon :
i) Carbon has ability to form longest chains with its own atoms. This special property of carbon is called catenation.
ii) Due to catenation property of carbon it can form largest chain containing millions of carbon atoms, branches and cyclic compounds.

Isomerism of carbon :
The phenomenon of possessing some molecular formula but different properties by the compounds is known as isomerism.

Molecular formula of above two molecules is C₄ H₁₀ but they have different structure. These two are isomers.

By there two special properties of carbon it can make number of compounds.

Question 11.

Observe the above table and answer the following questions. (TS March 2019)
1) Write the general formula of Alkanes.
2) Mention the names of unsaturated hydrocarbons.
3) Write the homologous series of Alkynes.
4) Write the formula of Hexyne.
Answer:
1) General formula for Alkanes : C_nH_2n+2.
2) Unsaturated Hydrocarbons in the list are :
Propene C₃H₆, Butene C₄H₆, Pentyne C₅H₈, Hexyne C₆H₁₀.

3) Homologous series of Alkynes is C₂H₂ (Ethyne), C₃H₄ (Propyne), C₄H6 (Butyne), C₅H₈ (Pentyne), C₆H₁₀ (Hexyne).

4) Formula of Hexyne is C₆H₁₀.

Question 12.
Complete the following table based on functional groups of organic compounds, their structural formulas and respective suffixes. (AP SCERT: 2019-20)

Answer:

Question 13.
Explain the occurrence of carbon.
Answer:
Carbon occurs in nature in free state as well as in combined state.

Question 14.
What is sp hybridisation? Explain.
Answer:

• Each carbon is only joining to two other atoms rather than four or three.
• Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise two of the orbitals.
• They use the ‘s’ orbital (2s) and one of the 2p orbitals, but leave the other 2p orbitals unchanged.
• The new hybrid orbitals formed are called sp-hybrid orbitals, because they are made by an s-orbital and a p-orbital reorganizing themselves.

Question 15.
Write the characteristics of homologous series of organic compounds.
Answer:
Characteristics of homologous series :

1. They have one general formula.
   e.g.: Alkanes (C_nH_2n+2)
2. Successive compounds in the series possess a difference of – CH₂ unit.
3. They possess similar chemical properties due to same functional group.
   e.g.: C – OH
4. They show a regular gradation in their physical properties.

Question 16.
What is sp³ hybridisation with diagram? Explain.
Answer:
The excited carbon atom allows its one s-orbital (2s) and three p-orbitals (2p_x, 2p_y, 2p_z) to intermix and reshuffle into four identical orbitals known as sp³ orbitals. Thus, carbon atom undergoes sp³ hybridization. The four electrons enter the new four identical hybrid orbitals known as sp³ hybrid orbitals, one each as per Hu nd’s rule.

1) Since carbon has four unpaired electrons, it is capable of forming bonds with four other atoms.

2) When carbon reacts with hydrogen, four hydrogen atoms allow their ‘s’ orbitals containing one electron each to overlap with four sp³ orbitals of carbon atom which are oriented at an angle of 109°. 28’.

3) Four orbitals of an atom in the outer shell orient along the four corners of a tetrahedron to have minimum repulsion between their electrons. ‘The nucleus of the atom is at the centre of the tetrahedron.

4) This leads to form four sp³ – s sigma bonds between carbon atom and four hydrogen atoms, All these bonds are of equal energy,

Question 17.
What is sp² hybridisation? Explain.
Answer:
Consider ethene molecule

