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NCERT Exemplar Class 10 Science Solutions Chapter 11 The Human Eye and the Colourful World

NCERT Exemplar Class 10 Science Solutions Chapter 11 The Human Eye and the Colourful World

Edited By Sumit Saini | Updated on Sep 09, 2022 02:59 PM IST | #CBSE Class 10th
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NCERT exemplar Class 10 Science solutions chapter 11 discusses the human eye in detail. The formation of rainbow and primary colours are also discussed in the chapter. The Class 10 Science NCERT exemplar chapter 11 solutions prepared by our experienced subject matter experts are designed keeping in mind the difficulties faced by the students while studying this chapter of NCERT Class 10 Science Book. These NCERT exemplar Class 10 Science chapter 11 solutions deliver comprehensive solutions for a sounder learning of concepts of the human eye and colourful world. The NCERT exemplar Class 10 Science solutions chapter 11 are compatible with the CBSE Syllabus for Class 10.

This Story also Contains
  1. NCERT Exemplar Class 10 Science Solutions Chapter 11-MCQ
  2. NCERT Exemplar Class 10 Science Solutions Chapter 11-Short Answer
  3. NCERT Exemplar Class 10 Science Solutions Chapter 11-Long Answer
  4. NCERT Exemplar Class 10 Science Solutions Chapter 11 - Human Eye and Colourful Important Topics:
  5. NCERT Class 10 Exemplar Solutions for Other Subjects:
  6. NCERT Class 10 Science Exemplar Solutions for Other Chapters:
  7. Features of NCERT Exemplar Class 10 Science Solutions Chapter 11 - Human Eye and Colourful :

NCERT Exemplar Class 10 Science Solutions Chapter 11-MCQ

Question:1

A person cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using a lens of power
(a) + 0.5 D
(b) – 0.5 D
(c) + 0.2 D
(d) – 0.2 D
Ans. (b)
As the person has the eye defect, myopia, therefore, a concave lens has to be used whose focal length will be, f = -2 m.
Thus,
Power
P = 1/f
P = 1/-2
P = -0.5 D
Therefore option (B) is correct

Question:2

A student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book. Which of the following statements is correct?
(a) The near point of his eyes has receded away
(b) The near point of his eyes has come closer to him
(c) The far point of his eyes has come closer to him
(d) The far point of his eyes has receded away
Ans. (a)

The student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book.
He is suffering from hypermetropia or far sightedness. He can see distant objects clearly but cannot see nearby objects distinctly.
Therefore option (a) is correct

Question:3

A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident on the prism as shown in Figure 11.1. In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky?

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Ans.
Given that third colour from the top corresponds to the colour of the sky, i.e., blue.
In normal orientation, the third colour from top is yellow (as shown in the figure).
The third colour from bottom is blue.

So we have to rotate the prism upside down so that blue comes at the third position from top. We can do this as follows:

Therefore option (ii) is correct

Question:4

At noon the sun appears white as
(a) light is least scattered
(b) all the colours of the white light are scattered away
(c) blue colour is scattered the most
(d) red colour is scattered the most
Ans. (a)
At noon the sun appears white because the light from the sun is directly overhead us. The light travels relatively shorter distance during the noon.
Scattering of light is reduced as the distance is reduced. So, the sun appears white as only a little of the blue and violet colours are scattered.
Therefore option (a) is correct

Question:5

Which of the following phenomena of light are involved in the formation of a rainbow?
(a) Reflection, refraction and dispersion
(b) Refraction, dispersion and total internal reflection
(c) Refraction, dispersion and internal reflection
(d) Dispersion, scattering and total internal reflectio
Ans. (c)
A rainbow is caused by dispersion, refraction and internal reflection of sunlight by tiny water droplets.
The tiny droplets of water present in the atmosphere act as a prism after rainfall, for the rays coming from the Sun. Subsequently, the sunlight after striking the surface of these droplets gets refracted and dispersed into its seven colours. After this, the light rays are subjected to total internal reflection (complete reflection of a ray of light within a medium from the surrounding surfaces back into the medium).
At that point, the rays are again refracted when they come out of the water droplet.
Hence, rainbow formation is the combined effect of the refraction, dispersion, and total internal reflection of light.

