KSEEB SSLC Class 10 Sanskrit नंदिनी Solutions Karnataka State Syllabus

Expert Teachers at KSEEBSolutions.com has created KSEEB SSLC Class 10 Sanskrit नंदिनी Solutions Pdf Free Download of 10th Standard Karnataka Sanskrit Textbook Solutions Answers, SSLC Sanskrit Guide Notes Pdf, Textbook Questions and Answers, Model Question Papers with Answers, Study Material, are part of KSEEB SSLC Class 10 Solutions. Here we have given 1st Language KTBS Karnataka State Board Syllabus for Sanskrit 10th Class Textbook Solutions of Samskrita Nandini 3 Guide.

Sanskrit Guide for Class 10 Karnataka State Syllabus (1st Language)

KSEEB Solutions for Class 10 Sanskrit 1st Language Gudie Notes Pdf Karnataka State Syllabus

Students can download Samskrita Nandini 3 Guide for Class 10 Karnataka State Syllabus of 1st Language Sanskrit.

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KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Students can download Class 10 Science Chapter 11 Human Eye and Colourful World Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Chapter 11 Human Eye and Colourful World

Question 1.
Draw a neat diagram of the human eye and label the parts.
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 1

Question 2.
What is the screen of the human eye called?
Answer:
The screen of the human eye is called retina.

Question 3.
What type of lens is present in the human eye? What is its function?
Answer:
In the human eye, there is a convex lens. The eye-lens provides the finer adjustment of focal length so that the light rays from objects get focussed on the retina of the eye.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 4.
What is cornea? What is its Junction?
Answer:
The transparent portion of the outermost coat of the human eye is called cornea. Cornea refracts most of the light rays that enter the eye and direct them to the retina.

Question 5.
What is pupil of the eye? Write its function.
Answer:
The opening of the iris through which light enters the eye is called pupil. The pupil regulates and controls the amount of light that enters the eye by adjusting its diameter. This is achieved by contraction and dilation of the iris.

Question 6.
What is the function of ciliary muscles of our eye?
Answer:
Ciliary muscles of our eye hold the eye lens in position. These muscles help to adjust the focal length of the eye lens for seeing both nearby and distant objects.

Question 7.
Name the diaphragm that controls the size of thepupiL
Answer:
The muscular diaphragm that controls the size of the pupil is called iris.

Question 8.
What is the nature of the image formed on the retina of the eye?
Answer:
The image formed on the retina is real, inverted and smaller than the object.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 9.
What is the function of the optic nerves?
Answer:
The optic nerves in our eyes transmit the image formed on the retina to the brain in the form of electrical signals.

Question 10.
What is meant by power of accommodation of the eve?
Answer:
Our eye can focus clearly all objects that are beyond 25 cm by adjusting the focal length of its lens. The ability of the eye lens to adjust its focal length is called power of accommodation.

Question 11.
What happens to the image distance in the eve when we increase the distance of an object from the eve?
Answer:
When the distance of an object from the eye is increased, the image distance remains same. The eye forms the image on the retina by adjusting the focal length of its lens.

Question 12.
How does the curvature of the eye lens change when eye is viewing

  1. Nearby objects
  2. Distant objects?

Answer:
The curvature of the eye lens will increase while viewing distant objects and the curvature of the eye lens will decrease while viewing nearby objects.

Question 13.
What is the near point of the human eve with normal vision?

OR

What is meant by least distance of distinct vision? What is its value?
Answer:
The minimum distance, at which objects can be seen most distinctly without strain is called the least distance of distinct vision. It is also called the near point of the eye. The value of least distance of distinct vision of the human eye is 25 cm.

Question 14.
Why is a normal eve not able to see objects placed closer than 25 cm clearly?
Answer:
Our eye can adjust the focal length of its eye lens to enable us to see both nearby and distant objects clearly. However, the focal length of the eye lens cannot be decreased below a certain minimum limit. Therefore, the object must be held at certain minimum distance to be seen clearly. This minimum distance is about 25 cm for a normal eye. Objects closer than this minimum distance cannot be seen without discomfort and strain to the eye.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 15.
What is far point of the human eve with normal vision?
Answer:
The farthest point up to which the eye can see objects clearly is called the far point of the eye. The far point of a normal human eye is infinity.

Question 16.
What is the range within which a normal human eye can see objects clearly?
Answer:
A. normal human eye can see all objects clearly which are between 25 cm and infinity.

Question 17.
What is cataract? How does it affect vision? How can it be corrected?
Answer:
Sometimes, the crystalline lens of our eye may lose its transparency and become milky and cloudy. This condition is called cataract. Cataract causes partial or complete loss of vision. Cataract is corrected and vision can be restored through a cataract surgery. In this surgery, the lens is removed and replaced by a synthetic lens.

Question 18.
Which are the common defects of vision that can be corrected by using spherical lenses?
Answer:
There are mainly three common defects of vision, which can be easily corrected with spherical lenses. They are:

  1. Myopia or near-sightedness,
  2. Hypermetropia or far-sightedness, and
  3. Presbyopia.

Question 19.
What is myopia? How is it caused?
Answer:
A defect of the eye that causes light to focus in front of the retina instead of directly on it, resulting in an inability to see distant objects clearly is called myopia. It is also called shortsightedness. A person with this defect has the far point nearer than infinity.
Myopia may arise due to any or both of the following reasons:

  1. Excessive curvature of the eye lens, and
  2. Elongation of the eyeball.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 20.
Draw a diagram showing the image formation in a normal eye and myopic eye while viewing distant objects.
Answer:
1. Normal eye:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 2

2. Myopia (short sighted):
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 3

Question 21.
How is myopia corrected? Show this by a suitable diagram.

OR

Name the lens used to correct myopia.

OR

Draw diagrams to show

  1. Far point of a myopic eye,
  2. Myopic eye and
  3. Correction for myopia.

Answer:
Myopia can be corrected by using spectacles fitted with concave lenses of suitable power.
1. Far point of a myopic eye:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 4

2. Myopic Eye:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 5

3. correction for myopia:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 6

Question 22.
A person is wearing spectacles with concave lenses. What is the defect of vision he is suffering from?
Answer:
The person who is wearing spectacles with concave lenses is suffering from a defect called myopia or short-sightedness. This person has difficulty in seeing distant objects clearly.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 23.
A person with a myopic eve cannot see objects beyond 1.2 m distinctly. What type of corrective lens should be used to restore proper vision?
Answer:
The person with myopic eye has inability to see distant objects clearly. This person has far point nearer than infinity. This defect can be corrected and vision can be restored by using spectacles fitted with concave lens of suitable power.

Question 24.
A student has difficulty readme the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
The student who has difficulty in reading the blackboard while sitting in the last row is having myopia or shortsightedness. This defect can be corrected by using spectacles fitted with concave lenses of suitable power.

Question 25.
The far point of a myopic person is 80 cm in front of the eve. What is the nature and power of the lens required to correct the problem?
Answer:
The person whose far point is 80 cm is having myopia as the far point of the eye is nearer than infinity. The person cannot see distant objects clearly. This problem can be corrected by using spectacles fitted with concave lenses (diverging lenses) of suitable power. Let us now calculate the power of the lens required to correct the problem.
Data: Object distance, u = -80 cm, image distance v = infinity (∞), focal length f = ?, power P = ?
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 7
Concave lens of power -1.25 D must be used to correct the defect.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 26.
How does the use of concave lenses in spectacles restore the vision of distant objects in a myopic person?
Answer:
A person having myopia has difficulty in seeing distant objects clearly as the light from distant objects is brought to focus not on the retina but in front of it. Using spectacles with concave lenses will correct this condition. A concave lens of suitable power will bring the image back on to the retina and thus the defect is corrected.

Question 27.
What is hypermetropia? How is it caused?
Answer:
A defect of the eye that causes light to focus behind the retina instead of directly on it, resulting in an inability to see nearby objects clearly is called hypermetropia. It is also called long-sightedness. A person with this defect has the near point farther than 25 cm.
Hypermetropia may arise due to any or both of the following reasons:

  1. The focal length of the eye lens is too long.
  2. Contraction of the eyeball.

Question 28.
Draw neat diagrams to show

  1. Near point of a hypermetropic eye, and
  2. Hypermetropic eye.

Answer:
1. Near point of a hypermetropic eye:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 8

2. Hypermetropic eye:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 9

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 29.
How is hypermetropia corrected? Show this by a suitable diagram.

OR

Make a diagram to show how hypermetropia is corrected.
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 10
Hypermetropia can be corrected by using spectacles fitted with convex lenses of suitable power.

Question 30.
Observe the diagram given below and answer the questions that follow:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 11

  1. Name the defect of vision represented in the diagram above. Justify your answer.
  2. Show by a diagram the correction of this defect.

Answer:
1. The defect of vision represented in the diagram is hypermetropia. This means the person has difficulty in seeing nearby objects clearly. The person is not able to see clearly an object kept at the normal near point of the eye.

2. Hypermetropia corrected:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 12

Question 31.
Observe the figure given. Name the eye defect indicated in the figure and also mention the lens
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 13
The defect of the eye shown in the figure given is called myopia or short-sightedness. It can be corrected by using spectacles with concave lenses of suitable power.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 32.
A man can read the route number of a distant bus clearly but he finds difficulty in reading a book. From which defect of the eye is he suffering from? What type of spectacles lens should be used to correct the defect?
Answer:
Since the man cannot see nearby objects (letters in the book) clearly, he is suffering from the defect of vision called hypermetropia. Hypermetropia is corrected by using spectacles containing convex lenses of suitable power.

Question 33.
The near point of a hypermetropic eve is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eve is 25 cm.
Answer:
The image of the object at 25 cm must be shifted to a distance of lm for the object to be seen clearly. Given: Object distance, u = -25 cm; image distance, v = -1 m = -100 cm; focal length, f = ?; power, P = ?
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 14
A convex lens of power +3 D must be used to correct the defect.

Question 14.
What is presbyopia? Explain. How can it be corrected?
Answer:
Long-sightedness caused by loss of elasticity of the lens of the eye, occurring typically in middle and old age is called presbyopia.

The power of accommodation of the eye usually decreases with ageing. For most people, the near point gradually recedes away. They find it difficult to see nearby objects comfortably and distinctly without corrective eyeglasses.

Thus, presbyopia arises due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens. Presbyopia can be corrected by using spectacles fitted with convex lenses of suitable power.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 35.
Some people in old age develop both myopia and hypermetropia. How can this problem be corrected?
Answer:
Sometimes, a person in old age may suffer from both myopia and hypermetropia. Such people require bifocal lenses for correcting their eye defect. A common type of bifocal lens consists of both concave and convex lenses. The upper portion consists of a concave lens. It facilitates distant vision. The lower part is a convex lens. It facilitates near vision.

Question 36.
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

  1. Distant vision
  2. Near vision?

Answer:
The power P of a lens of focal length/is given by the relation P = \frac{1}{f}.
1. Correction for distant vision:
The person needs a lens of power -5.5 dioptre to correct myopia.
Power of the lens P = \frac{1}{f}
f = \frac{1}{p}
f = \frac{1}{-5.5} = – 0.812 m = – 18.2 m.
The person needs a convex lens of focal length 18.2 cm for correcting distant vision.

2. Correction for near vision:
The person is using a lens of power +1.5 dioptre to correct hypermetropia.
Power of the lens P = \frac{1}{f}
f = \frac{1}{p}
f = \frac{1}{1.5} = 0.66 m = 66.66 cm.
The person needs a convex lens of focal length 66.66 cm for correcting near vision.

Question 37.
What is a prism?
Answer:
Prism:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 15
A transparent solid object, made of glass or any other transparent material having two identical triangular bases and bound by lateral surfaces in the form of parallelograms and usqd for separating white light passed through it into a spectrum is called a prism.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 38.
Show by a suitable diagram the path of a light ray through a triangular glass prism. Also show the angle of deviation in the prism.
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 16
∠i – Angie of incidence, ∠D = Angle of deviation

Question 39.
What is meant by angle of deviation?
Answer:
When a ray of light passes through a prism, the angle formed between the direction of the incident ray and the emergent ray is called the angle of deviation.

Question 40.
What is meant by dispersion of light?
Answer:
The phenomenon of breaking up of composite light into its constituent colours is called dispersion of light.

Question 41.
How do you demonstrate the dispersion of light through a triangular glass prism?
Answer:
Take a thick sheet of cardboard and make a small hole or a narrow slit in the centre. Allow sunlight to fall on the narrow slit. Now, we get a narrow beam of white light through the hole. Pass the beam of white light through a glass prism.
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 17
Turn the prism slowly until the light that comes out of it appears on a nearby screen. A beautiful band of seven colours appears on the screen. This is called visible spectrum. The sequence of colours from the bottom end of the spectrum is violet, indigo, blue, green, yellow, orange and red.

Question 42.
What causes dispersion of white light in a glass prism?
Answer:
Different colours present in composite light have different wavelengths. These colours travel at the same speed through air but travel at different speeds in a given refracting medium (glass). As a result, the refractive index of glass depends on the colour of light.

The various colours of white light undergo different angles of deviation as they pass through a glass prism. Lights of shorter wavelength deviate most while those of longer wavelengths deviate least. Therefore, the various components of white light separate as they pass through a glass prism.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 43.
Which colour of white light has the longest wavelength and which one has the shortest wavelength? Which colour deviates most and which colour deviates least on passing through a triangular glass prism?

OR

Mention the colour that bends the least and the colour that bends the most when light undergoes dispersion through a prism.
Answer:
Of the colours of white light, red has the longest wavelength and violet has the shortest wavelength. On passing through a prism, the red coloured light deviates least and violet coloured light deviates the most.

Question 44.
How can you show that white light consists of seven colours?

OR

Draw a neat diagram to show the recombination of the spectrum of white light.
Answer:
White light is composite light. This means that white light consists of more than one colour. When a ray of white light is passed through a triangular glass prism, we get a band of seven colours. When these lights are passed through another identical prism held in inverted position, we get back white light. This shows that white light is made up of seven colours.
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 18
This shows that white light is a combination of seven colours.

Question 45.
What is spectrum of light? Name the colours present in the spectrum of white light
Answer:
The band of colours obtained by the dispersion of composite light is called spectrum. The spectrum of white light consists of violet, indigo, blue, green, yellow, orange and red colours.

Question 46.
Mention any four phenomena that can be observed due to atmospheric refraction of light on the earth.
Answer:
The various phenomena that involve refraction of light caused by the atmosphere of the earth are:

  1. Early sunrise and delayed sunset
  2. Twinkling of stars
  3. Formation of rainbow
  4. The apparent random wavering or flickering of objects when seen through a turbulent stream of hot air rising above a fire or a radiator.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 47.
What is formation of rainbow due to? Why can’t we see all the colours in a rainbow?
Answer:
A rainbow is a natural spectrum appearing in the sky after a rain shower. It is formed due to dispersion of sunlight by droplets of water present in the atmosphere. These droplets of water act like tiny prisms and cause the splitting of white light into its constituent colours. We cannot see all the colours in a rainbow because colours overlap.

Question 48.
Name the three major phenomena that contribute to the formation of a rainbow. In which direction do we usually see a rainbow formed in the sky?
Answer:
Three phenomena such as refraction, dispersion and total internal reflection contribute to the formation of a rainbow. The sunrays on entering, water droplets, refract and disperse, then reflect back internally. Finally, the light rays refract again when they come out of the raindrop. Due to the combined effect of refraction, dispersion of light and internal reflection, different colours reach the observer’s eye.

Question 49.
Draw a simple ray diagram showing the refraction, dispersion and total internal reflection happening in a raindrop during the formation of rainbow.
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 19

Question 50.
Explain the phenomenon of formation of rainbow in the sky.
Answer:
A rainbow is a natural spectrum caused in the sky due to the combined effect of three phenomena namely refraction, dispersion and total internal reflection. The tiny droplets of water hanging in air act like tiny prisms. When sunlight falls on these water droplets they undergo refraction and break up the constituent colours of sunlight. The raindrops reflect the dispersed light internally. When this light enters our eyes the rainbow is seen in the sky.