• In the formation of CH₂ – CH₂ each carbon atom in its excited state undergoes sp² hybridisation by intermixing one s-orbital (2s) and two p-orbitals (say 2p_x and 2p_y) and reshuffling to form three sp² orbitals.
• Mow each carbon atom is left with one ‘p’ orbital (say 2p_z) unhybridised,
• The three sp² orbitals having one electron each get separated around the nucleus of carbon atoms at an angle of 120°.
• When carbon is ready to form bonds one sp² orbital of one carbon atom overlaps the sp² orbital of the other carbon atom to form sp² – sp² sigma (σ) bond,
• The remaining two sp² orbitals of each carbon atom get overlapped by ‘s’ orbitals of two hydrogen atoms containing unpaired electrons.
• The unhybridised p_z orbitals on the two carbon atoms overlap laterally as shown in figure to form a π (pi) bond.
• Hence, there exist a sigma (σ) bond and a pi π (pi) bond between two carbon atoms in ethene molecule. Hence, the molecule ethene (C₂H₄) is

Question 18.
The list of some organic compounds is given below.
Ethanol, ethane, methanol, methane, ethyne and ethene.
From the above list name the compound …………..
a) formed by the dehydration of ethanol by cone. H₂SO₄.
b) which forms methanoic acid on oxidation?
c) which forms chloroform on halogination in the presence of light?
d) which are unsaturated compounds?
e) which have compounds containing alcohol group?
Answer:
a) Dehydration ethanol in the presence of Cone. H₂SO₄ forms ethene,
b) Methanol on oxidation turns to methanoic acid,
c) Methane in the presence of light forms chloroform,
d) Unsaturated compounds are ethene and ethyne.
e) The compounds containing alcohol group are methanol, ethanol,

Question 19.
Give the IUPAC names of the following compounds.

Answer:

1. 1 – propyne
2. 3 – pentanel (or) pentamSml
3. 2 – methyl propane
4. 1, 2 dichloro ethane

Question 20.
Give one example of each of the following.
i) Saturated hydrocarbon
ii) Cyclic compounds
iii) Unsaturated hydrocarbon
iv) Functional group
v) Homologous series
Answer:
i) Saturated hydrocarbons are Alkanes, So the examples are methane (CH₄), Ethane, (C₂H₆).
ii) Cyclic compounds are cycle alkanes, eg : Cyclo propane (C₃H₆), Cycle butane (C₄H₆).
iii) Unsaturated hydrocarbons are Aikynes, eg : Ethene (C₂H₄), Propene
iv) The examples for functional groups are ‘ 1. Aldehyde – CHO, 2. Alcohol = OH
v) A series of carbon compounds that differ by – CH₂ with similar chemical properties is called homologous series.
eg: 1, Alkane, 2, Alkene

Question 21.
Write the differences between saturated and unsaturated hydrocarbons.
Answer:

Question 22.
Answer the following.
a) What are the first three members of carboxylic acid series?
b) Name the compounds which can be oxidised directly or in stages to produce ethanoic acid.
c) Write one equation each when acetic acid reacts with a metal, a base, and a carbonate.
d) Name the organic compound formed when acetic acid and ethanol react together.
Answer:
a) The first three members of carboxylic acids are :
i) Methanoic acid – HCOOH
ii) Ethanoic acid – CH₃COOH
iii) Propanoic acid – CH₃CH₂COOH

b) Ethanol in stages oxidises to acetic acid whereas ethanol directly oxidises to ethanoic acid.

c) i) 2 CH₃COOH + 2 Na → 2CH₃COONa + H₂
ii) CH₃COOH + NaOH → CH₃COONa + H₂O
iii) CH₃COOH + Na₂CO₃ → CH₃COONa + H₂O + CO₂

d) When ethanol reacts with ethanoic acid it forms an ester namely ethyl acetate.