Therefore option (c) is correct

Question:6

Twinkling of stars is due to atmospheric
(a) dispersion of light by water droplets
(b) refraction of light by different layers of varying refractive indices
(c) scattering of light by dust particles
(d) internal reflection of light by clouds
Ans. (b)
The light of stars undergoes refraction continuously on entering the earth's atmosphere, before it reaches the earth.
Now we know that the apparent position of the star changes or fluctuates, so the path of rays of light coming from the star goes on changing slightly. That is why the amount of starlight entering the eye flickers. The star sometimes appears brighter and at other times, it appears fainter. This makes the star twinkle for us.
Hence the twinkling of a star is due to atmospheric refraction of lights of stars.
Therefore option (b) is correct

Question:7

The clear sky appears blue because
(a) blue light gets absorbed in the atmosphere
(b) ultraviolet radiations are absorbed in the atmosphere
(c) violet and blue lights get scattered more than lights of all other colours by the atmosphere
(d) light of all other colours is scattered more than the violet and blue colour lights by the atmosphere
Ans. (c)
Amount of scattering is inversely proportional to fourth power of wavelength.
The wavelengths of colours are given below:

Colour

Approximate wavelength (nm)

Red

650

Orange

600

Yellow

580

Green

550

Cyan

500

Blue

450

Violet

400

Based on the above data, blue and violet colours scatter the most.
Hence, the clear sky appears blue due to scattering of sunlight.
Therefore option (c) is correct.

Question:8

Which of the following statements is correct regarding the propagation of light of different colours of white light in air?
(a) Red light moves fastest
(b) Blue light moves faster than green light
(c) All the colours of the white light move with the same speed
(d) Yellow light moves with the mean speed as that of the red and the violet light
Ans. (c)
The propagation of light of different colours of white light in air or vacuum move with the same speed but different wavelengths and frequencies.
Velocity of light is dependent on medium but it does not depend on the frequency and wavelength of light.
Therefore option (c) is correct

Question:9

The danger signals installed at the top of tall buildings are red in colour. These can be easily seen from a distance because among all other colours, the red light
(a) is scattered the most by smoke or fog
(b) is scattered the least by smoke or fog
(c) is absorbed the most by smoke or fog
(d) moves fastest in air
Ans. (b)
Amount of scattering is inversely proportional to fourth power of wavelength.
Wavelength of red colour is the largest so red colour is scattered the least by smoke or fog.
Hence the danger signals installed at the top of tall buildings are red in colour.
Thus it can be easily seen from a distance. This is the reason red is associated with danger and red is used as a stop signal as well.
Therefore option (b) is correct

Question:10

Which of the following phenomena contributes significantly to the reddish appearance of the sun at sunrise or sunset?
(a) Dispersion of light
(b) Scattering of light
(c) Total internal reflection of light
(d) Reflection of light from the earth
Ans. (b)
Amount of scattering is inversely proportional to fourth power of wavelength.
Wavelength of red colour is the largest so red colour is scattered the least by smoke or fog.
The reddish appearance of the sun at sunrise or sunset is due to least scattering of red light.
Therefore option (b) is correct

Question:11

The bluish colour of water in deep sea is due to
(a) the presence of algae and other plants found in water
(b) reflection of sky in water
(c) scattering of light
(d) absorption of light by the sea
Ans. (c)
Amount of scattering is inversely proportional to fourth power of wavelength.
Wavelength of blue colour is the lowest so blue colour is scattered the most.
Hence, the bluish colour of water in deep sea is due to scattering of light. The very fine particles in water scatter mainly blue light.
The red, green, orange and yellow having higher wavelengths get absorbed more easily by water than blue because of shorter wavelength.
Therefore option (c) is correct

Question:12

When light rays enter the eye, most of the refraction occurs at the
(a) crystalline lens
(b) outer surface of the cornea
(c) iris
(d) pupil
Ans. (b)
At the point when the light rays enters the eye through a thin membrane, it forms a transparent bulge on the front surface of the eyeball, called the cornea.
Most refraction in the eye happens when light rays travel through the cornea.
Therefore option (b) is correct

Question:13

The focal length of the eye lens increases when eye muscles
(a) are relaxed and lens becomes thinner
(b) contract and lens becomes thicker
(c) are relaxed and lens becomes thicker
(d) contract and lens becomes thinner
Ans. (a)
When eye muscles are relaxed and become thinner, the focal length of the eye lens increases.
Similarly when eye muscles are contracted, focal length decreases.
The sharp image of the distant object is formed at the retina by adjusting the focal length of eyelens. This enables us to focus accurately on distant objects.