Question 51.
The atmosphere surrounding the earth is composed of air. How does refraction occur within the atmosphere although there is only one medium such as air?
Answer:
The air just above the earth becomes hotter than the air further up. The hotter air is less dense than the cooler air above it. Therefore, these layers of air with different density have slightly different refractive index. This is why light gets refracted while passing through different layers of the atmosphere. This is known as atmospheric refraction.

Question 52.
We observe the apparent random wavering or flickering of objects when we see them through a turbulent stream of hot air rising above a fire or a radiator. How do you explain this?
Answer:
The air just above the fire or hot radiator becomes hotter than the air further up. The hotter air is less dense than the cooler air above it. Therefore, these layers of air with different density have slightly different refractive index. Since the physical conditions of the medium of air are not stationary, the apparent position of the object, as seen through the hot air, fluctuates. This wavering is thus an effect of atmospheric refraction.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 53.
Give two examples of optical phenomena which occur in nature due to atmospheric refraction.
Answer:
Two of the effects caused by atmospheric refraction are:

  1. Twinkling of stars, and
  2. Advance sunrise and delayed sunset.

Question 54.
What is the twinkling of stars due to?
Answer:
The twinkling of a star is due to atmospheric refraction of starlight.

Question 55.
Explain why the planets do not twinkle.
Answer:
Planets are very close to us as compared to the distance between stars and earth. They appear as a collection of large number of point-sized sources of light (the reflected light from the sun). These different points produce either brighter or dimmer effect in such a way that the total variation in amount of light entering the eye from all sources averages out to zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 56.
Why do stars twinkle?

OR

How do stars twinkle? Explain.
Answer:
Stars may be considered as point-sized sources of light. The starlight that enters the atmosphere undergoes refraction continuously before it reaches the earth. The atmospheric refraction is caused due to varying refractive index of air at different altitudes.

Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. The star appears slightly higher than its actual position when viewed near the horizon.

This shift in the apparent position of the star is continuous as the physical conditions on the earth’s atmosphere are not stationary. As a result, starlight entering the eye flickers. The star sometimes appears brighter, and at some other time, fainter. This gives the twinkling effect.
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 20

Question 57.
Stars appear to twinkle but planets do not appear to twinkle. Why?
Answer:
Planets, being larger in size, can be taken as a collection of large number of point-sized objects. The refraction effect of light rays coming from such an extended object will nullify the twinkling effect and hence planets do not twinkle.

Question 58.
How do you explain early sunrise and delayed sunset?
Answer:
Sunrise is defined as the moment that the Sun first appears just over the horizon. This means that we can’t see the Sun before it appears above the horizon. However, the Sun is visible to us about two minutes before the actual sunrise, and about two minutes after the actual sunset.

This is because of the atmospheric refraction of the light from the Sun by the Earth’s atmosphere. Earth’s atmosphere bends the path of the light so that we see the Sun in a position slightly above. Therefore, sun becomes visible before sunrise. Similar things happen during sunset. Therefore, we see the sun for about two minutes even after sunset.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 59.
The time difference between the actual subset and the apparent sunset is about two minutes. What is the reason for this? Explain with the help of a diagram.
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 21
When the sun has set, the position of the sun appears to have shifted to a slightly higher position due to the refraction of the sun’s light by the atmosphere of the earth. Therefore, the sun would still be visible to us even after it is below the horizon. It is the atmospheric refraction that causes a time difference of about two minutes between actual sunset and apparent sunset.

Question 60.
The sun becomes visible to us two minutes before actual sunrise. How do you account for this?
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 22
Sunrise is defined as the moment that the Sun first appears just over the horizon. Obviously we should not be able to see the sun before it actually rises above the horizon. However, due to refraction of light by the atmosphere, the sun’s apparent position is shifted to a slightly higher point. This makes the sun visible even before sunrise. This explains the time difference of sighting the sun two minutes earlier than actual sunrise.

Question 61.
What is meant by scattering of light? Give examples for scattering of light.
Answer:
The phenomenon by which a beam of light is redirected in many different directions when it interacts with a particle of matter is called scattering of light.

The blue colour of the sky, colour of water in deep sea, the reddening of the sun at sunrise and sunset are due to scattering of light.

Question 62.
What is Tyndall effect? Explain with an example.
Answer:
The phenomenon of scattering of light by colloidal particles is called Tyndall effect. The path of a beam of sunlight coming through a window becomes visible due to the scattering of light by the dust particles in the room. This is an instance of Tyndall effect. Similarly, the path of light becomes visible in a smoke-filled room due to Tyndall effect.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 63.
What is the relationship between the size of the scattering particles and the colour of the scattered light?
Answer:
Finer particles scatter light of shorter wavelengths (in the blue region) while larger particles scatter light of longer wavelengths (in the red region).

Question 64.
How do you explain the blue colour of the sky?
Answer:
The sky appears blue when seen from the earth. The blue colour of the sky can be explained on the basis of scattering of light. The molecules present in the atmosphere are ideally suited to scatter blue light. The sky appears blue because we see it through the blue light scattered by the atmospheric air.

Question 65.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears blue because we see it through the blue light scattered by the atmospheric air. However, there is no atmosphere in space to scatter blue light. Hence the sky appears dark to an astronaut.

Question 66.
What will be the colour of the sky when it is observed from a place in the absence of any atmosphere? Explain.
Answer:
We see the sky as coloured because the molecules present in our atmosphere interact with the sunlight passing through it and scatter light of certain wavelengths. The type of scattering responsible for blue sky is known as Rayleigh scattering. In the absence of any atmosphere, there will be no scattering of sunlight and hence the sky will appear dark.

Question 67.
Why does the sun appear reddish at sunrise and sunset?

OR

Why does the sun appear reddish early in the mornine?
Answer:
The reddish appearance of the sun at sunrise and sunset can be explained on the basis of scattering of light. During sunrise and sunset, the sun is near the horizon. Therefore, the light of the sun has to pass through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes.

The blue component of sunlight gets almost completely scattered in different directions and the light that reaches our eyes is of longer wavelengths. Therefore, the sun appears reddish during sunrise and sunset.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 68.
Why does the Sun appear reddish early in the morning? Explain with the help of a diagram.
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 23
The molecules in the earth’s atmosphere are ideally suited to scatter blue light. During early morning, the sun is near the horizon. Therefore, the light of the sun will have to pass through denser layers of the atmosphere and over long distances before reaching our eyes.

During this process, the blue component of sunlight is almost completely scattered away and the light that reaches our eyes is largely in the red region. Therefore, the sun appears reddish during early morning.

Question 69.
Why does the sun appear white during midday?
Answer:
During midday, the sun is almost right above us. Therefore, the sunlight will largely pass through less dense regions of the atmosphere and shorter distance before reaching our eyes. Therefore, less amount of blue light present in sunlight is scattered and the light that reaches our eyes has almost all the components of white light. Hence the sun appears white during midday.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 70
How do you show experimentally why the sky appears blue and the sun appears reddish during sunrise and sunset?
Answer:
KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World 24
Place a strong source (S) of white light at the principal focus of a converging lens L1. This lens provides a parallel beam of white light. Pass this beam through a sodium thiosulphate solution taken in glass beaker, B. Allow the beam that emerges out of the beaker to pass through a circular hole of a screen. Pass the parallel beam of light that emerges out of the hole through another converging lens L2.

Obtain the image of the source on a screen placed at the principal focus of the second lens. The image will be in the form of a whitish circular patch of light. Now add 3 – 5 drops of concentrated sulphuric acid to the sodium thiosulphate solution and stir.

Keep observing the solution and the image formed on the screen. The solution gradually turns blue due to the scattering of blue light by the particles of the solution. The whitish patch of light on the screen gradually turns to red because the light that reaches the screen is almost free of blue light. After some time, the image disappears from the screen. This explains the blue colour of the sky and the reddish appearance of the sun during sunrise and sunset.

Question 71.
Water mixed with milk is taken in beaker ‘A ’ and sugar solution is taken in beaker ‘B ’. Light is passed through both the beakers. In which beaker is the path of light visible? Why?
Answer:
The path of light is visible in beaker ‘A’ that contains milk. Milk is a colloidal solution that scatters light. When a beam of light strikes the particles of milk, the path of the beam becomes visible, i.e., scattering of light by colloidal particles gives rise to Tyndall effect.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Fill In The Blanks

1. The screen of the human eye is called retina
2. The ability of the eye to adjust the focal length of its lens is called accommodation
3. The near point of a normal adult eye is 25 cm
4. Breaking up of the constituent colours of composite light is called dispersion
5. The defect of the eye due to which nearby objects cannot be seen clearly, is called hypermetropia / longsightedness
6. Myopia can be corrected with spectacles fitted with suitable concave lens
7. The constituent colour of white light that has the highest wavelength is red

Multiple Choice Questions

Question 1.
The human eye can focus 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.
Answer:
(B) accommodation.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 2.
The human eye forms the image of an object at its
(A) cornea.
(B) iris.
(C) pupil.
(D) retina.
Answer:
(D) retina.

Question 3.
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.
Answer:
(C) 25 cm.

Question 4.
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.
Answer:
(C) ciliary muscles.

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 5.
Scattering of light by particles involves
(A) change in direction of light
(B) dispersion
(C) reflection
(D) refraction
Answer:
(A) change in direction of light

Question 6.
Which of the following phenomena of light are involved in the formation of a rainbow?
(A) Dispersion, scattering and total internal reflection
(B) Reflection, dispersion and total internal reflection
(C) Reflection, refraction and dispersion
(D) Refraction, dispersion and total internal reflection
Answer:
(D) Refraction, dispersion and total internal reflection

Question 7.
A person has an elongated eyeball and hence is suffering from shortsightedness. He requires spectacles fitted with
(A) convex lens to focus distant objects
(B) concave lens to focus distant objects
(C) convex lens to focus near objects
(D) concave lens to focus near objects.
Answer:
(B) concave lens to focus distant objects

Question 8.
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
Answer:
(A) light is least scattered

Question 9.
The bluish colour of water in deep sea is due to
(A) the presence of algae and other plants found in water
(B) scattering of light
(C) reflection of sky in water
(D) absorption of light by the sea
Answer:
(B) scattering of light

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 10.
Twinkling of stars is due to atmospheric
(A) dispersion of light by water droplets
(B) internal reflection of light by clouds
(C) refraction of light by different layers of atmospheric air
(D) scattering of light by dust particles
Answer:
(C) refraction of light by different layers of atmospheric air

Question 11.
The clear sky appears blue because
(A) blue light gets absorbed in the atmosphere
(B) lights of all other colours are scattered more than violet and blue colour lights by the atmosphere
(C) ultraviolet radiations are absorbed in the atmosphere
(D) violet and blue lights get scattered more than lights of all other colours by the atmosphere
Answer:
(D) violet and blue lights get scattered more than lights of all other colours by the atmosphere

Question 12.
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 the colours, 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
Answer:
(B) is scattered the least by smoke or fog

Question 13.
Dispersion is a phenomenon involving
(A) refraction of light
(B) reflection of coloured light
(C) combining of coloured lights into white light
(D) separation of light into its spectrum
Answer:
(B) reflection of coloured light

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 14.
Because of atmospheric refraction, the Sun actually sets
(A) after we see it disappear
(B) before we see it disappear
(C) when it just disappears
(D) hours before we see it disappear
Answer:
(B) before we see it disappear

Question 15.
Which of the following statements 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.
Answer:
(C) A person with myopia can see nearby objects clearly.

Question 16.
The colour of light that refracts most on entering a glass prism is
(A) yellow
(B) violet
(C) blue
(D) red
Answer:
(D) red

Question 17.
Which of the following colours of light is least scattered by fog, dust or smoke? .
(A) Violet
(B) Blue
(C) Red
(D) Yellow
Answer:
(C) Red

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Question 18.
A person cannot see distinctly objects kept beyond 2 nu This defect can be corrected by using a lens of power
(A) +0.2 D
(B) -0.2 D
(C) +0.5 D
(D) -0.5 D
Answer:
(D) -0.5 D

Question 19.
Red coloured light is used in traffic signals to indicate the vehicles to stop, because compared to other colours red light
(A) has high frequency
(B) scatters more
(C) has less wavelength
(D) scatters less
Answer:
(D) scatters less

Question 20.
The characteristic of the image of an object formed on the retina by the lens of the eye is
(A) real and inverted
(B) virtual and erect
(C) real and erect
(D) virtual and inverted
Answer:
(A) real and inverted

KSEEB Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Match The Following

Question 1.

Column A Column B
1. Rainbow a. Bouncing back of light from a surface
2. Tyndall effect b. Causes blue colour of the sky
3. Myopia c. Caused by dispersion of light
4. Hypermetropia d. Scattering of light by colloidal particles
5. Scattering of light e. Cannot see nearby objects clearly
f. Cannot see distant objects clearly

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

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Students can download Class 10 Science Chapter 10 Light Reflection and Refraction Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 1.
What is light?
Answer:
Light is a form of energy. It is a form of electromagnetic radiation of a wavelength that can be detected by the human eye.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 2.
How do we see objects?

OR

What makes things visible?
Answer:
An object reflects light that falls on it. This reflected light, when received by our eyes, enables us to see things.

Question 3.
Give one instance front our daily life, which shows that light travels along straight lines.
Answer:
We can easily see a bright object, such as a candle flame, through a straight tube. But we cannot see the same through a bent tube. This shows that light travels along straight lines.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img1

Question 4.
What is a mirror? List the various types of mirrors that you know.
Answer:
Any smooth highly polished surface that reflects most of the light that falls on it is called a mirror. There are several types of mirrors namely plane mirror, concave mirror, convex mirror, parabolic mirror and so on.

Question 5.
What are

  • convergent rays
  • divergent rays,
  • parallel rays of light? Show them by separate diagrams.

Answer:

  • Convergent rays: Light rays that are moving towards a point are called convergent rays of light.
  • Divergent rays: Light rays that are emerging from a point are called divergent rays of light.
  • Parallel rays: Light rays that are parallel to one another are called parallel rays of light. These rays neither diverge nor converge.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img2

Question 6.
What is reflection of light?
Answer:
When light falls on a surface, some or most of the light is turned back from the surface. This phenomenon of bouncing back of light from a surface is called reflection.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 7.
Show by a suitable diagram the reflection of a light ray from a mirror. Show in the diagram the incident ray, the reflected ray, normal at the point of incidence, angle of incidence and the angle of reflection.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img3
In the diagram, AO is the ray that is falling on a plane mirror. It is called the incident ray. O is the point of incidence. It is also the point of reflection. OB is the ray coming back from the mirror. It is called the reflected ray. ON is the normal drawn to the mirror at the point of incidence. ∠AON is the angle of incidence and ∠NOB is the angle of reflection.

Question 8.
Define

  1. angle of incidence
  2. angle of reflection.

Answer:

  1. The angle formed between the incident ray and the normal at the point of incidence is the angle of incidence.
  2. The angle formed between the reflected ray and the normal at the point of incidence is the angle of reflection.

Question 9.
State the laws of reflection of light.
Answer:
There are two laws of reflection of light. They are

  1. Angle of incidence is always equal to the angle of reflection.
  2. The incident ray, the reflected ray and the normal drawn to the mirror at the point of incidence all lie in the same plane.

Question 10.
For what type of surfaces do the laws of reflection apply?
Answer:
The laws of reflection are applicable to all types of surfaces including plane, concave and convex surfaces.

Question 11.
What is an image? Mention its types.
Answer:
If light rays coming from a point after reflection meet at another point or appear to meet at another point, then the second point is called the image of the first point. Images are classified into two types

  • virtual image
  • real image.

Question 12.
Distinguish between virtual image and real image.
Answer:
Virtual image:

  1. Virtual images cannot be caught on a screen.
  2. Virtual images are erect with respect to the object.
  3. Virtual images are formed by divergent rays of light.

Real image:

  1. Real images can be caught on a screen.
  2. Real images are inverted with respect to the object.
  3. Real images are formed by convergent rays of light.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 13.
List the properties of the image formed by a plane mirror.
Answer:
The images formed by a plane mirror are

  • Erect and virtual. They cannot be caught on a screen.
  • Equal to the size of the object.
  • Formed as far behind the mirror as the object is in front of it.
  • Laterally inverted.

Question 14.
What are spherical mirrors? Mention their types.
Answer:
A mirror whose reflecting surface forms a part of a sphere is called a spherical mirror. There are two types of spherical mirrors

  • concave mirror
  • convex mirror.