Question 23.
What are the rules to be followed to name a carbon compound?
Answer:
Rules to be followed
i) Longest carbon chain is selected,
ii) Chain is numbered in such a way that the branched chain or substituent gets the smallest number,
iii) If the functional group is present, it is given the. lowest number,
iv) Substituents are named in the alphabetical order,
v) The position of substituents are prefixed with hyphen,
vi) Multiple substituents are written with numerical prefixes such as di or tri,

Question 24.
Write suffixes and prefixes for some important characteristic functional group in a tabular form.
Answer:

Question 25.
Correct the following statements.
1) Alkenes undergo substitution reactions.
2) Alkanes are polar in nature.
3) When sodium piece is added to ethanol oxygen gas liberates.
4) On complete combustion of carbon compound it gives carbon monoxide and water.
Answer:

1. Alkenes are unsaturated hydrocarbons. So they undergo addition reactions.
2. Alkanes are covalent compounds. So they are non-polar in nature.
3. When sodium piece is added to ethanol it releases hydrogen gas.
4. On complete combustion of carbon compound it forms carbon dioxide and water.

Question 26.
Copy and complete the following table which relates to three homologous series of hydrocarbons.

Answer:

Question 27.
Draw the structures of isomers of butane.
Answer:
Isomers of butane are n-butane, iso butane and cyclo butane :
Structures :

Question 28.
Draw the structures of the following.
a) Ethanoic acid
b) Propanal
c) Propene
d) Chloro propene
Answer:
Structures:

Question 29.
Draw the structures of the following compounds
a) 2 – bromo pentane
b) 2 – methyl propane
c) butanal
d) 1 – hexyne
Answer:

Question 30.
Write the molecular formula of the first four compounds of the homologous series of aldehydes.
Answer:
Homologous series of aldehydes ate Formaldehyde, Acetaldehyde, Propionaldehyde and Butanaldehyde.

Question 31.
How many isomers can be drawn for pentane with molecular formula C-H(2? What are they? Draw their structures and mention theii common names.
Answer:
Isomers of pentane are three. These are
1) Pentane
2) Iso pentane
3) Neo pentane.
Structures :

Question 32.
Draw the Allotropes of Carbon. Diamond
Answer:

Question 33.
Draw the Graphite.
Answer:

Question 34.
Draw the Buckminsterfullerene (₆₀C).
Answer:

Question 35.
Draw the Nanotubes. A.
Answer:

Question 36.
Draw the structures of Methane :
Answer:
Methane :

Question 37.
Draw the structures of Ethyne :
Answer:

Question 38.
Draw structures of the Ethane and electron dot structure of Chlorine.
Answer:
Ethane:

Chlorine:

Question 39.
Draw the electron dot structures of Ethanoic acid arid Ethyne (Acetylene).
Answer:
Ethanoic acid (Acetic acid) :

Structure of Ethyne (Acetylene) :

Question 40.
Draw the electronic dot structure of ethane molecule.
Answer:

Question 41.
Write the structures of the following compounds.
a) prop-l-ene
b) 2, 3-dimethyl butane
c) 3-hexene
d) 2-methyl prop-l-ene
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 5 Refraction of Light at Plane Surfaces.

AP State Syllabus SSC 10th Class Physics Important Questions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces 1 Mark Important Questions and Answers

Question 1.
Take a bright metal ball and make it black with soot on a candle flame. Immerse it in the water. Mention one observation. (AP June 2015)
Answer:
1) The ball shines.
2) The ball appears to raise up in water.

Question 2.
What is critical angle? (AP March 2015)
Answer:
The angle of incidence at which the light ray propagates from denser to rarer graze along interface is called critical angle of denser medium.

Question 3.
If refractive index of glass is \(\frac{3}{2}\), then what is speed of light in glass? (AP June 2016)
(OR)
Find the speed of light in a transparent medium, whose refractive index is 3/2.
Answer:
The refractive index of glass or transparent medium = \(\frac{3}{2}\)

Question 4.
Write any two questions about the ‘formation of mirages’. (AP June 2017)
Answer:

1. When does a mirage form?
2. How does a mirage form?

Question 5.
Optical Fibre Cable (OFC) are oftenly used in tele-communications. What is the working principle behind the OFC? (AP March 2017)
Answer:
Total Internal Reflection.

