Questions Related to NCERT

Updated on May 31, 2025 | By Learnzy Academy

Q201. Why does the sky appear dark instead of blue to an astronaut?

The sky appears dark to an astronaut because there is no atmosphere in space to scatter sunlight. On Earth, the atmosphere scatters blue light in all directions, making the sky look blue. But in space, without air or atmosphere, there is no scattering, so the sky looks black or dark.

Q202. Explain why the planets do not twinkle.

Planets do not twinkle because they are closer to the Earth and appear as small discs, not points of light like stars. When their light passes through the atmosphere, the bending of light is averaged over the whole disc, so their brightness and position stay steady. That is why planets shine with a steady light and do not twinkle.

Q203. Why do stars twinkle?

Stars twinkle because their light passes through the Earth’s atmosphere. The atmosphere has layers of air with different temperatures and densities, which bend the light in different directions. This makes the stars look like they are changing brightness and position, causing them to twinkle.

Q204. What happens to the image distance in the eye when we increase the distance of an object from the eye?

When we increase the distance of an object from the eye, the image distance inside the eye stays almost the same because the retina is fixed. The eye changes the shape of the lens to focus the image properly on the retina.

Q205. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

A normal eye cannot see objects clearly if they are closer than 25 cm because the eye lens cannot become thick enough to focus the image on the retina. This 25 cm is called the near point, which is the shortest distance at which the eye can see clearly. Objects closer than this appear blurry.

Q206. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

The far point of the myopic person is 80 cm.
To correct myopia, a concave lens is used.
Power of the lens = –100 / 80 = –1.25 dioptres.

Hence, the lens needed is a concave lens with power –1.25 dioptres.

Q207. A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Given - 
Power for distant vision = –5.5 dioptres
Power for near vision = +1.5 dioptres

Using the formula:
Power (P) = 100 / focal length (f in cm)
So, focal length (f) = 100 / Power (P)

(i) For distant vision:
f = 100 / (–5.5) = –18.18 cm

(ii) For near vision:
f = 100 / (+1.5) = 66.67 cm

Hence the focal length of the lens required for correcting:

Distant vision = –18.18 cm  &  Near vision = +66.67 cm

Q208. The change in focal length of an eye lens is caused by the action of the (a) pupil. (b) retina. (c) ciliary muscles. (d) iris

The correct answer is (c) ciliary muscles.

The ciliary muscles control the change in the focal length of the eye lens.

Q209. The least distance of distinct vision for a young adult with normal vision is about (a) 25 m. (b) 2.5 cm. (c) 25 cm. (d) 2.5 m

The correct answer is (c) 25 cm.

The least distance of distinct vision for a young adult with normal vision is about 25 centimeters.

Q210. The human eye forms the image of an object at its (a) cornea. (b) iris. (c) pupil. (d) retina.

The correct answer is: (d) retina

The human eye forms the image of an object on the retina.

Q211. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to (a) presbyopia. (b) accommodation. (c) near-sightedness. (d) far-sightedness

The correct answer is: (b) accommodation

The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This ability is called accommodation.

Q212. A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

If a student has difficulty reading the blackboard while sitting in the last row, the student is suffering from myopia (short-sightedness).

In myopia, distant objects appear blurry because their images are formed in front of the retina. This defect can be corrected by using a concave lens, which helps to focus the image on the retina and makes distant objects appear clear.

Q213. What is the far point and near point of the human eye with normal vision?

For a human eye with normal vision:

  1. Far point is infinity. This means a normal eye can see distant objects clearly, no matter how far they are.
  2. Near point is about 25 centimeters from the eye.This is the minimum distance at which the eye can see objects clearly without strain.

Q214. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

A person with a myopic eye cannot see objects beyond 1.2 m clearly. To correct this defect, a concave lens should be used.

A concave lens helps to diverge the light rays so that they appear to come from the person's far point (1.2 m) and can be focused properly on the retina.

Q215. What is meant by power of accommodation of the eye?

Power of accommodation of the eye is the ability of the eye to adjust the focal length of its lens to see both nearby and distant objects clearly.

This is done by the ciliary muscles, which change the shape of the eye lens:

  • For distant objects, the lens becomes thin.
  • For nearby objects, the lens becomes thick.

A normal human eye can see objects clearly from about 25 cm to infinity.

Q216. What determines the rate at which energy is delivered by a current?

The rate at which energy is delivered by a current is called power.

Power (P) = Voltage (V) × Current (I)

So, the rate of energy delivery depends on both the voltage and the current.

Q217. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Given:
Resistance (R) = 20 ohms
Current (I) = 5 amperes
Time (t) = 30 seconds

Formula:
Heat developed (H) = I² × R × t
H = 5² × 20 × 30 = 25 × 20 × 30 = 15000 joules

Hence heat developed in 30 seconds = 15000 joules

 

Q218. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V

Given:
Charge (Q) = 96000 coulombs
Potential difference (V) = 50 volts

Formula:
Heat generated (H) = Voltage (V) × Charge (Q)
H = 50 × 96000 = 4800000 joules

Hence the heat generated is 4800000 joules (4.8 million joules).

Q219. Why does the cord of an electric heater not glow while the heating element does?

The cord of an electric heater does not glow because it is made of a material with low resistance, so it does not heat up much.

The heating element glows because it is made of a material with high resistance, which causes it to get very hot and produce light when current passes through it.

Q220. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Advantages of connecting electrical devices in parallel instead of in series:

  1. Each device gets the full voltage of the battery:
    In parallel, every device gets the same voltage as the battery voltage, so all devices work properly.
  2. Devices work independently:
    If one device stops working or is switched off, the others continue to work because they have their own separate paths.
  3. Current is divided among devices:
    Total current is split according to resistance, preventing any one device from getting too much current.
  4. Adding more devices does not reduce voltage:
    You can add more devices without reducing the voltage across each device.

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