Question 15.
What is a concave mirror? How is a concave mirror represented in a diagram?
Answer:
A spherical mirror whose reflecting surface is curved inwards is called a concave mirror.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img4

Question 16.
What is a convex mirror? How is a convex mirror represented in a diagram?
Answer:
A spherical mirror whose reflecting surface is curved outwards (bulged) is called a convex mirror.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img5

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 17
Define the following with respect to a spherical mirror

  1. Pole
  2. Centre of curvature
  3. Radius of curvature
  4. Aperture
  5. Principal axis.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img6
Answer:

  1. Pole: The centre or midpoint of the reflecting surface of a spherical mirror is called its pole. It is represented by the letter P.
  2. Centre of curvature: The centre of the sphere of which the given spherical mirror forms a part is called its centre of curvature. It is represented by the letter C.
  3. Radius of curvature: The radius of the sphere of which the given spherical mirror forms a part is called its radius of curvature. It is represented by the letter R. Thus, R = PC.
  4. Aperture: The diameter of the reflecting surface of a spherical mirror is called its aperture.
  5. Principal axis: An imaginary straight line passing through the pole (P) and centre of curvature (C) of a spherical mirror is called its principal axis.

Question 18.
What is meant by principal focus of a concave mirror? Show by a diagram.
Answer:
Principal focus of a concave mirror is defined as the point at which incident rays parallel to principal axis converge after reflection from the mirror. It is represented by the letter F.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img7

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 19.
Why are concave mirrors called converging mirrors?
Answer:
A concave mirror redirects parallel incoming rays that fall on it and makes them meet at a point. This is why concave mirrors are called converging mirrors.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img8

Question 20
What is meant by principal focus of a convex mirror? Show by a diagram.
Answer:
When rays of light close and parallel to the principal axis are made to fall on a convex mirror, after reflection, all the light rays appear to diverge from a fixed point on its principal axis. This point is known as the principal focus of the convex mirror. It is represented by the letter F.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img9

Question 21.
Why are convex mirrors called diverging mirrors?
Answer:
A convex mirror redirects parallel incoming rays that fall on it and makes them appear to diverge from a point. This is why convex mirrors are called diverging mirrors.

Question 22.
What is meant by focal length of a spherical mirror?
Answer:
The distance between the pole (P) and the principal focus (F) of a spherical mirror is called its focal length. It is represented by the letter f

Question 23.
What is the relationship between focal length and radius of curvature of a spherical mirror?
Answer:
In a spherical mirror, the principal focus (F) lies exactly at the centre of the line joining P and C.
PC = PF + FC
R = f + f = 2f
Therefore, the radius of curvature of a spherical mirror is equal to twice its focal length.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 24.
What is the focal length of a concave mirror whose radius of curvature is 40 cm?
Answer:
The focal length of a concave mirror is equal to half its radius of curvature.
f=\frac{R}{2}=\frac{40}{2}=20 \mathrm{cm}

Question 25.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
The focal length of a spherical mirror is equal to half its radius of curvature.
f=\frac{R}{2}=\frac{20}{2}=10 \mathrm{cm}

Question 26.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
f=\frac{R}{2}=\frac{32}{2}=16 \mathrm{cm}

Question 27.
Describe an experiment to deternune the focal length of a concave mirror.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img10
Take a concave mirror, a mirror stand, and a screen. Place the given concave mirror on the mirror stand. Direct the mirror towards a bright distant object, say, a tree. Place the screen in front of the mirror.

Move the screen back and forth until a sharp, well-defined image is formed on the screen. Measure the distance between the mirror and the screen. This distance gives the approximate focal length of the concave mirror.

Question 28.
Name a mirror that can give an erect and enlarged image of an object Where should the object be placed to get this type of image?
Answer:
A concave mirror can give an erect and enlarged image of an object. To get this type of image, the object must be placed between the pole (P) and the principal focus (F).

Question 29.
For what positions of the object does a concave mirror give a real image?
Answer:
A concave mirror gives a real image when the object is placed beyond its principal focus. The various positions of the object for which a concave mirror gives real image are:

  • when the object is between F and C
  • when the object is at C
  • when the object is beyond C
  • when the object is far beyond C.

Question 30.
For what position of the object does a concave mirror give a virtual, enlarged image?
Answer:
A concave mirror gives a virtual, enlarged image when the object is placed between P and F.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 31.
Where should an object be placed in front of a concave mirror to get a diminished image on a screen?
Answer:
The object should be placed beyond C of a concave mirror in order to get a diminished image on a screen.

Question 32.
For what position of the object does a concave mirror give an image equal to the size of the object?
Answer:
A concave mirror gives a real image equal to the size of the object when the object is placed at C.

Question 33.
Where should a source of light be placed in front of a concave mirror to get parallel beams of light?
Answer:
A source of light should be kept at the principal focus (F) of a concave mirror to get parallel beams of light.

Question 34.
Show by separate diagrams the reflected ray for the following incident rays in a concave mirror:

  1. A ray parallel to the principal axis.
  2. A ray passing through the principal focus.
  3. A ray passing through the centre of curvature, and
  4. A ray incident obliquely to the principal axis arid moving towards the pole.

Answer:
1. A ray parallel to the principal axis:
A ray parallel to the principal axis of a concave mirror, after reflection, will pass through the principle focus(F).
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img11

2. A ray passing through the principal focus:
A ray passing through the principal focus of a concave mirror, after reflection, will emerge parallel to the principal axis.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img12

3. A ray passing through the centre of curvature:
A ray passing through the centre of curvature (C) of a concave mirror, after reflection, will get reflected back along the same path.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img13

4. A ray incident obliquely to the principal axis and moving towards the pole:
A ray incident obliquely to the principal axis and moving towards the pole of a concave mirror, after reflection, will be reflected obliquely in accordance with the laws of reflection i.e., Li = Lx.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img14

Question 35.
Explain why a ray of light passing through the centre of curvature of a concave mirror gets reflected back along the same path.
Answer:
A ray of light that is passing through the centre of curvature and incident on a concave mirror is actually travelling and striking the mirror along the normal to the surface at that point. Therefore, the angle of incidence is equal to zero and the angle of reflection is also zero. Therefore, the light ray gets reflected back along the same path.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 36.
Draw ray diagrams to show the image formation in a concave mirror when the object is placed

  1. at infinity
  2. beyond C
  3. at C
  4. between F and C
  5. at F
  6. between P and F.

Describe the nature, position and relative size of the image in each case.
Answer:
1. When the object is placed at infinity:
When an object is placed at infinity in front of a concave mirror, we get a real, inverted, highly diminished image at the principal focus (F).
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img15

2. When the object is placed beyond C:
When an object is placed beyond the centre of curvature of a concave mirror but at finite distance, we get a real, inverted and diminished image between F and C.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img16

3. When the object is placed at C:
When an object is placed at the centre of curvature (C) of a concave mirror, we get a real, inverted image equal to the size of the object.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img17

4. When the object is placed between F and C:
When an object is placed between F and C of a concave mirror, we get a real, inverted and enlarged image beyond the centre of curvature (C) of the mirror.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img18

5. When the object is placed at F:
When an object is placed at the principal focus (F) of a concave mirror, we get a real, inverted and a highly enlarged image at infinity.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img19

6. When the object is placed between P and F:
P and F of a concave mirror, we get a virtual, erect and enlarged image behind the mirror.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img20

Question 37.
Mention the points with respect to a spherical mirror between which an object should be placed to obtain on a screen an image twice the size of the object
Answer:
An object must be placed at an appropriate point between F and C of a concave mirror in order to obtain on a screen an image equal to twice the size of the object.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 38.
List the uses of concave mirrors.
Answer:
The following are some of the uses of concave mirrors

  1. Concave mirrors are commonly used in torches, search-lights and vehicle headlights to get powerful parallel beams of light.
  2. They are often used as shaving mirrors to see a larger image of the face.
  3. Dentists use concave mirrors to see large images of the teeth of patients.
  4. Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Question 39.
Show by separate diagrams the reflected ray for the following incident rays in a convex mirror:

  1. A ray parallel to the principal axis.
  2. A ray passing through the principal focus.
  3. A ray passing through the centre of curvature, and
  4. A ray incident obliquely to the principal axis and moving towards the pole.

Answer:
1. A ray parallel to the principal axis:
A ray of light parallel to the principal axis of a convex mirror, after reflection, will appear to diverge from the principal focus (F).
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img21

2. A ray passing through the principal focus:
A ray of light incident on a convex mirror in the direction of the principal focus (F), after reflection, will emerge parallel to the principal axis.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img22

3. A ray of light passing through the centre of curvature:
A ray of light that is incident on a convex mirror in the direction of centre of curvature (C) gets reflected back along the same path.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img23

4. A ray incident obliquely to the principal axis and moving towards the pole:
A ray of light incident obliquely on a convex mirror in the direction of P, after reflection, will emerge obliquely in accordance with the laws of reflection, making equal angles with the principal axis.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img24

Question 40.
Draw ray diagrams to show the image formation in a convex mirror when the object is placed

  1. at a finite distance in front of the mirror
  2. at infinity.

Describe the nature, position and relative size of the image in each case.
Answer:
1. When the object is placed at a finite distance in front of the mirror:
When an object is placed at a finite distance in front of a convex mirror, we get a virtual, erect and diminished image between principal focus F and pole P behind the mirror.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img25

2. When the object is placed at infinity:
When an object is placed at infinity in front of a convex mirror, we get a virtual, erect and highly diminished image (point image) at its principal focus (F), behind the mirror.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img26

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 41.
Summarise in a table the nature, position and relative size of the image formed by a convex mirror for various positions of the object.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img27

Question 42.
Listfour properties of the image formed by a convex mirror.
Answer:
The following are the properties of the image formed by a convex mirror

  1. Image is always virtual and erect.
  2. Image is always smaller than the object.
  3. Image is formed between P and F of the mirror.
  4. As we move the object away from the mirror, the image moves towards F.

Question 43.
List the uses of convex mirrors.
Answer:
The following are some of the uses of convex mirrors

  1. Convex mirrors are used as rear view mirrors in vehicles.
  2. They are used as reflector in street lamps to spread the light over a larger area.

Question 44.
Why are convex mirrors used as rear-view mirrors in vehicles?

OR

Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Convex mirrors are commonly used as rear-view mirrors in vehicles for the following reasons

  1. Convex mirrors always give an erect though diminished image of objects.
  2. Convex mirrors have a wider field of view as they are curved outwards. Therefore, convex mirrors provide the driver a view of a much larger area behind the vehicle.
  3. Images of nearing objects grow in size while those of receding objects diminish. These factors enable the driver to drive safely.

Question 45.
Name the type of mirror used in the following situations

  1. Headlights of a car.
  2. Side/rear-view mirror of a vehicle.
  3. Solar furnace.

Support your answer with reason.
Answer:
1. Concave mirror is used in the headlights of a car. This is because, a concave mirror produces a powerful parallel beam of light when the light source is placed at the principal focus.

2. Convex mirror is used in side/rear-view mirror of a vehicle. This is because, a convex mirror always gives an erect, diminished image and has a wider range of view.

3. A concave mirror is used in solar furnace. This is because, a concave mirror focuses the heat from the sun in a narrow region and thus helps to create higher temperatures.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 46.
You are given a plane mirror, a concave mirror and a convex mirror. How do you find out which of these can give a full length image of a tall object such as a distant tree for all positions?
Answer:
Observe the image of a distant object (say a distant tree) in a plane mirror. The full image is not seen in the mirror. When we try with plane mirrors of different sizes, the image is not seen in full. To see a full image in a plane mirror, we need a plane mirror which is at least half the size of the object. Now direct a concave mirror towards the distant tree and observe the image.

The concave mirror gives full-size image of the object only for certain positions of the object. A convex mirror, however, gives full-size image for all positions of the object. This activity shows that a plane mirror and a concave mirror cannot give full-length image for all positions of the object. A convex mirror always gives full-size image irrespective of the position of the object.

Question 47.
Compare the size of the image formed by a plane mirror, a concave mirror and a convex mirror.
Answer:
A plane mirror will always give an image equal to the size of the object. A concave mirror gives an image which is larger, smaller or equal to the size of the object depending on the position of the object. A convex mirror always gives a diminished image irrespective of the position of the object.

Question 48.
Distinguish between concave and convex mirrors.
Answer:
Concave mirror:

  1. The reflecting surface is curved inwards.
  2. Converges light rays.
  3. Gives either real or virtual image depending on the position of the object.
  4. Image may be enlarged, diminished or of same size as that of the object.

Convex mirror:

  1. The reflecting surface is curved outwards.
  2. Diverges light rays.
  3. Always gives virtual image irrespective of the position of the object.
  4. Image is always diminished irrespective of the position of the object.

Question 49.
What is meant by ‘object distance’ and ‘image distance’ with respect to a Spherical mirror?
Answer:
1. Object distance:
The distance of the object from the pole of the mirror is known as ‘object distance’. It is represented by the letter u.

2. Image distance:
The distance of the image from the pole of the mirror is known as ‘image distance’. It is represented by the letter v.

Question 50.
Write the sign conventions for reflection by spherical mirrors.
Answer:
The following are the sign conventions applicable to reflection by spherical mirrors

  1. All distances parallel to the principal axis are measured starting from the pole of the mirror.
  2. Distances measured in the direction of the incident ray are taken as positive and those measured against it are taken as negative.
  3. The focal length of a concave mirror is always taken to be negative and the focal length of a convex lens is always taken to be positive.
  4. Distances measured perpendicular to and above the principal axis are taken as positive and the distances measured perpendicular to and below the principal axis are taken as negative.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 51.
Write the mirror formula with respect to spherical mirrors.
Answer:
The mirror formula is \frac{1}{f}=\frac{1}{u}+\frac{1}{v}
where u = distance of the object from the pole of the mirror
v = distance of the image from the pole of the mirror
f = focal length of the mirror.

Question 52.
Define magnification with respect to spherical mirrors. Give a suitable equation for it.
Answer:
Magnification produced by a spherical mirror is defined as the ratio of the height of the image to the height of the object. It is usually represented by the letter m. If h is the height of the object and h’ is the height of the image, then the magnification m produced by a spherical mirror is given by
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img28

Question 53.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
The statement above means that the plane mirror produces an image equal to the size of the object. This means that the ratio between the height of the image and the height of the object in a plane mirror is always equal to 1. Since m is positive, the image is erect with respect to the object.

Question 54.
A convex mirror used for rear-view in an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.
Answer:
Given:
Radius of curvature, R = + 3.00 m; Object-distance, u = – 5.00 m; Image-distance, v = ?, Height of the image, h’ = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img29
The image is virtual, erect and diminished. The size of the image is 0.23 times the size of the object.

Question 55.
‘A concave mirror of focal length ‘f conform a magnified, erect as well as an inverted image of an object placed in front of it’. Justify this statement stating the position of the object with respect to the mirror in each case for obtaining these images.
Answer:
A concave mirror of focal length/can form

  1. a magnified erect image when the object is placed between its pole (P) and principal focus (F)
  2. an inverted image of an object when the object is placed
    • at F
    • between F and C. Thus, the given statement is justified.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 56.
A concave mirror produces three times magnified (enlarged) real image of an object placed 10 cm in front of it. Where is the image located?
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img30
Image distance v = 3 × -10 cm = -30 cm. The image is located at a distance of 30 cm from the mirror.

Question 57.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the ranee of distance of the object from the mirror? What is the nature of the image? Is the image larser or smaller than the object? Draw a ray diasram to show the image formation in this case.
Answer:
In order to obtain an erect image in a concave mirror, the object must be placed between P and F. Therefore, the range of ‘object distance’ must be 0-15 cm. The image formed is virtual and erect. The image is larger than the object. The ray diagram for getting an erect image in a concave mirror is given below
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img31

Question 58.
An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.
Answer:
Given: Object-size, h = +4 cm; Object-distance, u = -25 cm; Focal length, f= -15 cm; Image distance, v = ?; Image size, h’ = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img32
The screen should be placed at 37.5 cm from the mirror. The image is real
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img33
Height of the image, h’ = -6.0 cm. The image is inverted and enlarged. Therefore, a screen must be placed at a distance of 37.5 cm from the mirror to get a sharp image. The height of the image is 6 cm. The image is inverted and enlarged.