Therefore option (a) is correct

Question:14

Which of the following statement is correct?
(a) A person with myopia can see distant objects clearly
(b) A person with hypermetropia can see nearby objects clearly
(c) A person with myopia can see nearby objects clearly
(d) A person with hypermetropia cannot see distant objects clearly
Ans. (c)
Hypermetropia - far sightedness.
Myopia - short sightedness.
A person with myopia can see nearby objects clearly.
A person with hypermetropia can see distant objects clearly.
Therefore option (c) is correct

NCERT Exemplar Class 10 Science Solutions Chapter 11-Short Answer

Question:15

Draw ray diagrams each showing
(i) myopic eye and
(ii) hypermetropic eye.
Ans:

Question:16

A student sitting at the back of the classroom cannot read clearly the letters written on the blackboard. What advice will a doctor give to her? Draw ray diagram for the correction of this defect.
Ans.
Myopia - short sightedness.
In myopia, a person can see nearby objects clearly but cannot see distant objects clearly.
The doctor advices a concave lens of suitable power to bring the image back onto the retina. This is shown as follows:

Question:17

How are we able to see nearby and also the distant objects clearly?
Ans.
Human eye has an eye lens to adjust its focal length. This is known as power of accommodation.
We are able to see nearby and also the distant objects clearly due to this power of accommodation of the eye.
When the muscles are relaxed, the lens becomes thin and its focal length increases.
When the muscles are contracted, the lens becomes thick and its focal length decreases.
This enables us to see the distant as well as nearby objects clearly.
This is what makes eye to see different objects at different distances clearly.

Question:18

A person needs a lens of power –4.5 D for correction of her vision.
(a) What kind of defect in vision is she suffering from?
(b) What is the focal length of the corrective lens?
(c) What is the nature of the corrective lens?
Ans.
(a) Power of lens is given as negative, so she is suffering from myopia.
(b) Power, P = -4.5 D, focal length, f =?
P = 1/f
f = 1/P
f = 1/-4.5 = -0.222m = -22.2cm
(c) If focal length is positive, then lens to be used is convex lens.
If focal length is negative, then lens to be used is concave lens.
The nature of the corrective lens required here is concave.

Question:19

How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.

Ans.
In normal orientation of a prism, we have:

If we place another prism in front of this prism, we get:

Angle of deflections of both the prisms is equal and opposite. The first prism splits the white light in seven constituent colours, and the second prism combines these colours back to a single ray to give white light as the emergent light.

Question:20

Draw a ray diagram showing the dispersion through a prism when a narrow beam of white light is incident on one of its refracting surfaces. Also indicate the order of the colours of the spectrum obtained.
Ans.
The ray diagram is as follows:

Order of colours from bottom to top: remembering keyword VIBGYOR
V - Violet, I - Indigo, B - Blue, G - Green, Y - Yellow, O - Orange and R - Red.

Question:21

Is the position of a star as seen by us its true position? Justify your answer.
Ans.
The light of stars undergoes refraction continuously on entering the earth's atmosphere, before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index.
The atmosphere bends starlight towards the normal, so the apparent position of the star is slightly different from its actual position. So the apparent position of the star changes or fluctuates.
So, the position of a star as seen by us is not its true position.
Thus, star appears slightly higher than its actual position

Question:22

Why do we see a rainbow in the sky only after rainfall?
Ans.
A rainbow is caused by dispersion, refraction and internal reflection of sunlight by tiny water droplets.
We see a rainbow in the sky only after rainfall due to dispersion of sunlight by tiny water droplets, present in the atmosphere due to rainfall. These droplets act as a prism after rainfall, for the rays coming from the Sun. Subsequently, the sunlight after striking the surface of these droplets gets refracted and then gets dispersed into its seven constituent colours. After this, the light rays are subjected to total internal reflection (complete reflection of a ray of light within a medium from the surrounding surfaces back into the medium).
At that point, the rays are again refracted when they come out of the water droplet.
Hence, the rainbow is formed.

Question:23

Why is the colour of the clear sky blue?
Ans.
This is due to the Rayleigh scattering of sunlight.
Amount of scattering is inversely proportional to fourth power of wavelength.
Wavelength of blue colour is the lowest so blue colour is scattered the most.
The molecules of air and other fine particles in the atmosphere have smaller size than the wavelength of visible light. These particles are more effective in scattering light of shorter wavelengths (blue) than light of longer wavelengths (red).