Question 6.
Among objects made of glass and diamond, which one shines more? Why? (AP June 2015)
Answer:
Diamond shines more because of low conical angle of 24.4°.

Question 7.
Suggest reasons for the phenomenon associated with the following : Twinkling of stars. (TS March 2015)
Answer:
Refraction of light is the reason for the twinkling of stars.

Question 8.
Draw the diagram showing the path of the ray when it travels from denser medium to rarer medium when the incident angle is more than the critical angle. (TS June 2016)
Answer:

Question 9.
Why does the light ray deviate in refraction? (AP SA-1:2019-20)
Answer:
Light ray always chooses the path of least time to travel. Hence speed of light changes at interface of two media. So, the light ray deviate in refraction.

Question 10.
Name the phenomenon involved in the function of optical fibre. (AP SCERT : 2019-20)
Answer:
Total Internal Reflection.

Question 11.
What is Fermat’s principle?
Answer:
The light ray always travels in a path which needs shortest possible time to cover the distance between the two given points.

Question 12.
What happens when light travels from one medium to another medium?
Answer:
It bends towards or away from normal.

Question 13.
When does speed of light decrease?
Answer:
When it travels from rarer to denser medium.

Question 14.
What do you mean by denser medium?
Answer:
The medium which has more optical density.

Question 15.
What is refraction?
Answer:
The process of changing speed when light travels from one medium to another is called refraction of light.

Question 16.
Which quantity will compare the refractive indices of two media?
Answer:
Relative refractive index.

Question 17.
What is relative refractive index?
Answer:
It is the ratio of refractive index of second medium to refractive index of first medium.

Question 18.
When does a light ray bend away from normal?
Answer:
When a light ray moves from denser to rarer medium it bends away from normal.

Question 19.
When will angle of refraction be equal to 90°?
Answer:
When angle of incidence is equal to critical angle then angle of refraction will be equal to 90°.

Question 20.
When angle of incidence of light ray is greater than critical angle, what happens?
Answer:
Light ray undergoes total internal reflection.

Question 21.
What happens to refractive index of air with height?
Answer:
Refractive index of air increases with height.

Question 22.
Which has greater refractive index between these?
1) cool air at the top
2) hotter air just above the road
Answer:
Cooler air has greater refractive index due to more density.

Question 23.
What is mirage?
Answer:
The virtual images of distant high objects cause the optical illusion called mirage.

Question 24.
What happens to a light ray when it falls perpendicular to one side of the slab surface?
Answer:
It comes out without any duration.

Question 25.
What are the conditions for total internal reflection?
Answer:

1. The rays of light must travel from denser to rarer medium.
2. The angle of incidence of denser medium must be greater than critical angle.

Question 26.
What is meant by a vertical shift?
Answer:
When a ray emerges out of a glass slab, it is parallel to the incident ray but is displaced laterally relative to incident ray. This shift of emergent ray is called vertical shift.

Question 27.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
It bends towards the normal. This is because it travels from an optically rarer to , optically denser medium.

Question 28.
When does Snell’s law fail?
Answer:
Snell’s law fails when light is incident normally on the surface of a refracting medium.

Question 29.
Why does a ray of light bend when it travels into another medium?
Answer:
It bends because its velocity changes when it moves from one medium to the other.

Question 30.
A pencil when dipped in a glass tumbler containing water appears to be bent at the interface of air and water. Explain why.
Answer:

• When light travels obliquely from one transparent medium to another, the direction of propagation of light changes due to refraction of light.
• In this case, light travels from denser medium to rarer medium, hence it bends away from the normal and the pencil appears to be bent.

Question 31.
Why does light travel in vacuum?
Answer:
Light travels in vacuum because it does not require medium for its propagation.

Question 32.
What are the factors on which refractive index depends?
Answer:

1. Nature of material
2. Wavelength of light used.

Question 33.
What does the ratio of sine of angle of incidence and sine of angle of refraction give?
Answer:
The ratio of sine of angle of incidence and sine of angle of refraction gives refractive index.