Question 59.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Given
Object distance, u = -10 cm; Focal length, f = +15 cm; Image distance, v = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img34
A virtual, erect and diminished image is formed 6 cm behind the mirror.

Question 60.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Given
Object distance, u = -20 cm; Focal length, f=\frac{R}{2}=\frac{30}{2}=+15 \mathrm{cm}
Height of the object, h – 5 cm; Image distance, v = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img35
A virtual, erect and diminished image of height 2.15 cm is formed 8.6 cm behind the mirror.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 61.
The focal length of a converging mirror is 4 cm. If an object of height 2 cms is placed 9 cms from the mirror, find the image distance, nature and size of the image.
Answer:
Given
Object distance, u = -9 cm, Focal length, f= -4, Height of the object h = 2 cm Image distance v = ?, Height of the image h’ = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img36
Thus, a real, inverted and diminished image of height 1.6 cm is formed at a distance of 7.2 cm between F and C.

Question 62.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Given
Object distance, u = -27 cm; Size of the object, h = 7 cm; Focal length, f = -18 cm; Image distance, v = ? ; Height of the image, h’ = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img37
So, image is formed in front of the mirror at a distance of 54 cm from the mirror. The size of the image is 14 cm and it is real and inverted.

Question 63.
What is refraction of light?
Answer:
A ray of light that enters obliquely from one medium into another, changes its direction on entering the second medium. This phenomenon of bending of light is called refraction of light.

Question 64.
How will a ray of light that travels obliquely, bend

  1. when it travels from a rarer medium to a denser medium
  2. when it travels from a denser medium to a rarer medium?

Answer:
1. A ray of light that enters obliquely from a rarer medium to a denser medium, on entering the second medium, will bend towards the normal. In this case, the angle of refraction is smaller than the angle of incidence.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img38

2. A ray of light that enters obliquely from a denser medium to a rarer medium, on entering the second medium, will bend away from the normal. In this case, the angle of refraction is larger than the angle of incidence.

Question 65.
When will a ray of light not change its direction despite changing the medium of propagation?
Answer:
A light that enters from one medium to another along the normal to the interface of the two media will not change its direction.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img39

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 66.
A ray of light travelline in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
A ray of light that travels obliquely from air to water will bend towards the normal on entering water. This is because water is an optically denser medium than air.

Question 67.
Mention any two applications of refraction of light in daily life.
Answer:
Two instances of refraction of light in daily life are:
1. A coin placed inside a bucket filled with water appears to be raised. This is due to refraction of light. Similarly, when a thick glass slab is placed over some printed matter, the letters appear raised when viewed through the glass slab. This is due to refraction of light.

2. A pencil partially immersed in water appears bent due to refraction of light.

3. A lemon kept in water in a glass tumbler appears to be bigger than its actual size, when viewed from the sides. This is due to refraction of light.

Question 68.
Describe a simple activity to show refraction of light.
Answer:
Place a coin at the bottom of a bucket filled with water. With your eye to a side above the bucket, try to pick up the coin in one go. Try this a few times. We may not succeed in picking up the coin on most occasions.

This is because the position of the coin appears to have shifted due to refraction of light. The actual position of the coin and its apparent position are different. This is why it is difficult to pick up the coin in one swift effort.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img40

Question 69.
Why does a pencil partially immersed in water taken in a glass tumbler appear to be bent?
Answer:
A pencil kept immersed partially in water appears bent due to refraction of light. The light coming from the part of the pencil inside water will bend away from the normal on entering air. The light reaching us from the portion of the pencil inside water seems to come from a different direction, compared to the part above water.

This makes the pencil appear to be displaced at the interface. As a result, the position of the pencil inside water appears to have shifted to a new position. This causes the pencil to appear bent.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img41

Question 70.
Why does a swimming pool appear shallower than what it actually is?
Answer:
The bottom of a swimming pool appears to be raised due to refraction of light. When light rays enter from water to air, they bend away from the normal. This causes an apparent upward shift of the bottom of the pool. This is why swimming pool appears shallower.

Question 71.
Draw a neat labelled diagram showing the path of a light ray through a rectangular glass slab.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img42

  • PQRS – Rectangular glass slab
  • AB – Incident ray
  • BC – Refracted ray
  • CD – Emergent ray
  • d = Lateral shift
  • N1N2 = Normal drawn to surface
  • PQ N’1N’2 = Normal drawn to surface
  • ∠ABN1 = i = Angle of incidence
  • ∠N2BC= r = Angle of refraction
  • ∠DCN’2 = e = Angle of emergence

Question 72.
Draw a neat diagram showing the path of a light ray through a parallel-sided glass slab.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img43

Question 73.
What is meant by lateral shift with reference to the refraction of a ray of light through a parallel-sided glass slab?
Answer:
When a ray of light is incident obliquely on a parallel-sided glass slab, the emergent ray shifts laterally. The perpendicular distance between the direction of the incident ray and the emergent ray is called‘lateral shift’.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 74.
What are the important characteristics you observe when a ray of light travelling obliquely enters a rectangular glass slab?
Answer:
When a ray of light travelling obliquely enters into a rectangular glass slab, the following observations are made

  1. The ray of light on entering from air to glass is found to bend towards the normal.
  2. The ray of light bends away from the normal when it enters from glass to air.
  3. The emergent ray gets shifted laterally. This means that the emergent ray is parallel to the direction of the incident ray.
  4. The angle of incidence (when the ray enters from air to glass) is equal to the angle of emergence (when the ray moves out of glass to air).

Question 75.
Why does refraction occur when a ray of light travels obliquely from one medium to another?
Answer:
Light travels at different speeds through different media. Refraction of light occurs due to the change in the speed of light as it enters from one transparent medium to another.

Question 76.
State the laws of refraction of light.
Answer:
The following are the laws of refraction of light
1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

2. The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the given pair of media and for the light of a given colour. This law is also known as Snell’s law of refraction. (This is true for angle 0 < i < 90°)

Question 77.
State Snell’s law of refraction. Express it in the form of an equation.
Answer:
Snell’s law of refraction states that ‘when a ray of light travels obliquely from one transparent medium to another, the ratio between the sine of the angle of incidence and the sine of the angle of refraction is a constant for the given pair of media for the light of a given colour.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img44
Consider a ray of light travelling obliquely from medium 1 into medium 2. Let i be the angle of incidence and r be the angle of refraction. Now, according to Snell’s law of refraction, \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\text { Constant }

Question 78.
When a ray of light travels obliquely from one medium to another, what is the ratio between sine of the angle of incidence and sine of the angle of refraction called?
Answer:
When a ray of light travels obliquely from one medium to another, the ratio between the sine of the angle of incidence and the sine of the angle of refraction is a constant and it is called the ‘refractive index’ of the second medium with respect to the first.

Question 79.
Define ‘refractive index’ of a medium. What is its symbol? Give an equation for the same.
Answer:
When a ray of light enters obliquely from one medium to another, the ratio between the sine of the angle of incidence and the sine of the angle of refraction is called the refractive index of the second medium with respect to the first. This is also called the relative refractive index of medium 2 with respect to medium 1. It is represented by the symbol n2 1.

Relative refractive index of medium 2 with respect to medium 1, n2 1=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}

Similarly, refractive index of medium 1 with respect to medium 2 is denoted by n1 2=\frac{\text { Speed of light in medium } 2}{\text { Speed of light in medium } 1}=\frac{v_{2}}{v_{1}}

Question 80.
Define relative refractive index in terms of speed of light. Give an equation for the same.
Answer:
The relative refractive index of medium 2 with respect to medium 1 is defined as the ratio between the speed of light in medium 1 and the speed of light in medium 2. Let the relative refractive index of medium 2 with respect to medium 1 be n2 1=\frac{\text { Speed of light in medium } 1}{\text { Speed of light in medium } 2}

If v1 is the speed of light in medium 1 and v2 is the speed of light in medium 2, then, the refractive index of medium 2 with respect to medium 1 is given by
n_{2 l}=\frac{v_{1}}{v_{2}}

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 81.
What does the refractive index of a medium actually indicate?
Answer:
The refractive index actually indicates the extent of the change in direction of a ray of light that takes place in a given pair of media.

Question 82.
Define absolute refractive index of a medium both in terms of Snell’s law of refraction and speed of light.
Answer:
Refractive index in terms of Snell’s law: When a ray of light enters obliquely from air into a medium, the ratio between the speed of light in air and the speed of light in the medium is called the absolute refractive index of the medium. It is represented by the symbol nm
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img45
Refractive index in terms of speed of light: The absolute refractive index of a medium is defined as the ratio between the speed of light in air (vacuum) and the speed of light in the given medium.
n_{m}=\frac{\text { Speed of light in air (vacuum) }}{\text { Speed of light in the medium }}
If va is the speed of light in air (vacuum) and vm is the speed of light in the given medium, then the absolute refractive index of the medium, nm is given as,
n_{m}=\frac{\mathbf{v}_{\mathbf{a}}}{\mathbf{v}_{\mathrm{m}}}

Question 83.
The refractive index of glass is 1.5. Explain the meaning of this statement in terms of Snell’s law.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img46
When a ray of light enters from air to glass obliquely, the ratio between the sine of angle of incidence and the sine of angle of refraction is a constant and its value is equal to 1.5. Refractive index of glass, ng is given by the equation,
n_{g}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}

Question 84.
What is the meaning of ‘the refractive index of crown glass is 1.52’?
Answer:
The refractive index of crown glass is 1.52. This means that the ratio of the speed of light in air and the speed of light in crown glass is equal to 1.52.

Question 85.
The velocity of light in water is \frac{3}{4} t h times the velocity of light in vacuum. Find the refractive index of water.
Answer:
Let the velocity of light in vacuum be x. Now the velocity of light in water is \frac{3 x}{4}
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img47
The refractive index of water is 1.33.

Question 86.
The refractive index of water is 1.33. Explain the meaning of this statement In terms of speed of light
Answer:
The statement above means that the ratio between the speed of light in air (vacuum) and the speed of light in water is 1.33.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 87.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
This statement means that the ratio between the speed of light in air and the speed of light in diamond is equal to 2.42.

Question 88.
Light enters from air to glass having refractive index 1.50. What is the speed of light in glass? The speed of light in vacuum is 3 × 108 ms-1.
Answer:
Given
Refractive index of glass, ng= 1.5; Speed of light in vacuum, C = 3 × 108 ms-1; Speed of light in glass, vg = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img48

Question 89.
The velocity of light in water and kerosene is respectively 2.25 × 108 ms-1 and 2.08 × 108 ms-1. Which material has the highest refractive index? Prove your answer by calculation.
Answer:
Speed of light in water Vw = 2.25 × 108 ms-1
Speed of light in kerosene Vk = 2.08 × 108 ms-1
Refractive index of water nw = ?, Refractive index of kerosene nk = ?
Let us take the value of the speed of light in vacuum to be 108 ms-1.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img49
The refractive index of water is 1.33.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img50
The refractive index of kerosene is 1.44.

Question 90.
The refractive index of water, glass, diamond and alcohol are 1.33,1.5, 2.4 and 1.36 respectively. Compare the optical density of these substances. Also, find the medium with lowest optical density.
Answer:
The substances in the decreasing order of optical density are diamond, glass, alcohol and water. The substance with lowest optical density is water.

Question 91.
Find out from the table below, the medium havine highest optical density. Also find the medium with lowest optical density.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img51
From among the substances given in the table, diamond has highest optical density while air has lowest optical density. It may be noted here that a medium with highest refractive index will have the highest optical density and vice-versa.

In the table, diamond and air have the highest and lowest refractive index respectively. Therefore, diamond has the highest optical density and air has the lowest optical density.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 92.
You are given kerosene, turpentine and water. In which of these does light travel fastest? The refractive index of kerosene, turpentine and water are 1.44, 1,47 and 1.33 respectively.
Answer:
The speed of light is lower in that medium which has higher refractive index. Therefore, the speed of light in water is higher than the speed of light in kerosene; the speed of light in kerosene is higher than the speed of light in turpentine. Thus the speed of light is highest in water and lowest in turpentine.

Question 93.
Refractive index of media A, B, C and D are given in the table below.

Medium Refractive index
A 1.33
B 1.52
C 1.44
D 1.65

In which of these four media is the speed of light

  • highest
  • lowest?

Answer:
Of the given media, the refractive index of medium A has the lowest value of 1.33. Therefore, the speed of light is higher in medium A. The refractive index of medium D has the highest value of 1.65. Therefore, the speed of light is lowest in medium D.

Question 94.
What is a lens? Mention various types of lenses.
Answer:
Any transparent material bound by two surfaces, of which one or both surfaces are spherical, is called a lens. The various types of lenses are biconvex lens, biconcave lens, plano-concave lens, plano-convex lens and concavo-convex lens.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img52

Question 95.
Which are the two major types of lenses?
Answer:
The two major types of lenses are

  • convex lens
  • concave lens.

Question 96.
What is a convex lens? Show by a diagram the convergence of light by a convex lens.
Answer:
A lens that is bound by two bulged spherical surfaces is called a convex lens. It is also called a converging lens. A convex lens is thicker in the middle and thinner at the edges.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img53

Question 97.
What is a concave lens? Show by a diagram diverging action of a concave lens.
Answer:
A lens that is bound by two shallow spherical surfaces is called a concave lens. It is also called diverging lens. A concave lens is thinner in the middle and thicker at the edges.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img54

Question 98.
Define the following terms with reference to a lens

  1. centre of curvature
  2. principal axis
  3. optical centre
  4. aperture
  5. radius of curvature.

Answer:
1. Centre of curvature of a lens:
The centre of the sphere of which the surface of a lens forms a part is called its centre of curvature. A double convex lens and a double concave lens have two centres of curvature.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img55

2. Principal axis of a lens:
An imaginary straight line passing through the two centres of curvature of a lens is called its principal axis.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img56

3. Optical centre of a lens:
The geometric centre of a lens through which a ray of light passes through without undergoing any deviation is called the optical centre of the lens. It is represented by the letter O.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img57

4. Aperture of a lens:
The diameter of the refracting surface of a lens is called its aperture.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img58

5. Radius of curvature of a lens:
The radius of the sphere of which the refracting surface of a lens forms a part is called its radius of curvature. It is the distance between the optical centre and the centre of curvature of the given lens.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 99.
With the help of a suitable diagram explain the principal focus of a convex lens.
Answer:
When rays of light close and parallel to the principal axis are made to fall on a convex lens, after refraction, all the light rays actually pass through a fixed point on its principal axis. This point is known as the principal focus of the convex lens. It is represented by the letter F. A double convex lens has two principal foci, one on each side.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img59

Question 100.
How do you show that a convex lens converges light rays?
Answer:
Hold a convex lens in your hand. Direct it towards the Sun. Do not look at the sun through the lens. Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun on the paper. The rays from the sun are parallel. These parallel rays are brought to focus by the convex lens.

We get a bright tiny spot where the image of the sun is formed on the paper. Now, hold the paper and the lens in the same position -for a while. The paper begins to bum producing smoke. It may even catch fire after a while. This shows that a convex lens converges parallel rays of light.

Question 101.
With the help of a suitable diagram, explain the principal focus of a concave lens.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img60
When rays of light close and parallel to the principal axis are made to fall on a concave lens, after refraction, all the light rays appear to diverge from a fixed point on its principal axis. This point is known as the principal focus of the concave lens. It is represented by the letter F. A double concave lens has two principal foci.

Question 102.
Define focal length of a lens. Show this by suitable diagram.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img61
The distance between the optical centre (O) and the principal focus (F) of a spherical lens is called its focal length. It is represented by the symbol f

Question 103.
Describe a simple experiment to find out the focal length of a convex lens.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img62
Take a convex lens, a lens stand, and a screen. Place the convex lens on the lens stand. Direct the lens towards a bright distant object, say a tree. Place the screen on the other side of the lens. Move the screen back and forth until a sharp, well-defined image of the tree is formed on the screen. Measure the distance between the lens and the screen using a scale. This distance gives the approximate focal length of the convex lens.

Question 104.
Show by separate diagrams the refracted ray in a convex lens for the following incident rays:

  1. A ray parallel to the principal axis.
  2. A ray passing through the principal focus.
  3. A ray incident obliquely in the direction of optical centre.