Question:24

What is the difference in colours of the Sun observed during sunrise/sunset and noon? Give explanation for each.
Ans.
At noon the sun appears white because the light from the sun is directly overhead us. The light travels relatively shorter distance during the noon. Scattering of light is reduced as the distance is reduced.
Amount of scattering is inversely proportional to fourth power of wavelength.
So, the sun appears white as only a little of the blue and violet colours are scattered.
The sun looks reddish during sunrise/sunset, because at this stage, rays from the sun have to travel a much larger part of atmosphere. As the distance increases in this case, scattering of light increases. So, red colour having largest wavelength is scattered the least.

NCERT Exemplar Class 10 Science Solutions Chapter 11-Long Answer

Question:25

Explain the structure and functioning of Human eye. How are we able to see nearby as well as distant objects?
Ans.
Structure of eye:

Cornea is transparent. It is responsible for most of the refraction of light rays towards the retina.
Iris is a dark, circular muscular diaphragm behind the cornea.
It controls the pupil size which in turn controls the amount of light entering the eye.
Pupil is surrounded by the iris. It regulates and controls the amount of light entering the eye.
Retina is the innermost layer of the eye. It contains an outer pigmented layer and an inner nervous layer which contains the photo receptors (rods
and cones).
Cone cells are coned shaped cells which detect colour
Rod cells are rod shaped cells which are light sensitive.
Optic nerve connects the eye to the brain.
Nerve fibres carry impulses from the retina to the visual cortex.
Eye lens is made up of a fibrous, jelly-like material. It is transparent and has biconvex structure. It forms an inverted real image of the object on
the retina and also separates the aqueous and the vitreous humours.
Vitreous humour is clear, semi-solid supporting the eye ball
Human eye has an eye lens to adjust its focal length. This is known as power of accommodation.
We are able to see nearby and also the distant objects clearly due to this power of accommodation of the eye.
When the muscles are relaxed, the lens becomes thin and its focal length increases.
When the muscles are contracted, the lens becomes thick and its focal length decreases.
This enables us to see the distant as well as nearby objects clearly.
This is what makes eye to see different objects at different distances clearly.

Question:26

When do we consider a person to be myopic or hypermetropic? Explain using diagrams how the defects associated with myopic and hypermetropic eye can be corrected?
Ans.
Myopia - short sightedness.
A person with myopia can see nearby objects clearly.
Light from a distant object forms an image before it reaches the retina
This is because the eye is too long or the cornea or crystalline lens is too short.

To correct this defects a concave lens is placed in front of a myopic eye, moving the image back to the retina and clarifying the image as shown below:

Hypermetropia - far sightedness.
A person with hypermetropia can see distant objects clearly.
Image of a nearby object is formed behind the retina.
This is because the eye is too short or the cornea or crystalline lens does not refract the light enough.

To correct this defect, a convex lens is placed in front of a hypermetropic eye.
The image is moved forward and focuses correctly on the retina.

Question:27

Explain the refraction of light through a triangular glass prism using a labelled ray diagram. Hence define the angle of deviation.
Ans.
The refraction of light through a triangular glass prism is shown below

A – Angle of prism
D – Angle of deviation
i – incident angle
r – angle of refraction
e – emergent angle
A ray of light PE enters at the first surface AB (from air to glass).
As light travels from rarer to denser medium, it bends towards the normal.
Hence, EF bends towards the normal.
At the second surface AC, the light enters from denser to rarer medium, so it bends away from the normal.
Hence, FS bends away from the normal.
The angle made by extending incident ray with the emergent ray is called angle of deviation

Question:28

How can we explain the reddish appearance of sun at sunrise or sunset? Why does it not appear red at noon?
Ans.
Amount of scattering is inversely proportional to fourth power of wavelength.
Wavelength of red colour is the largest so red colour is scattered the least by smoke or fog.
The reddish appearance of the sun at sunrise or sunset is because the wavelength of visible light from the sun near the horizon is greater than the scattering of light by the molecules of air and other fine particles in the atmosphere.
It passes through larger distance in the earth's atmosphere and thicker layers of air before reaching our eyes. Most of the blue light (shorter wavelengths) are scattered away by the particles.
So, only red light, being of highest wavelength reaches us which gives reddish appearance of sun at sunrise or sunset.
At noon the sun appears white because the light from the sun is directly overhead us. The light travels relatively shorter distance during the noon.
Scattering of light is reduced as the distance is reduced. So, the sun appears white as only a little of the blue and violet colours are scattered.