Question 34.
What is the relationship between critical angle and refractive index?
Answer:
The relationship between critical angle and refractive index

Question 35.
What is the relationship between angle of incidence and shift?
Answer:
As the angle of incidence increases, the shift also increases.

Question 36.
Why does a coin placed in a water appear to be raised?
Answer:
It is due to refraction of light.

Question 37.
Can you guess what happens when light travels from denser medium to rarer medium?
Answer:
The light ray bends away from normal.

Question 38.
A ray of light falls normally on a face of a glass slab. What are the values of angle of incidence and angle of refraction of this ray?
Answer:
Both angles are zero.

Question 39.
When does light ray from slab not undergo any deviation?
Answer:
The light ray that incidents perpendicular to one side of the slab surface comes out without any deviation.

Question 40.
What is the factor on which refraction depends?
Answer:
Refraction depends on optical density.

Question 41.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in vacuum to speed of light in medium.
\(\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}\)

Question 42.
In water filled vessel, the coin of the bottom can be seen at a height. Give reasons.
Answer:
Rising of coin when water is poured in a cylindrical transparent vessel.

Question 43.
Write one activity in showing the process of ‘total internal refraction’.
Answer:
Due to refraction of light speed of light changes when it travels from one medium to another medium.

Question 44.
Define “Glass Slab”.
Answer:
A thin glass slab is formed when a medium is isolated from its surroundings by two plane surfaces parallel to each other.

Question 45.
A ray of light is incident normally on a plane glass slab. What will be the angle of refraction and angle of deviation for the ray?
Answer:
The ray is incident normally on a plane glass slab. So there is no deviation of light ray. Therefore the angle of refraction and angle of deviation both have 0° values.

Question 46.
A light ray in passing from water to a medium (a) speeds up, (b) slows down. In each case get one example of the medium.
Answer:
a) Air, because its optical density is less than water,
b) Glass, because its optical density is more than water.

Question 47.
If an angle of refraction is 90°, what is the corresponding angle of incidence called?
Answer:
The angle of incidence is called critical angle.

Question 48.
If the angle of incidence is more than critical angle, what happens to light ray if the light ray travels from denser to rarer medium?
Answer:
The light ray undergoes total internal reflection.

Question 49.
The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?
Answer:
It means that light travels 2.42 times faster in vacuum than in diamond.

Question 50.
For the same angle of incidence 45°, the angle of refraction in two transparent media, I and II is 20° and 30° respectively. Out of I and II, which medium is optically denser and why?
Answer:
Medium I is optically dense as angle of refraction is lesser in it, hence light bends towards normal.

Question 51.
For which colour of white light is the refractive index maximum and for which colour of white light is the refractive index minimum?
Answer:
The refractive index is maximum for violet because its wavelength is least.
The refractive index is minimum for red because its wavelength is maximum.

Question 52.
Correct the statement. “If the angle of incidence is greater than the critical angle the light is refracted when it falls on the surface from a denser medium to rarer medium”.
Answer:
If the angle df incidence is greater than the critical angle the light undergoes total internal reflection when it falls on the surface from a denser medium to rarer medium.

Question 53.
A light ray passes from medium 1 to medium 2. Which of the following quantities of refracted ray will differ from that of the incident ray?
Speed, intensity, frequency, wavelength.
Answer:
Speed, intensity and wavelength will differ from that of incident ray.

Question 54.
The refractive indices of alcohol and turpentine oil with respect to air are 1.36 and 1.47 respectively. Find the refractive index of turpentine oil with respect to alcohol. Which one of these permits the light to travel faster?
Answer:
The refractive index of turpentine oil with respect to alcohol = \(\frac{1.47}{1.36}\) = 1.08.

The refractive index increases when the speed of light decreases. So light travels faster in alcohol as its refractive index is less.