Answer:
1. A ray parallel to the principal axis:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img63
A ray of light incident on a convex lens parallel to the principal axis, after refraction, will actually pass through the principal focus (F) on the other side of the lens.

2. A ray passing through the principal focus:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img64
A ray of light passing through or directed to the principal focus, will emerge parallel to the principal axis.

3. A ray incident obliquely in the direction of optical centre:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img65
A ray of light passing through the optical centre of a convex lens will emerge out without any deviation.

Question 105.
Draw ray diagrams to show the image formation by a convex lens when the. object is placed

  1. at infinity
  2. beyond 2F
  3. at 2F
  4. between F and 2F
  5. at F and
  6. between O and F.

Describe the nature, position and relative size of the image in each case.
Answer:
1. When the object is placed at infinity:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img66
When the object is placed at infinity, a convex lens forms a real, inverted and highly diminished image at its principal focus on the other side of the lens.

2. When the object is placed beyond 2F:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img67
When the object is placed at finite distance beyond 2F of a convex lens, we get a real, inverted and diminished image between F and 2F on the other side of the lens.

3. When the object is placed at 2F:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img68
When the object is placed at 2F of a convex lens, we get a real, inverted image equal to the size of the object at 2F on the opposite side of the lens.

4. When the object is placed between F and 2F:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img69
When the object is placed between F and 2F of a convex lens, we get a real, inverted and magnified image beyond 2F on the other side of the lens.

5. When the object is placed at the principal focus (F):
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img70
When the object is placed at the principal focus of a convex lens, a real, inverted and a highly magnified image is formed at infinity.

6. When the object is placed between optical centre (O) and principal focus (F):
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img71
When an object is placed between O and F of a convex lens, we get a virtual, erect and magnified image on the same side of the lens as the object.

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 106.
Show by separate diagrams, the refracted ray produced by a concave lens for the following incident rays

  1. A ray of light parallel to the principal axis.
  2. A ray of light passing through the principal focus.
  3. A ray of light passing through the optical centre.

Answer:
1. A ray of light parallel to the principal axis:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img72
A ray of light parallel to the principal axis, after refraction from a concave lens, appears to diverge from the principal focus (F) located on the same side of the lens.

2. A ray of light passing through the principal focus:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img73
A ray of light appearing to meet at the principal focus of a concave lens, after refraction, will emerge parallel to the principal axis.

3. A ray of light passing through the optical centre:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img74
A ray of light passing through the optical centre of a concave lens, after refraction, will move without any deviation.

Question 107.
Draw ray diagrams to show the image formed by a concave lens when the object is placed

  1. at infinity
  2. at a finite distance in front of the lens.

Describe the nature, position and relative size of the image in each case.

OR

Draw ray diagrams to show the nature, position and relative size of the image formed by a concave lens
Answer:
1. When the object is placed at infinity:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img75
When the object is placed at infinity, a concave lens gives a virtual, erect and highly diminished image (point image) at its principal focus on the same side of the lens as object.

2. When the object is placed at a finite distance in front of the lens:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img76
When the object is placed at a finite distance from a concave lens, we get a virtual, erect ar:’ diminished image between focus and optical centre on the same side of the lens.

Question 108.
Summarize in a table the nature, position and relative size of the image formed by concave lens for various position of the object.
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img77

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 109.
Distinguish between a convex lens and a concave lens.
Answer:
Convex lens:

  1. Converges light rays.
  2. Lens is thicker in the middle and thinner at the edges.
  3. Gives either a real image or a virtual image depending on the position of the object with respect to the lens.
  4. Gives either a diminished image, enlarged image or image of the same size depending on the position of the object.

Concave lens:

  1. Diverges light rays.
  2. Thinner in the middle and thicker at the edges.
  3. Always gives virtual image irrespective of the position of the object with respect to the lens.
  4. Gives only diminished image of the object irrespective of the position of the object with respect to the lens.

Question 110.
State the sign convention for lenses.
Answer:
The following sign conventions must be followed for lenses

  1. All distances along the principal axis must be measured from the optical centre of the lens.
  2. Distances measured in the direction of the incident ray are taken as positive and those measured against it are taken as negative.
  3. The focal length of a convex lens is taken as positive and that of a concave lens as negative.
  4. Distances measured perpendicular to and above the principal axis are taken as positive and distances perpendicular to and below the principal axis are taken as negative.

Question 111.
Write the lens formula and give the meaning of the symbols used.
Answer:
The lens formula is
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
where, v = distance of image from the opucal centre of lens
u = distance of object from the optical centre of lens
f = focal length of lens.

Question 112.
Define magnification produced by a tens. Give suitable equations for it.
Answer:
The magnification produced by a lens is defined as the ratio of the height of the image and the height of the object. It is represented by the letter m.
If h is the height of the object and h’ is the height of the image given by a lens, then the magnification produced by the lens is given by,
Magnification, m =\frac{\text { Height of the image }}{\text { Height of the object }}=\frac{h^{\prime}}{h}

Magnification produced by a lens is also defined as the ratio between object-distance, u and the image-distance, v. This relationship is given by
Magnification, m =\frac{\text { Image distance }}{\text { Object distance }}=\frac{\mathbf{v}}{\mathbf{u}}

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 113.
A concave lens has focal length of 15 cm. At what distance should the object be placed from the lens so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.
Answer:
Given
Image distance, v = -10 cm
Focal length,f = -15 cm
Object distance, u = ?
For a lens,
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img78
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img79
Thus, the object must be placed 30 cm from the lens,
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img80
The positive sign shows that the image is erect and virtual. The image is one-third the size of the object.

Question 114.
A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is IS cm. Find the nature, position and size of the image. Also find its magnification.
Answer:
Given
Height of the object, h = + 2.0 cm
Focal length,f = + 10 cm
Object distance, u = -15 cm
Image distance, v = ?
Height of the image, h’ = 1
For a lens,
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img81
The image is formed at a distance of 30 cm on the other side of the optical centre and the image is real, inverted and enlarged.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img82
The image is twice the size of the object.

Question 115.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Given
Focal length,f= -15 cm
Image distance, v = -10 cm
Object distance, u = ?
For a lens,
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img83
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img84

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 116.
The focal length of a concave lens is 30 cm. At what distance should the object be placed from the lens so that it forms an image at 20 cm from the lens?
Answer:
Given
Image distance, v = -20 cm
Focal length,f= -30 cm
Object distance, u = ?
We know that for a lens,
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img85
The object must be kept at a distance of 60 cm from the lens.

Question 117.
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:
Yes, the lens will produce a complete image. Even if half of the convex lens is covered with black paper, the lens will produce a complete image. However, the intensity of the image gets reduced. This can be verified experimentally by using a lighted candle and a convex lens.

Take a lighted candle and place it in front of a convex lens mounted on a stand. Move the candle back and forth to obtain the full image on the screen. Once the full image is observed, mark the position of the candle. Now cover the lower half of the lens with a black opaque paper. Do not change the position of the candle.

Now, we observe a full image of the candle, but the intensity is reduced. This is because the covered part of the lens does not allow light to pass through it. So the amount of light reaching the screen is reduced.

Question 118.
An object 5 cm in length is held 25 cm away from a conversine lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed. Also find the height of the image.
Answer:
Given
Object size, h’ = 5 cm
Object distance, u = -25 cm
Focal length of the lens,f= 10 cm
Image distance, v = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img86
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img87
A real image is formed on the other side of the lens at 16.66 cm.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img88
A real, inverted and diminished image 3.3 cm tall is formed below the principal axis.

Question 119.
What is meant by power of a lens? Mention its S.I. unit
Answer:
The degree of convergence or divergence of light rays achieved by a lens is expressed in terms of its power. The power of a lens is defined as the reciprocal of its focal length, when focal length is expressed in metre.
Power (P)=\frac{1}{\text { focal length (in metre) }}=\frac{1}{f}
The S.I. unit of power of a lens is called ‘dioptre’ (D).

Question 120.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of a lens whose focal length is 1 metre.

Question 121.
What is the meaning of “The power of a lens is 1 dioptre”?
Answer:
One dioptre is the power of a lens whose focal length is 1 metre.

Question 122.
Find the power of a concave lens of focal length 2 m.
Answer:
\text { Power }=\frac{1}{f}=\frac{1}{-2 m}=-0.5 D

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 123.
A convex lens forms a real and inverted image of a needle at a distance of SO cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Given
Image distance, v = 50 cm
Magnification, m = 1
Object distance, u = ?
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img89
The needle is placed 50 cm from the lens.
Focal length of the convex lens f=\frac{50}{2}=25 \mathrm{cm}=0.25 \mathrm{m}
Power of the lens P=\frac{1}{f}=\frac{1}{0.25 m}=+4 D

Question 124.
Find the focal length of a lens of power -2.0 D. What type of lens is this?

OR

If the power of a lens is -2.0 D, then what type of lens is it?
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img90
Since f is negative, the lens is concave.

Question 125.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or conversine?
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img91
Since power is positive, lens is converging or convex

Question 126.
Name the type of mirror that always produces virtual image. Mention one use of this mirror.
Answer:
A concave lens always gives a virtual image. Concave lens is used to correct myopia.

Question 127.
An object is kept at a distance of 20 cm from a lens of power +5 D. State the nature, position and relative size of the image formed in this situation. Draw a suitable ray diagram for this.
Answer:
Focal length (in metre), f=\frac{1}{\text { Power }}=\frac{1}{+5 D}=+0.2 \mathrm{m} .=+20 \mathrm{cm}
This is a convex lens of focal length 20 cm. The object is kept at F. Therefore, a real, inverted and highly enlarged image is formed at infinity. The ray diagram for this is given below
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img92

Question 128.
The object distance of a lens is -30 cm and image distance is -10 cm. Find the magnification of the lens. With the help of this, decide whether the size of the image is smaller or bigger than the size of the object
Answer:
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img93

Fill In The Blanks

1. A lens that converges light rays is convex lens
2. The centre of the reflecting surface of a spherical mirror is called pole
3. The radius of curvature of a spherical mirror of focal length 20 cm is 40 cm
4. The magnification in a plane mirror is always equal to 1
5. The mirror used as rear-view mirror in vehicles is convex mirror
6. The reciprocal of the focal length of a lens is called its power
7. According to Snell’s law, the ratio between sine of the angle of incidence and the sine of the angle of refraction is always constant
8. The centre of the sphere of which the refracting surface of a lens forms a part is called its centre of curvature

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Multiple Choice Questions

Question 1.
Which one of the following materials cannot be used to make a lens?
(A) Water
(B) Glass
(C) Plastic
(D) Clay
Answer:
(D) Clay

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the position of the object be?
(A) Between the principal focus and the centre of curvature
(B) At the centre of curvature
(C) Beyond the centre of curvature
(D) Between the pole of the mirror and its principal focus.
Answer:
(D) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(A) At the principal focus of the lens
(B) At twice the focal length
(C) Between the optical centre of the lens and its principal focus.
(D) At infinity
Answer:
(B) At twice the focal length

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of-15 cm. The mirror and the lens are likely to be
(A) both concave.
(B) both convex.
(C) the mirror is concave and the lens is convex.
(D) the mirror is convex, but the lens is concave.
Answer:
(A) both concave.

Question 5.
No matter how far you standfrom a mirror, your image appears erect. The mirror is likely to be
(A) plane.
(B) concave.
(C) convex.
(D) either plane or convex.
Answer:
(D) either plane or convex.

Question 6.
The image formed by a plane mirror is always
(A) virtual and erect
(B) real and erect
(C) real and inverted
(D) virtual and inverted
Answer:
(A) virtual and erect

Question 7.
A concave mirror gives a real, inverted image equal to the size of the object when the object is placed
(A) at F
(B) at infinity
(C) at C
(D) beyond C
Answer:
(C) at C

Question 8.
If the power of a lens is -40 D, its focal length is
(A) 4 m
(B) -40 m
(C) -0.25 m
(D) -0.025 m
Answer:
(D) -0.025 m

Question 9.
QuestA concave mirror gives a virtual, erect and enlarged image of the object when the object is
(A) at infinity
(B) between F and C
(C) between P and F
(D) beyond C
Answer:
(C) between P and F

Question 10.
In optics, an object which has higher refractive index, is called
(A) optically denser
(B) optically rarer
(C) optical density
(D) refractive index
Answer:
(A) optically denser

Question 11.
The optical phenomena, twinkling of stars, is due to
(A) Atmospheric reflection
(B) Atmospheric refraction
(C) Total reflection
(D) Total refraction
Answer:
(B) Atmospheric refraction

Question 12.
Convex lens gives a real, point sized image at the principal focus when the object is placed
(A) at 2F
(B) at focus
(C) at infinity
(D) between F and 2F
Answer:
(C) at infinity

Question 13.
The unit of power of lens is
(A) Metre
(B) Centimetre
(C) Dioptre
(D) joule per second
Answer:
(C) Dioptre

Question 14.
The radius of curvature of a mirror is 20 cm. The focal length is
(A) 10 cm
(B)20cm
(C).40cm
(D) 5 cm
Answer:
(A) 10 cm

Question 15.
The rear view mirror in motor vehicles is a
(A) plane mirror
(B) convex mirror
(C) concave mirror
(D) convex lens.
Answer:
(B) convex mirror

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 16.
A concave mirror forms a magnified inverted image when the object is placed at
(A) F
(B) C
(C) beyond C
(D) between F and C
Answer:
(D) between F and C

Question 17.
The laws of reflection hold good for
(A) plane mirrors only
(B) convex mirrors only
(C) concave mirrors only
(D) all types of mirrors
Answer:
(D) all types of mirrors

Question 18.
The mirror formula for a spherical mirror is
(A) \frac{1}{f}=\frac{1}{u}+\frac{1}{v}
(B) \frac{1}{f}=\frac{1}{2 u}+\frac{1}{v}
(C) \frac{1}{f}=\frac{1}{u}-\frac{1}{v}
(D) \frac{1}{f}=\frac{1}{u}+\frac{1}{2 v}
Answer:
(A) \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Question 19.
Negative value offocal length of a spherical mirror indicates that it is
(A) concave mirror
(B) convex mirror
(C) plane mirror
(D) convex mirror of small focal length
Answer:
(A) concave mirror

Question 20.
A convex mirror always gives
(A) a real, inverted, enlarged image of the object
(B) a real, diminished, erect image of the object
(C) a virtual, erect, diminished image of the object
(D) a virtual, inverted, enlarged image of the object.
Answer:
(C) a virtual, erect, diminished image of the object

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 21.
On a new stainless steel spoon, one can see the image of one’s face upside down. The part of the spoon acts like a
(A) convex lens
(B) concave mirror
(C) convex mirror
(D) concave lens
Answer:
(B) concave mirror

Question 22.
According to Cartesian Sign Convention,
(A) object distance is always negative.
(B) object distance is always positive
(C) image distance is always negative.
(D) image distance is always positive.
Answer:
(A) object distance is always negative.

Question 23.
Power of a lens is
(A) equal to its focal length
(B) reciprocal of the focal length (in metres)
(C) square of its focal length (in metres)
(D) reciprocal of its radius of curvature.
Answer:
(B) reciprocal of the focal length (in metres)

Question 24.
An object placed between F and 2F of a convex lens will produce
(A) a virtual image
(B) a diminished image
(C) a real and inverted image
(D) an erect image
Answer:
(C) a real and inverted image

Question 25.
Lens formula is expressed as
(A) \frac{1}{v}-\frac{1}{u}=\frac{1}{f}
(B) \frac{1}{v}+\frac{1}{u}=\frac{1}{f}
(C) \frac{1}{u}-\frac{1}{v}=\frac{1}{f}
(D) u + v = f
Answer:
(A) \frac{1}{v}-\frac{1}{u}=\frac{1}{f}

KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction

Question 26.
Magnitude of magnification less than 1 indicates that the
(A) size of image > size of object
(B) size of image = size of object
(C) size of image < size of object
(D) size of image is independent of size of object.
Answer:
(C) size of image < size of object Question

27. According to the law of refraction (Snell’s law),
(A) \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{constant}
(B) angle i = angle r
(C) sin i = sin r
(D) sin i > sin r
Answer:
(A) \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{constant}

Question 28.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(A) A convex lens of focal length 50 cm.
(B) A concave lens of focal length 50 cm.
(C) A convex lens of focal length 5 cm.
(D) A concave lens of focal length 5 cm.
Answer:
(C) A convex lens of focal length 5 cm.