Question:29

Explain the phenomenon of dispersion of white light through a glass prism, using suitable ray diagram.
Ans.
When the white light is passed through a pin, it get reduced to a narrow beam.
When this narrow beam is made to fall on a triangular glass prism and a white screen is held on the other side of the prism, we observe a band of seven colours on the screen. This phenomenon is called dispersion of white light.
Order of colours from bottom to top: remembering keyword VIBGYOR
V - Violet, I - Indigo, B - Blue, G - Green, Y - Yellow, O - Orange and R - Red.

Amount of scattering is inversely proportional to fourth power of wavelength.
The red colour (with maximum wavelength) bends the least on passing through the prism and violet colour (with minimum wavelength) bends through maximum angle on passing through the prism. This is why red is at top and violet is at bottom.
Thus, the phenomenon of splitting of white light into its constituent seven colours on passing through a glass prism is known as dispersion of light.

Question:30

How does refraction take place in the atmosphere? Why do stars twinkle but not the planets?
Ans.
The refraction take place in the atmosphere because of the gradually changing refractive index of the different layers of atmosphere. The layers at the top is optically rare, while the layers at the bottom is optically denser.
When light travels through different layers of the atmosphere, refraction takes place. Since light passes from rarer to denser mediums, it tends to bend towards the normal.
So, the light of stars undergoes refraction continuously on entering the earth's atmosphere, before it reaches the earth. The atmosphere bends starlight towards the normal, so the apparent position of the star is slightly different from its actual position. So the apparent position of the star changes or fluctuates.
So, the position of a star as seen by us is not its true position.
Thus, star appears slightly higher than its actual position because of atmospheric refraction.

As discussed above, the light of stars undergoes refraction continuously on entering the earth's atmosphere, before it reaches the earth.
Now we know that the apparent position of the star changes or fluctuates, so the path of rays of light coming from the star goes on changing slightly. That is why the amount of starlight entering the eye flickers. The star sometimes appears brighter and at other times, it appears fainter. This makes the star twinkle for us.
Hence the twinkling of a star is due to atmospheric refraction of light of stars.

NCERT Exemplar Class 10 Science Solutions Chapter 11 - Human Eye and Colourful Important Topics:

  • The reflection of light through a prism and the method of dispersion.
  • NCERT exemplar class 10 Science solutions chapter 11 discusses about the formation of rainbow.
  • Human eye in detail.
  • The defects in vision of human eye like near-sightedness and farsightedness.
  • Choice of lens to remove these defects.
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NCERT Class 10 Exemplar Solutions for Other Subjects:

NCERT Class 10 Science Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Science Solutions Chapter 11 - Human Eye and Colourful :

These class 10 science NCERT exemplar chapter 11 solutions provide a basic understanding of the human eye. We know that white light comprises infinite colours. In this chapter, we will understand the reflection of light through a prism and learn the method of dispersion. The books such as NCERT Class 10 Science, Physics question bank, S. Chand by Lakhmir Singh and Manjit Kaur, etc. can be easily attempted for Human Eye and Colourful World related practice problems once the students have gone through NCERT exemplar Class 10 Science chapter 11 solutions Human Eye and Colourful World.

Careers 360 has kept in mind the difficulties related to the internet connectivity faced by students and has introduced the feature of NCERT exemplar Class 10 Science solutions chapter 11 pdf download to explore NCERT exemplar Class 10 Science chapter 11 in offline mode.

Check NCERT Solutions for questions given in the book

Must check NCERT Solution Subject Wise

Read more NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. What is farsightedness and how is it removed?

It is a defect of the eye, in which any human cannot see the objects near the eye. It can be removed by using a converging lens which will have a positive focal length.

2. What is near-sightedness and how is it removed?

It is a defect of the eye, in which any human cannot see the object far from the eye. It can be removed by using diverging lens which will have negative focal length

3. How many colours are there in white light?

There are infinite colours in white lights which are divided in seven bands of colour named violet, indigo, blue, green, yellow, orange and red.

4. What is dispersion?

 Fragmentation of white colour in all different constituents it’s called dispersion.

Phenomenon of dispersion can be easily demonstrated with the help of a prism and it is one of the reasons for rainbow formation

5. Will these questions of exemplar require a supplementary understanding of Human Eye and Colourful World?

NCERT exemplar Class 10 Science solutions chapter 11 is sufficient to understand and practice the questions based on the concept of Human Eye and Colourful World.

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Questions related to CBSE Class 10th

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

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Read Complete Answer
sai raja 01 September,2024
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024
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Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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