Question 55.
Light enters from air to diamond which has refractive index of 2.42. Calculate the speed of light in diamond, if speed of light in air 3 × 10⁸ ms^-1.
Answer:
Absolute refractive index = \(\frac{c}{v}\)
2.42 = \(\frac{3 \times 10^{8}}{\mathrm{v}}\) ⇒ v = 1.24 × 10⁸ ms^-1.

Question 56.
A glass block 3.0 cm thick is placed over a stamp. Calculate the height through which image of stamp is raised. Refractive index of glass is 1.54.
Answer:

Question 57.
The refractive index of water is 4/3. Calculate the critical angle for water – air interface (sin 49 = 3/4).
Answer:

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 2 Marks Important Questions and Answers

Question 1.
A ray of light enters from air to a medium X. The speed of light in the medium is 1.5 × 10⁸ m/s and the speed of light in air is 3 × 10⁸ m/s.
Find the refractive index of the medium X. (TS March 2015)
Answer:

Question 2.
What are the applications of optical fibres?
(OR)
Write two uses of fibre optics in daily life. (TS June 2016)
Answer:
Applications oruses of optical fibres :

1. Light pipes using optical fibres may be used to see places which are difficult to reach things such as inside of a human body.
2. The other important application of fibre optics is to transmit communication signals through light pipes.

Question 3.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

• When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
• Lens do not diverge or converge the light.
• Light do not get refracted when it passes through that lens.

Question 4.
Why does the light ray travel slowly in diamond when compared to vacuum? (AP SA-I : 2019-20)
Answer:

• Refractive index of diamond (2.42) is greater than that of vacuum (1).
• Speed of light is inversely proportional to refractive index of substances.
• Hence, light ray travel slowly in diamond when compared to vacuum.

Question 5.
Write about laws of refraction.
(OR)
Write the laws of refraction.
Answer:
Laws of refraction :

1. The incident ray, the refractive ray and the normal to interface of two transparent media at a point of incidence lie in the same plane.
2. During refraction light follows Snell’s law, i.e., the ratio of sine of angled of incidence to sine of angle of refraction is constant.
   n₁ sin i = n₂ sin r (OR) \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = constant.

Question 6.
What is total internal reflection ? What are the applications of total internal reflection?
Answer:
When the angle of incidence is greater than critical angle, the light ray is reflected into denser medium at interface i.e., light never enters rarer medium. This phenomenon is called total internal reflection.
1) Brilliance of diamonds :
Total internal reflection is the main cause for brilliance of diamonds. The critical angle of diamonds is very low (24.4°). So if a light ray enters a diamond it is very likely to get total internal reflection which makes the diamond shine brilliant.

2) Optical fibres:
Total internal reflection is the basic principle for working of optical fibre.

Question 7.
What are optical fibres? How do they work?
Answer:

• An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer.
• A bunch of such thin fibres forms a light pipe.

Working :

1. Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall.
2. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
3. The light is thus transmitted along the fibre.

Question 8.
How can a patient’s stomach be viewed by using optical fibres?
(OR)
How do you observe patient’s stomach by using a light pipe?
Answer:

• The patient’s stomach can be viewed by inserting one end of a light pipe into the stomach through the mouth.
• Light is sent down through one set of fibres in the pipe.
• This illuminates the inside of the stomach.
• The light from the inside travels back through another set of fibres in the pipe and the viewer gets the image at the outer end.

Question 9.
State four differences between reflection and total internal reflection.
Answer:

Question 10.
The figure shows refraction and emergence of a ray of light incident on a rectangular glass slab. Copy the diagram and mark the lateral displacement of the incident ray. Name the two factors on which the lateral displacement depends.

Answer:
The lateral displacement depends on

1. The angle of incidence of the incident ray PQ, on the slab and
2. The thickness of the glass slab.

The perpendicular distance between the emergent forward.