Question 29.
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img94
Observe the figure above. The correct figure indicating the direction of the light ray FG after refraction is
(A) KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img95
(B) KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img96
(C) KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img97
(D) KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img98
Answer:
(D)
KSEEB Class 10 Science Important Questions Chapter 10 Light Reflection and Refraction img98

Match The Following

Column A Column B
Position of the object with respect to a convex lens Position of the image
1. Between O and F (a) At F
2.  At F (b) At 2F
3.  Between F and 2F (c) Between F and 2F
4.  At 2F (d) Beyond 2F
5.  Beyond 2F (e) Same side of lens as objec
(f) At infinity -c.

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

Column A Column B
Position of the object with respect to a concave mirror Properties of the image
1. Between P and F (a) Real and same size as the object
2. At F (b) Virtual and enlarged
3. Between C and F (c) Real and highly enlarged
4. At C (d) Real and enlarge
5. Beyond C (e) Real and point sized
6. At infinity (f) Real and diminished
(g) Real point image – f, 6 – e.

Answer:
1 – b, 2 – c, 3 – d, 4 – a, 5 – f, 6 – e.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Students can download Class 10 Science Chapter 9 Heredity and Evolution Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 1.
What are inherited traits?
Answer:
The traits (characteristics) that are transferred from the parents to their offspring through genes are called inherited traits.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 2.
What is heredity?
Answer:
The transmission of characteristics (traits) from the parents to their offspring through genes is known as heredity.

Question 3.
What is genetics?

OR

Define genetics.
Answer:
The branch of biology that deals with the systematic study of heredity, variation and factors responsible for these is known as genetics.

Question 4.
Who is considered as the father of modern genetics?
Answer:
Gregor Johann Mendel is considered as the father of modem genetics.

Question 5.
Name the plant on which Gregor Mendel did his experiments on genetics.
Answer:
Mendel performed his experiments on heredity and genetics on garden pea plants.

Question 6.
What is the contribution of Gregor Mendel to our understanding of inheritance of characteristics?
Answer:
Mendel did his experiments on garden pea (Pisum sativum) and discovered the scientific principles which govern patterns of inheritance, i.e., the principle of inheritance. He explained that contrasting characters are controlled by units which he called factors. Today these factors are called genes.

Question 7.
Why did Mendel choose pea plants for his experiments on inheritance?
Answer:
Mendel chose pea plants for his experiments for the following reasons

  1. Pea plants could be grown easily either in an open field or in pots.
  2. They have short growth period and short lifecycle.
  3. They bear self-pollinating flowers, which could be cross-pollinated artificially.
  4. They produce large number of seeds.
  5. They show a fairly large number of contrasting heritable characters.
  6. They produce fertile hybrids on cross-pollination.

Question 8.
What is variation? Why are hereditary variations important?
Answer:
The differences in the characters among the offspring compared to that of their parents is called variation. Variation is necessary for organic evolution.

Question 9.
Give an example for the variation found in human population.
Answer:
If we Observe the ear lobes of a set of people, some have free ear lobes while a few have attached ear lobes. This is an example of variation in human population.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
Sexual reproduction involves two parents and hence produces relatively large number of variations. The two parents already have variations accumulated from the previous generations and hence their progeny would be novel as they inherit traits from both parents. In asexual reproduction, however, the organisms have fewer variations due to errors in DNA copying.

Variations in organisms are the key for evolution. Since sexually reproducing organisms have more variations, the process of evolution gains speed. Asexual reproduction takes longer time for the evolution of newer species.

Question 11.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, the variations that give advantage to an individual organism will increase its survival in the population. This is because the favourable variations enable the organism to survive even in unfavourable conditions and hence increase the chances of its survival.

Such organisms reproduce and increase their number in the population. Those with unfavourable variations diminish in number and eventually perish. Thus more offspring and population with genetic variations will survive.

Question 12.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier? Give reason.
Answer:
Asexually reproducing species reproduce by cell division creating two identical organisms (same DNA). In asexual reproduction, the chances of variation due to inaccuracies in DNA copying are quite small.

If a trait is found in only some members of a species then it originally occurred due to a random mutation. This random mutation then is passed down to future generations every time the organism with the trait replicates itself.

Thus, if 60% of a population contains this trait, it means that its members have been replicating themselves for a longer period of time than those in the population where the trait is seen in only 10% of the population. Hence it is more likely that trait B has emerged earlier than trait A.

Question 13.
How does the creation of variations in a species promote survival?
Answer:
Variations occur in species due to errors in DNA copying. Some of these variations may be favourable to the organism while others may not be. Depending on the nature of the variations, different individuals of a species get different kinds of advantages and enable them to adjust and adapt to changes in the environment.

Organisms with favourable variations survive even under changed conditions and reproduce and continue their species. This is how variations help species to adapt and hence promote their survival.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 14.
When a pesticide is sprayed on a population of insects, all insects do not get killed and a few of them survive. Give reason.
Answer:
When a pesticide is sprayed on a field, most of the insects are killed because they are sensitive to the pesticide. A few of the insects survive because they have variations, which have made them resistant to the pesticide. Hence the pesticide has no effect on these insects.

Question 15.
What is pollination? What is its importance? Mention its types.
Answer:
The transfer of pollen from the anther of a flower to the stigma of the same flower or of another flower is called pollination. Pollination is an essential prerequisite for fertilization.
There are two types of pollination namely

  • self-pollination
  • cross-pollination.

Question 16.
Differentiate between self-pollination and cross-pollination.
Answer:
Self-pollination:

  1. The transfer of pollen grains from the anther to the stigma of the same flower or another flower of the same plant is called self¬pollination.
  2. Self-pollinating flowers usually are small and less attractive without coloured petals, nectar, scent or long stamens and pistils.
  3. It results in uniform progeny. It allows plant to be less resistant as a whole to diseases. However, if does not need to expend energy to attract pollinators.
  4. Self-pollinating flowers usually have fewer pollen grains.

Cross-pollination:

  1. The transfer of pollen grains from the anther of a flower to the stigma of a flower of a different plant of the same species is called cross-pollination.
  2. Cross-pollination is seen in plants with brightly coloured petals, nectar and scent, long stamens and pistils.
  3. It allows for diversity in the species. However, it relies on pollinators that travel from plant to plant.
  4. Cross-pollinating flowers have large number of pollen grains.

Question 17.
Make a list offour observable traits in pea plants with their contrasting forms of characters.
Answer:

  1. The length of the stem has two contrasting forms namely ‘tall’ and ‘dwarf.
  2. The colour of the pod has two contrasting forms namely ‘yellow’ and ‘green’.
  3. Colour of the seed coat has two contrasting forms namely ‘grey’ and ‘white’.
  4. Position of the flower has two contrasting forms namely ‘axial’ or ‘terminal’.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 18.
Which were the seven pairs of contrasting traits in pea used by Mendel for his experiments on heredity?
Answer:

  • Character:
  • Stem height
  • Flower colour
  • Flower position
  • Pod shape
  • Pod colour
  • Seed shape
  • Seed colour

Contrasting traits:

  • Tall/dwarf
  • Red/white
  • Axial/terminal
  • Full/constricted
  • Green/yellow
  • Round/wrinkled
  • Yellow/green

Question 19.
Define the following terms: Character, Trait, Gene, Hybrid, Phenotype, Genotype, Homozygous, Heterozygous, Dominant, Recessive, Monohybrid cross, and Dihybrid cross.
Answer:

  1. Character: Any inheritable feature of an organism is called character.
  2. Trait: Any detectable contrasting variant of a character is called a trait.
  3. Gene: A gene is a unit of heredity that determines particular traits.
  4. Hybrid: A plant or an animal produced by parents that have contrasting characteristics of a trait is called a hybrid.
  5. Phenotype: Observable characteristics of an organism, which are genetically controlled, are called phenotype.
  6. Genotype: The genetic constitution of an organism is called its genotype.
  7. Allele: An alternative form of the same gene found at the same place on a chromosome is called allele.
  8. Homozygous: Two alleles of a gene are said to be homozygous if they are similar and hence two copies of the same allele exist. (Eg. RR or rr.)
  9. Heterozygous: Two alleles of a gene are said to be heterozygous if they are different. (Eg. Rr.) Dominant: A gene in a pair is said to be dominant if it masks or hides the expression of the other.
  10. Recessive: A gene in a pair is said to be recessive if it expresses itself only when there are two of them. Recessive genes do not express in the presence of a dominant allele.
  11. Monohybrid cross: Mendelian cross in which only one pair of contrasting characteristics is taken into consideration at a time is called monohybrid cross.
  12. Dihybrid cross: Mendelian cross in which two pairs of contrasting characteristics are taken into consideration at a time is called dihybrid cross.

Question 20.
Describe Mendel’s monohybrid cross experiments.
Answer:
Mendel took pure varieties of pea plants with contrasting characteristics. For instance, he took pure tall varieties and pure dwarf varieties and cross-pollinated them. He collected the seeds of these plants and sowed them in the next season. All the seeds bore tall varieties of plants. He called these hybrids ‘the first filial generation’ (F1 generation).

In the next season, Mendel sowed seeds obtained in the F1generation and self-pollinated them. He collected the seeds of these plants and sowed them in the next season. He found that these seeds produced tall and dwarf varieties in the ratio 3:1. He called these plants ‘the second filial generation’ or F2 generation.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 21.
How many factors will be there for each trait in sexually reproducing organisms? What are they?
Answer:
In any sexually reproducing organism, there will be two factors for each trait. One is the dominant factor and the other is the recessive factor.

Question 22.
What is monohybrid cross? What is its ratio? Show with the help of a checker board.
Answer:
Mendelian crosses, in which only one pair of contrasting characteristics is taken into consideration at a time, is called monohybrid cross. Monohybrid ratio obtained when only one pair of contrasting characteristics is taken at a time is 3:1.
KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img1
Phenotypic ratio: 3 tall: 1 dwarf
Genotypic ratio: 1 pure tall: 2 impure tall: 1 pure dwarf.

Question 23.
State the principle of dominance with an illustration.
Answer:
For any given trait, there are two forms of factors. One of these factors is dominant and the other is recessive. This idea is known as the principle of dominance. Consider a situation where tall pea plants are cross-pollinated with dwarf pea plants. When seeds obtained from them are sown again, we get all tall plants.

Here, tallness is a dominant factor and dwarfness is a recessive factor. When the plants grown from these seeds are self-pollinated and the next generation (F2 generation) seeds are obtained and sown again, dwarfness appears in some plants. This shows that the factor for dwarfness was masked in the F1 generation by the factor of tallness. We consider the factor for dwarfhess as the recessive trait.

Question 24.
Give the pair of contrasting traits of the following characters in pea plant and mention which is dominant and which is recessive

  1. Yellow seed
  2. Round seed.

Answer:

  1. Seed colour: Yellow (Dominant); Green (Recessive).
  2. Seed shape: Round (Dominant); Wrinkled (Recessive).

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 25.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
In his experiment on pea plants, Mendel cross-pollinated plants with contrasting pair (say tall and dwarf) of characteristics and obtained the seeds of F1 generation. All the plants grown from the F1 generation seeds showed only one trait of the pair (tall) and the other trait (dwarfhess) was missing in all the plants.

He self-pollinated these plants and collected the seeds ef this generation and grew plants of the F2 generation. A trait that appeared in all the members of the F1 generation (tallness) appeared only in 75% of members of F2 generation and the other 25% were dwarf. This means a trait that did not appear in the Fj1generation had reappeared in 25% plants of F2 generation.

This indicated that the trait that appeared in F1 generation in all plants was the dominant trait. A trait that did not appear in F1 but reappeared in F2 generation was the recessive trait. This established that traits may be dominant or recessive.

Question 26.
Genes related to one character have two contrasting traits. But only one among them is considered as dominant. Why?
Answer:
Of the two contrasting traits of an inherited character, the one that expresses itself in the next generation and determines the phenotype of the individual is called the dominant trait. This trait determines the character of the individual despite the presence of a contrasting allele for the same character.

Question 27.
In Mendel’s experiments on pea plants, were the tall plants in the Ft generation exactly the same as the tall plants of the parent generation? Explain.
Answer:
In Mendel’s experiment, the tall pea plants in the F, generation were not the same as the tall plants of the parent generation. The parent generation had pure tall varieties with homozygous alleles. Their genotype can be represented by TT.

This means the seeds obtained by self-pollination of these plants would yield only tall plants. In the Fi generation, the tall plants were heterozygous with their genotype being Tt. The seeds obtained by self-pollination of these plants’would yield some tall plants and a few dwarf plants.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 28.
The findings of Mendel’s experiment with pea plants with pure purple and pure white flowers are shown in the diagram given below. State the dominant trait and the recessive trait in this.
Answer:
KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img2
In this experiment, the trait for purple colour flowers is dominant. The trait for white flowers is recessive.

Question 29.
Outline a project that aims to find the dominant coat colour in doss.
Answer:
Dogs have a variety of genes that govern coat colour. There are at least eleven identified gene series that influence coat colour in dog. In order to find the dominant coat colour, we may choose one pair of contrasting traits. A dog inherits one gene for its coat colour from each of its parents. The dominant gene gets expressed.

For example, let us consider the factors that make a dog genetically black or brown. Let us assume that one parent is homozygous black (BB), while the other parent is homozygous brown (bb). By cross breeding these pure varieties we get the following genotypes:
KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img3
In this case, all the offsprings will be heterozygous black (Bb). B is the dominant factor while b is the recessive factor. If such heterozygous dogs are crossed again, they will produce 25% homozygous black (BB), 50% heterozygous black (Bb), and 25% homozygous brown (bb) offsprings.
KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img4
This shows that the factor for black is dominant while the factor for brown is recessive.

Question 30.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel crossed pure breeds of tall plants having round seeds with pure breeds of short plants having wrinkled seeds. The plants of Fj generation were all tall with round seeds indicating that the traits of tallness and roundness of seeds were dominant.

Self-breeding of Ft yielded plants with characters of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and 1 short wrinkled seeded. Tall wrinkled seeded and short round seeded plants are new combinations which could develop only when the traits are inherited independently.

Question 31.
How does inheritance work with asexually reproducing organisms? How is it different from the , inheritance in sexually reproducing organisms?
Answer:
The asexually reproducing organisms receive the complete set of chromosomes from a single parent but in sexually reproducing organisms, one set is received from each parent through gametes. These gametes then fuse and restore the chromosome number similar to that of parents.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 31.
How does inheritance work with asexually reproducing organisms? How is it different from the inheritance in sexually reproducing organisms?
Answer:
The asexually reproducing organisms receive the complete set of chromosomes from a single parent but in sexually reproducing organisms, one set is received from each parent through gametes. These gametes then fuse and restore the chromosome number similar to that of parents.

Question 32.
A study found that children with tight-coloured eyes are likely to have parents with lieht- coloured eves. On this basis, can we say whether the lisht eve colour trait is dominant or recessive? Why or why not?
Answer:
On the basis of the information provided, it is difficult to say if light colour of eye is a dominant trait or a recessive trait. However if the eye colours of both the parents were known along with the knowledge of whether the genes contributing for it are dominant or recessive type, then the nature of the trait could have been predicted.

Question 33.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
The information provided here is not enough to decide whether the trait for blood group A is dominant or the trait for blood group O is dominant. Either is possible as the individual carries two alleles for a trait. Recessive trait appears only when the two alleles are similar (homozygous). In the present case, there can be two possibilities

Possibility 1: When blood group A is dominant trait but blood group O is recessive trait: When father’s blood group A is dominant trait, it can have two genotypes: IAIA and IAI°. And when mother’s blood group O is recessive trait it can have only one genotype: I°I° (because it should have two recessive alleles). Now, if one recessive allele 1° comes from father and one recessive allele 1° comes from mother, then the daughter can also have the genotype I°I° which can give her blood group O.

Possibility 2: When blood group A is recessive trait but blood group O is dominant trait: When father’s blood group A is recessive trait, it can have only one genotype: IAIA (because it should have two recessive alleles). And when mother’s blood group O is dominant trait, then it can have two genotypes: I°I° and I°IA. Now, if one dominant allele 1° comes from the mother and one recessive allele IA comes from the father, the daughter will have the genotype I°IA which will again give her blood group O.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 34.
What is dihybrid cross? What is the dihybrid ratio obtained in the F2 generation by Mendel in his experiments on inheritance of characteristics?
Answer:
Show this with the help of a checker board. Mendelian crosses in which two pairs of contrasting characteristics are taken into consideration at a time are called dihybrid cross. The dihybrid ratio obtained by Mendel in the F2 generation was 9:3:3:1.

This means that for every nine tall plants with red flowers, there were three tall plants with white flowers and three dwarf plants with red flowers. There was one dwarf plant with white flowers for every nine tall plants with red flowers.
KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img5

  • Dihybrid ratio of the phenotype is 9:3:3:1
  • Tall plants with red flowers: 9
  • Tall plants with white flowers: 3
  • Dwarf plants with red flowers: 3
  • Dwarf plants with white flowers: 1.

Question 35.
What are chromosomes?
Answer:
A thread-like structure of nucleic acids and protein found in the nucleus of most living cells, carrying genetic information in the form of genes is called chromosome. Chromosomes bear genes, which are composed of DNA. Genes are responsible for the transmission of hereditary characteristics from one generation to another.

Question 36.
What is DNA?
Answer:
DNA stands for Deoxyribonucleic Acid. It is a macromolecule found in chromosomes of living beings and carries the genetic information of the organism.

Question 37.
Define gene.
Answer:
A segment of DNA that forms the hereditary unit is called a gene.

Question 38.
How is the sex of the child determined in human beings?

OR

Explain the process of sex determination in human beings.
Answer:
The sex of a child in humans is determined by the kind of male gamete (sperm) that fertilizes the female gamete (ovum). If a sperm carrying X chromosome fertilizes the ovum, the child will be a girl. If a sperm carrying Y chromosome fertilizes the ovum, the child will be a boy.
KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img6

question 39.
Write the differences between the sex chromosomes of man and the sex chromosomes of woman.
Answer:
A woman has a pair of identical sex chromosomes namely XX. A man has a normal sized chromosome called X and a shorter chromosome called Y. Thus the sex linked chromosome in a man are XY.

Question 40.
Sex of a child is determined by the father. How?
Answer:
A mother can give only X chromosome to her progeny. The father can give either X or Y chromosome. A child who gets X chromosome from the father will be a girl and a child who gets Y chromosome from the father will be a boy. Thus the sex of a child is determined by the father.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 41.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
In human beings, equal genetic contribution of male and female parents is ensured in the progeny through inheritance of equal number of chromosomes from both parents. There are 23 pairs of chromosomes. All human chromosomes are not paired. Out of these 23 pairs, the first 22 pairs are known as autosomes and the remaining one pair is known as sex chromosome represented as X and Y.

Females have a perfect pair of two X sex chromosomes and males have a mismatched pair of one X and one Y sex chromosome. During the course of reproduction, as fertilisation takes place, the male gamete (haploid) fuses with the female gamete (haploid) resulting in formation of the diploid zygote.

The zygote in the progeny receives an equal contribution of genetic material from the parents. Out of 23 pairs of chromosomes in progeny, male parent contributes 22 autosomes and one X or Y chromosome and female parent contributes 22 autosomes and one X chromosome.

Question 42.
What are acquired traits? Give an example.
Answer:
A trait of an organism, which is not inherited from parents but develops in response to the environment, is called an acquired trait. For example, a person may learn the skills of skating during her lifetime. This trait is not passed on to the person from her parents through genes. Thus, acquired traits are not inherited.

Question 43.
What are inherited traits? Give an example.
Answer:
Those traits that come to an individual from parents through gene transfer and hence can be passed on to the next generation through genes are called inherited traits. For example, colour of eyes, hair texture, etc., are inherited traits.

Question 44.
Why are traits acquired during the lifetime of an individual not inherited?
Answer:
Traits acquired during the lifetime of an individual just enable an individual to adapt itself to the surrounding and do not change the genetic make up of the organism. An acquired trait involves change in non-reproductive tissues, which cannot be passed on to germ cells or the progeny. Therefore, these traits cannot be inherited.

Question 45.
Rana’s father is a wrestler and has a robust body. His son is thin and has a relatively weak body structure.

  1. Is it true that a wrestler’s son should also have heavy muscles?
  2. What type of character is it – acquired or inherited?
  3. If you are Rana’s friend, how will you convince him that his son is normal?

Answer:
1. It is not necessary that a wrestler’s son should also have heavy muscles.

2. Having heavy muscles in the body is an acquired trait. They are not inherited through genes. Heavy muscles can be acquired by regular exercising and good diet.

3. We can convince Rana by telling him that traits like heavy muscles are not inherited but are acquired in one’s lifetime. His son can start exercising on a regular basis and take nutritious food and develop muscles just like his father. If he is not as strong as his father, there is also a possibility that he has inherited more traits from his mother.

Question 46.
What is DNA copying? What is its importance?
Answer:
The process by which a double-stranded DNA molecule is copied to produce two identical DNA molecules is called DNA copying. DNA copying gains tremendous importance during cell division. DNA copying during cell division ensures the restoration of the same genetic information in the next generation of cells.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 47.
What are the different wavs in which individuals with a particular trait may increase in a population?
Answer:
Traits arise due to variations, which occur due to sexual reproduction of inaccuracies during DNA copying or environmental factors. Individuals with a particular trait may increase in a population due to the following factors:

  1. Natural selection: Those variations, which give survival advantage to an organism, are selected in nature and such traits increase in the population.
  2. Genetic drift: It occurs due to change in gene frequency due to accumulation of particular type of genes.
  3. Geographical isolation: It leads to change in gene frequency leading to expression of one type of trait in a eographically isolated population.
  4. Migration: It leads to flow of a particular type of gene in a specified population.

Question 48.
Why are the small number of surviving tigers a cause of worry from the point of view of genetics?
Answer:
The small number of tiger population does not allow large number of variations to occur. Wider range of variations are however essential for the survival of the species. A deadly disease or calamity may cause death of all of the few remaining tigers. The small number of tigers also indicates that existing tiger variants are not well adapted to the existing environment and hence may be facing threat of extinction.

Question 49.
What is speciation ? How does it happen? Explain.
Answer:
The evolutionary process that leads to the formation of new but distinct species from the main population of an existing species is known as speciation. The new species that are formed have a distinct genetic composition that is different from the main population and therefore interbreeding between the members of the main population and those of the new population is not possible.

Speciation occurs when a population exhibits a continuous change due to changes in the environment. These unit changes get accumulated over a period of time in a section of the population which slowly alter the individual’s genetic structure and functions to give rise to a new species.

Question 50.
With the help of the following pattern briefly explain the reason for

  • DNA stability,
  • Variation phenomenon found in successive generations of species.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution img7
Answer:
A close observation of the given diagram reveals that the DNA molecule has a double helical structure with two strands twisted in a twisted ladder shape. All molecules of DNA have the same twisted ladder structure, which is highly stable. The double helix structure of DNA gives the DNA molecule its physical and chemical properties as well as stability. The stability of DNA structure helps the strands to hold together.

Sexual reproduction involves two parents and hence produces relatively large number of variations due to DNA crossing over. A few variations will also occur due to errors in DNA copying. The two parents pass on the accumulated variations to their progeny.

Hence their progeny would be unequal as they inherit traits from both parents. In asexual reproduction, however, the organisms have only fewer variations due to errors in DNA copying. Variations in organisms are the key for evolution.

Question 51.
What is genetic drift? Explain with an example.
Answer:
A random change (happens strictly by chance) in the gene pool of a small population is called genetic drift. Consider the example of an exploding volcano that destroys almost all of the most common trees on a small island. Over time, the types of trees that were not affected by the volcano continue to flourish, while the population of the tree that was once common in the area would dwindle.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 52.
What factors could lead to the rise of a new species?
Answer:
The factors that could lead to the rise of a new species are
1. By natural selection:
Organisms with a particular trait may be naturally selected because it provided a survival advantage. That particular trait may thus increase in the population. It may direct evolution of species population by adaptations to fit their environment better.

2. Genetic drift and inheritance:
An accident in small populations may result in surviving of organisms with a particular variant only. This can increase the frequency of some genes in that population, even if they give no survival advantage. This is the notion of genetic drift, which provides diversity without any adaptations.

Question 53.
Read the following information and answer the given questions:

  • Thousands of years ago only one species of squirrels was there and were evolved from a common ancestor.
  • At present there are two species of squirrels; though they have similarities among them, they cannot perform reproduction between them.
  1. Analyze the factor responsible for this change.
  2. How can this kind of changes be considered as beneficial for a species?

Answer:
1. In this case, two species of squirrels have emerged from a single common ancestral species. This has happened through a process called speciation. Speciation is an evolutionary process that has led to the formation of new but distinct species of squirrels from the main population of the pre-existing species of squirrels.

The new species of squirrels has a distinct genetic composition that is different from that of the main population. Therefore interbreeding between the members of the main population of squirrels and those of the new population is not possible. Speciation occurs due to the variations that exist among the members of the species.

These variations might have occurred due to natural selection, geographical isolation or genetic drift. Any one or a combination of these factors might have contributed to the evolution of a new species of squirrels over several generations. Speciation occurs because the individual members of a species (squirrels in this case) show lot of variation.

Some of these variations are favourable to the organisms. These favourable variations are transmitted to the progeny. These unit changes (variations) get accumulated over a period of time in a section of the population, which slowly alter the individual’s genetic structure and give rise to a new species of squirrels.

2. There is competition amongst the members of a given species for resources leading to struggle for existence. In this struggle, only the most adaptable will survive and reproduce. This produces young one’s that are better adapted for the next phase of struggle for existence.

These variations get accumulated over several generations leading to the emergence of new species that are distinct from the original species. This is called descent with modification. Thus the variations favour the survival of species and the emergence of better-adapted new species.

Question 54.
What is geographical isolation? How does it occur?
Answer:
The separation of two populations of the same species by a physical barrier, such as a mountain or body of water, is known as geographical isolation. Geographical isolation of a species occurs as a result of physical changes in the natural environment.

Any physical barrier such as a mountain or a water body that separates two populations of the same species results in geographical isolation and may lead to speciation.

Question 55.
Will geographical isolation be a major factor in the speciation of a self-oollinatine plant species? Why or why not?
Answer:
Geographical isolation can prevent cross-pollination with flowers of different plants. Since the plants are capable of self-pollination, the pollens are transferred from the anther of one flower to the stigma of the same flower or of another flower of the same plant and its distance from other plants hardly affects its reproduction.

Moreover, self-pollinated plants rarely show variations in characters. Therefore, geographical isolation cannot prevent speciation in the given case.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 56.
Will eeoeraphical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, geographical isolation cannot be a major factor in the speciation of an organism that reproduces asexually. This is because, asexually reproducing organisms pass on the parent DNA almost entirely to the offspring. This leaves no chance for speciation. However, geographical isolation works as a major factor in sexual reproduction involving cross-pollination.

Question 57.
Give an example of characteristics beine used to determine how close two species are in evolutionary terms.
Answer:
Presence of some common characteristics in two different species indicates a close relationship among them. For example, fossil studies indicate the presence of feathers in some ancient reptiles like dinosaurs.

However these animals used feather only to get some protection from cold and could not use them to fly. Birds evolved further and adapted the feathers for flight. This means that birds are very closely related to reptiles. In fact, birds are known to have evolved from reptiles.

Question 58.
What is organic evolution?
Answer:
The slow emergence of well adapted newer forms of organisms from pre-existing simpler organisms through modifications that occurred over millions of years is known as organic evolution.

Question 59.
What evidence do we have for the orisin of life from inanimate matter?
Answer:
The experiment conducted by Miller and Urey provides evidence for the origin of life from inanimate matter. These scientists took molecules of compounds that existed on early earth such as methane (CH4), ammonia (NH3), hydrogen (H2), and water (H2O) in a specially designed glass jar.

They provided energy for the interaction of these molecules in the form of an electric arc to simulate lightning storms. At the end of one week, they observed that as much as 10-15% of the carbon was now in the form of organic compounds. Two percent of the carbon had formed some of the amino acids which are used to make proteins.

This experiment showed that organic compounds such as amino acids, which are essential to cellular life, could be made easily under conditions that scientists believed to be present on early earth. This also gave support to the idea that life evolved from inanimate matter.

Question 60.
What is the theory proposed by Charles Darwin to explain organic evolution?
Answer:
The theory developed by Charles Darwin is called ‘theory of natural selection’. It is also known as Darwinism.

Question 61.
What are the main ideas of Darwin’s theory of evolution by natural selection?
Answer:
The following are the key points of Darwin’s theory of evolution

  • Existence of variation: There is variation in characteristics in every population.
  • Overproduction: All organisms tend to produce more offspring than can survive.
  • Struggle for existence: Organisms compete for limited resources.
  • Survival of the fittest: Only those individuals, which have most favourable variations, will survive and reproduce. Darwin called this ‘natural selection’.
  • Inheritance: Organisms with favourable adaptive characteristics will pass on their traits to their offspring.
    Origin of species: The accumulated changes over a long period of time lead to the emergence of new species.

Question 62.
What is natural selection? What is its consequence?
Answer:
The process by which nature selects individuals with the most favourable traits to survive and reproduce is known as natural selection.
As a consequence of natural selection, individuals in a given species with most favourable adaptations will survive.

Those with less favourable adaptations perish. As a result, individuals with favourable characteristics become more common in successive generations of a population and those with unfavourable characteristics become less common.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 63.
According to Darwin’s theory, how do new species originate?
Answer:
Darwin proposed in his theory that new species originate by natural selection. In the struggle for existence, only the fittest will survive and reproduce. This produces young ones which are better adapted for the next phase of struggle for existence. Darwin called it descent with modification. As the environment is ever changing, it demands new variations in the organisms.

Gradually new characters are accumulated in the offspring across generations. These accumulated changes over a long period of time will produce offspring, which vary in a major way from their original ancestors and hence will form new species.

Question 64.
What is artificial selection? Give an example.
Answer:
The process in which human beings artificially produce new plants or animals with improved characteristics through selective breeding is known as artificial selection.

The wild cabbage plant is a good example for artificial selection. Humans have, over more than two thousand years, cultivated wild cabbage as a food plant. They have generated different vegetables from it by artificial selection.

Question 65.
Mention some of the evidences that support evolution of life.
Answer:
The evidences that support evolution of life include the following

  • Fossil evidences
  • Presence of homologous organs
  • Presence of analogous organs.

Question 66.
How are the areas of study – evolution and classification – interlinked?
Answer:
The classification of organisms is a reflection of their evolutionary relationships. Classification is based on similarities and differences amongst organisms the more characteristics two organisms have in common, the more closely they are related and the more recently thpy will have had a common ancestor in the evolutionary chain the more different characteristics two organisms have, the more remotely they are related and they will have had a common ancestor in the more remote past.

Question 67.
What are fossils? Where are fossils usually found?
Answer:
The naturally preserved remains or impressions of a prehistoric plant or animal that had lived in the geological past are called fossils. Fossils are usually found between layers of rocks in petrified (rocky) form.

Question 68.
Which are the three basic types of fossils?
Answer:
Fossils are commonly found in the following forms

  • Fossils in the form of actual remains,
  • Petrified fossils, and
  • Moulds and casts.

Question 69.
Name the fossil bird that had characteristics of both reptiles and birds.
Answer:
The fossil bird that had characteristics of both reptiles and birds was archaeopteryx.

Question 70.
What do fossils tell us about evolution of life on earth?

OR

What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are the dead and decayed remains of organisms that lived millions of years ago. Fossils indicate that the present day organisms have evolved from previously existing ones. Thus fossils help us to reconstruct the evolutionary history of an organism.

The distribution pattern of fossils shows that the ancient fossils present in the bottom rocks are simpler, while the most recent fossils found in the upper strata are more highly evolved. It means, fossils form and become more and more complex as we proceed from earliest to recent rocks.