Question 11.
1) What happens to a ray of light when it travels from one equal refractive indices?
2) State the cause of refraction of light.
Answer:
1) No refraction or bending would take place. The light will travel in a straight line.

2) The refraction occurs due to change in speed of light as it enters from one medium to another.

Question 12.
A coin placed at the bottom of a tank appears to be raised when water is poured into it. Explain.
Answer:

• It happens due to the phenomenon of refraction of light.
• When the rays of light from the coin, in the denser medium fall on the interface separating the two media, the rays of light move away from the normal after refraction.
• The point from which the refracted rays appear to come gives the apparent position of the coin.
• As the rays appear to come from a point above the coin, therefore, the coin seems to be raised.

Question 13.
Define refractive index. Explain the relationship between the refractive index of the medium and to the speed of light in the medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in that medium is defined as refractive index ‘n’ with respect to the vacuum. It is also called absolute refractive index.

When refractive index of a medium is high, then the speed of light is low and vice-versa.

Question 14.
Explain lateral shift and vertical shift.
Answer:
Lateral shift:
The distance between incident and emergent ray is called lateral shift.

Vertical shift :
The perpendicular distance between object and its image is called vertical shift.

Question 15.
During refraction of light, which of the following quantities does not change.
(1) velocity,
(2) wavelength,
(3) frequency,
(4) amplitude.
Answer:
During refraction of light velocity of light changes and also wavelength and amplitude. Frequency does not change during refraction.

Question 16.
The upper surface of water contained in a beaker and held above the eye level appears silvery. Why?
Answer:
Critical angle for water is 48°. The rays of light entering in water from below, suffer refraction. If these rays strike the water-air surface at an angle which is greater than 48°, they get totally internally reflected. These rays on emerging out of water, appear to come from the upper surface of water, which in turn appear silvery.

Question 17.
Why don’t the planets twinkle?
Answer:

• The planets are much closer to the earth, and are thus seen as extended sources.
• We can consider a planet as a collection of a large number of point-sized sources of light.
• The total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying twinkling effect

Question 18.
Why did an empty test tube placed obliquely in water, appears filled with mercury, when seen from above?
Answer:
When the rays of light travelling through water they strike the water glass interface of test tube at an angle, which is more than critical angle for water, they suffer total internal reflection. When these totally reflected rays reach eye, then to the eye they appear as they come from surface of test tube, which in turn appears filled with mercury.

Question 19.
Why are the bubbles rising up the fish tank appear silvery?
Answer:
When the rays of light travelling through water they strike the water air interface of the bubble at an angle, which is greater than critical angle for water, they get totally internally reflected. These reflected rays on reaching the eye appear to come from air bubble, which in turn appears silvery.

Question 20.
Why does a crack in a window pane appear silvery?
Answer:
There is always some amount of air present in the crack. When the rays of light travelling through glass, strike the glass, the glass air interface at an angle, greater than critical angle of glass, they are totally internally reflected. When these reflected rays reach eye, then to the eye they appear to come from the crack, which in turn appears silvery.

Question 21.
Explain why a straight stick appears to be bent when dipped in water.
Answer:

• Suppose two rays originate from the end of the stick in water.
• As these rays get refracted into the air, they bend away from the normal.
• When these two refracted rays are produced backwards they seem to meet at a point higher than the end of stick.
• This point gives the apparent position of the end of the stick. Thus, the stick appears to be bent.

Question 22.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

• Suppose two rays originate from the bottom of the pond. As these rays get refracted into the air, they bend away from the normal.
• When these two refracted rays are produced backwards they seem to meet at a point higher than the bottom of the pond.
• This point gives the apparent position of the bottom of the pond.
• Thus, the pond appears to be shallower.
• This effect is absent if the pond is viewed normally.

Question 23.
Frame some questions to know about the formation of mirages.
Answer:

1. What are mirages?
2. What is the principle involved in mirages?
3. Can mirages be photographed?
4. Where does the water on the road go?