They give us an idea of time in history when different species were formed or became extinct. Fossils also help to trace the evolutionary history of some organisms.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 71.
Explain the importance of fossils in decidtne evolutionary relationships.
Answer:
Fossils are the remains of organisms formed over millions of years ago. Evolutionary relationship tells us how closely two organisms are related in the evolutionary tree. The fossils are an easy way of understanding the ancient species and also it helps us to find the missing link in the evolutionary tree.

By knowing more about the organism that lived so many centuries ago, we can understand the changes in the body designs and functions that led to the creation of species in due course of evolution and also evolution itself.

We get to understand the prehistoric period and the geographical changes that might have given rise to the organisms by analysing the structure of the organisms in fossils. Fossils are the key to understand the similarity between organisms and create a hierarchy that would help to classify the organisms and learn them easily without having learnt about all the organisms that ever survived.

A study of fossils helps us to know about the evolution of species. Fossils tell us how new species are developed from the old. Therefore, fossils have an importance in deciding evolutionary relationships.

Question 72.
Explain the terms analoeous and homologous organs with examples.
Answer:
Homologous organs:
The organs in different groups of organisms, which have the same basic structural design and origin but perform different functions, are called homologous organs.

For example, the forelimbs of a frog, a bird and a man have the same basic design of bones, but they perform different functions (frogs use them to jump, birds use them to fly and man use them to grasp).

Analogous qrgans:
The organs in different groups of organisms, which have different basic structural design and origin but have similar functions, are called analogous organs.

For example, the wings of birds and insects have different structural design but perform the same function, that is, assisting in flight. Therefore, they are analogous organs.

Question 73.
Write the differences between homologous organs and analogous organs.
Answer:

Homologous organs Analogous organs
1. Organs of different organisms have common origin. Organs of different organisms have different origins.                                                 •
2. They have similar structures but perform different functions. They have different structures but perform similar functions.
3. Ex: Forelimbs of frog and forelimbs of bird. Ex: Wings of bird and wings of bat.

Question 74.
Can the wine of a butterfly and the wine of a bat be considered homologous organs? Why or why not?
Answer:
Homologous organs are those that have the same basic structural design but perform different functions. Analogous organs are those organs that have different basic structural design but perform similar functions.

The structure of the wings of a butterfly and that of a bat are different but perform the same function. They help in flying. Therefore the wings of a butterfly and that of a bat are analogous organs and not homologous organs.

Question 75.
Is it necessary that organisms with homologous organs always have a common ancestor?
Answer:
Yes, organisms with homologous organs will have a common ancestor. This is because these organs have similar basic structure and similar embryonic origin. However, the organs are modified to perform different functions in different organisms.

Question 76.
Justify the following statements with one illustration each:

  1. Though organs of different organisms have more similarities in the shapes of organs, they need not be evolved from a common ancestor.
  2. Though variations are more between organisms, they might have evolved from a common ancestor.

Answer:
1. We see instances of different organisms having organs that are similar in shape but not evolved from a common ancestor. These are known as analogous organs. For example, birds, insects and bats have wings, and they all perform the same function, but they did not evolve from a common ancestor. Their wings work on different principles.

2. We see instances of more variations between the organisms but they have evolved from a common ancestor. These are called homologous organs. For example, the forelimbs of a frog, a bird and a man have the same basic design but perform different functions. Frogs use them to jump, birds use them to fly and humans use them to grasp.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 77.
Why are human beines who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
All human beings, even though they look different from each other in terms of size, colour and looks, belong to the same species because they have similar DNA sequences and have descended from the same ancestors. Further, human beings are capable of reproducing among themselves. These variations that we find in humans have arisen due to environmental factors, mutation or mixing of characters during reproduction.

Question 78.
In evolutionary terms, can we say which among bacteria, spider, fish and chimpanzee have a *better’ body design? Why or why not?
Answer:
Evolution is the generation of diversity due to environmental selection. Evolution has led to the emergence of more and more body designs over time. Among bacteria, spider, fish and chimpanzee, we can say that the chimpanzees have a better body design because their body design is highly complex with specialized body tissues and organs.

Chimpanzees are better adapted than others to survive in the present day environment. Therefore, we can say that from among the given set of organisms, chimpanzees are. better evolved and have a ‘better’ body design.

Question 79.
Is it true that human beings have evolved from chimpanzees? Justify your answer.
answer:
It is not true that human beings have evolved from chimpanzees. However, both human beings and chimpanzees have descended from a common ancestor. It is quite unlikely that the common ancestor is neither humans nor chimpanzees.

Also, the first step of separation from that ancestor is unlikely to have resulted in modem chimpanzees and human beings. Instead, the two resultant species have probably evolved in their own separate ways to give rise to the current forms.

Question 80.
Cat’s paw, human hand and horse’s legs – are these organs homologous or analogous? Give reason.
Answer:
Cat’s paw, human hand and horse’s legs are all homologous organs because they are of same origin (modified forearm) but perform different functions.

Question 81.
Is it fair to say that the body structure of the organisms that have emerged recently is more efficient than that of those that evolved earlier? Explain.
Answer:
It is not true that the body structure of recently evolved organisms is more efficient than that of those which evolved earlier. Many of the organisms that evolved earlier and have simpler body design are still well adapted to their conditions and are surviving in the environment.

Many of the simpler organisms are best adapted to survive in most inhospitable conditions. Therefore, there is no justification to argue that recently evolved organisms are better than those that evolved earlier. In other words, human beings are not spread across the world in stages.

Question 82.
What are the techniques commonly used to study human evolution?
Answer:
The tools such as excavating, time dating, studying fossils and determining DNA sequences are commonly used to study human evolution. These are the same techniques that are used to trace evolutionary relationships.

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 83.
What is the scientific name of human beings? Where did they originate first?
Answer:
The scientific name of humans is Homo sapiens. The earliest members of Homo sapiens have been traced to Africa. Our genetic footprints can be traced back to our African roots.

Question 84.
How and why did human race spread from Africa to other parts of the world?
Answer:
The earliest members of the human species can be traced back to our African roots. A couple of hundred thousand years ago, some of the early humans left Africa while others stayed on. While the residents spread across Africa, the migrants slowly spread across the planet. Their movement however was not linear.

They went forwards and backwards, with groups sometimes separating from each other, sometimes coming back to mix with each other, even moving in and out of Africa. Thus, study of the evolution of human beings indicates that all of us belong to a single species that evolved in Africa and

Fill In The Blanks

1. The plant used by Mendel to study the inheritance of characteristics was pea plant
2. The heredity material in all eukaryotes is DNA
3. The scientist who is called the father of modem genetics is Gregor Mendel
4. The scientist who gave the most acceptable theory about the origin of life was J.B.S. Haldane
5. The units of inheritance are called genes
6. Darwin’s theory of organic evolution is known as natural selection
7. The monohybrid ratio in Mendel’s experiment was 3:1
8. The gene with two identical alleles is said to be homozygous

Multiple Choice Questions

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short This suggests that the genetic make-up of the tall parent can be depicted as
(A) TTWW
(B) TTww
(C) TtWW
(D) TtWw
Answer:
(C) TtWW

Question 2.
An example of homologous organs is
(A) our arm and a dog’s foreleg.
(B) our teeth and an elephant’s tusks.
(C) potato and runners of grass.
(D) all of the above.
Answer:
(D) all of the above.

Question 3.
The critical determinant of the sex of a male child is the
(A) X-chromosome in the zygote.
(B) Y-chromosome in the zygote.
(C) Cytoplasm of germ cell.
(D) Environmental factor.
Answer:
(B) Y-chromosome in the zygote.

Question 4.
Exchange of genetic material takes place only in
(A) vegetative reproduction
(B) asexual reproduction
(C) sexual reproduction
(D) budding
Answer:
(C) sexual reproduction

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 5.
A cross between a tall plant (TT) and a short pea plant (tt) resulted in progeny that were all tall plants because
(A) tallness is the dominant trait
(B) shortness is the dominant trait
(C) tallness is the recessive trait
(D) height of pea plant is not governed by gene ‘T’ or ‘t’
Answer:
(A) tallness is the dominant trait

Question 6.
Father of Human genetics is
(A) Gregor Mendel
(B) Charles Darwin
(C) Sir Archibald Garrod
(D) J. B. S. Haldane
Answer:
(A) Gregor Mendel

Question 7.
The tendency of offspring to differ from parents is called
(A) heredity
(B) inheritance
(C) resemblance
(D) variation
Answer:
(D) variation

Question 8.
The character, which predominates and is clearly seen in F1 generation, is said to be
(A) acquired
(B) dominant
(C) recessive
(D) inherited
Answer:
(B) dominant

Question 9.
In peas, a pure tall plant (TT) is crossed with a short plant (tt). The ratio of pure tall plants to short plants in F2 is
(A) 1: 3
(B) 1 : 1
(C) 2:1
(D) 3: 1
Answer:
(D) 3: 1

Question 10.
If a round, green seeded pea plant (RR yy) is crossed with wrinkled, yellow seeded pea plant, (rr YY) the seeds produced in F1 generation are
(A) wrinkled and yellow
(B) round and green
(C) round and yellow
(D) wrinkled and green
Answer:
(C) round and yellow

Question 11.
How life might have originated on earth was experimentally shown by
(A) Gregor Mendel
(B) Oparin and Haldane
(C) Urey and Miller
(D) Watson and Crick
Answer:
(C) Urey and Miller

Question 12.
A gamete contains
(A) all alleles of an organism
(B) two alleles
(C) three alleles
(D) only one allele of a gene
Answer:
(D) only one allele of a gene

Question 13.
The two versions of a trait (character), which are brought in by the male and female gametes, are situated on
(A) copies of the same chromosome
(B) two different chromosomes
(C) sex chromosomes
(D) any chromosome
Answer:
(A) copies of the same chromosome

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 14.
Surgically removing tails of mice over several generations does not yield mice without tails. This proves that
(A) cutting the tail does not cause genetic change. So it is not inherited.
(B) characters acquired during one’s lifetime are not inherited.
(C) neither A nor B is correct
(D) both A and B are correct
Answer:
(D) both A and B are correct

Question 15.
The variation in the relative frequency of different genotypes in a small population, owing to the chance disappearance of particular genes as individuals die or do not reproduce is called
(A) Gene flow
(B) Genetic drift
(C) Genetic error
(D) Genetic inheritance
Answer:
(B) Genetic drift

Question 16.
Select the correct statement from among the following
(A) The eyes of octopus and vertebrates are homologous organs.
(B) The limbs of amphibians and mammals are analogous organs.
(C) The wings of birds and limbs of lizards are analogous organs.
(D) The wings of birds and wings of bats are homologous organs.
Answer:
(D) The wings of birds and wings of bats are homologous organs.

Question 17.
According to evolutionary theory, formation of a new species is generally due to
(A) sudden creation by nature
(B) accumulation of variations over several generations
(C) clones formed during asexual reproduction
(D) movement of individuals from one habitat to another
Answer:
(B) accumulation of variations over several generations

Question 18.
In evolutionary terms, we have more in common with
(A) a Chinese schoolboy
(B) chimpanzee
(C) spider
(D) bacterium.
Answer:
(A) a Chinese schoolboy

KSEEB Class 10 Science Important Questions Chapter 9 Heredity and Evolution

Question 19.
Identify the correct pair of analogous organs among the following
(A) The forelimb of man and the forelimb of a frog
(B) The wing of a butterfly and the wing of a bat
(C) The wing of a bird and the wing of a bat
(D) The forelimb of a lizard and the forelimb of a frog.
Answer:
(C) The wing of a bird and the wing of a bat

Match The Following

Column A Column B
1. Heredity (a) Development of new organism by modifications in pre-existing ones
3.  Genetics (b) Transmission of characters from parent to offspring
4.  Variation (c) Alleles representing the same trait
4.  Evolution (d) A fossil bird
5. Archaeopteryx (e) Branch of science that deals with heredity and variation
(f) Differences among the individuals of same species

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

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Karnataka 2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

1. सन्धिं विभजत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 1
उत्तरम्
बाल्यादेव = बाल्यात् + एव
एतन्नयेत् = एतत् + नयेत्
भिक्षाटनम् = भिक्षा + अटनम्

सन्धिं योजयत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 2
उत्तरम्
वयसि + एव = वयस्येव
कृतयः + तु = कृतयस्तु
कियत् + धनम् = कियद्धनम्

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

2. विग्रहवाक्यं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 3
उत्तरम्
अनवश्यकम् = न अवश्यकम्।
चिन्तनपद्धतिः = चिन्तनस्य पद्धतिः।
ज्ञानदाहः = ज्ञानस्य दाहः।

समस्तपदं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 4
उत्तरम्
धैर्य च स्थैर्यंच स्वाभिमानः च = धैर्यस्थैर्यस्वाभिमानाः।
राष्ट्रस्य कविः = राष्ट्रकविः
ज्योष्ठश्चासौ पुत्रश्चं = ज्येष्ठपुत्रः

3. लिङ्ग-विभक्ति-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 5
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 12

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

4. लकार-पुरुष-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 6
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 13

5. पदपरिचयं कुरुत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 7
उत्तरम्
सम्पाद्य – ल्यबन्ताव्ययम
अध्येतव्यः – तव्यत्प्रत्ययः
विदित्वा – क्त्वान्ताव्ययम्

6. प्रयोगं परिवर्तयत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 8
उत्तरम्
पत्रिकाम् आरब्धवान्।
पत्रिकाम् आरब्धम्।

कृष्णरावेण इयं घटना स्मर्यते।
कृष्णरावः इयं घटनां स्मरति।

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

योग्यताविस्तारः-
1. पाठे विद्यमानानि अनुनासिकसन्धेः उदाहरणानि चित्त्वा लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 9
उत्तरम्
1. मुखान्निसरन्ति

2. पाठे विद्यमानानि लोट् लकारस्य उदाहरणानि चित्त्वा लिखत –
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 10
उत्तरम्

  1. भवतु
  2. स्वीकरोतु
  3. जाग्रत
  4. सञ्चारयत
  5. पूरयत
  6. चिन्तयत
  7. पठत
  8. निबोदत

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

3. कन्नडभाषया – आङ्ग्लभाषया वा अनुवदत।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 11
उत्तरम्
अ) कृष्णशास्त्रीमहोदयः आदर्शः अध्यापकः अभवत् ।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 14
Krishnashastri became an ideal teacher.

आ) एतेषां बङ्किमचन्द्रः कृतिः केन्द्र साहित्य – अकादमी द्वारा पुरस्कृता।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 15
His work called ‘Bankimchandra’ has been honoured by the Kendra Sahitya Akademi.

इ) विद्यार्थिनां विद्यार्जनमेव तपः ।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 16
Learning itself is a penance for students.

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

ई) ध्वन्यालोकः इति कृतेः कन्नडभाषानुवादः डा. के. कृष्णमूर्ति महोदयेन कृतः।
2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः 17
The Kannada translation of the work called ‘Dhvanyaloka’ was done by Dr. K. Krishnamurthy.

व्याकरणदोषं परिहरत

भाषायां दोषमुक्ततया व्यवहारं कर्तुं व्याकरणम् अत्यवश्यकम्। तत्रापि साधुप्रयोगः अपेक्षितः। साधुप्रयोगार्थं केचननियमाः परिशिष्टभागे (P. 158) प्रदत्ताः। तत्साहाय्येन अत्र प्रदत्तान् दोषान् दूरीकृत्य साधुप्रयोगं ज्ञातुम् अर्हन्ति विद्यार्थिनः।

1. अम्बा सह स्नातुम् अभ्यागमम् ।
अम्बया सह स्नातुम् अभ्यागमम्।

2. देवाः अपि अस्त्रेण विभ्यति।
देवाः अपि अस्त्रात् विभ्यति।

3. भवान् गन्तुम् अर्हसि।
भवान् गन्तुम् अर्हति।

4. मोगल चक्रवर्तिणां सह अयुध्यत।
मोगल् चक्रवर्तिभिः अयुध्यत ।

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

5. भवान् यथाकामं गच्छत।
भवान् यथा कामं गच्छतु।

6. शालायाः परितः वृक्षाः सन्ति।
शाला परितः वृक्षाः सन्ति।

7. जलात् अन्तरेण मीना न वसन्ति।
जलम् अन्तरेण मीना न वसन्ति ।

8. व्रतभङ्गेन भीतेव अलक्षत।
व्रतभङ्गात् भीतव अलक्षत।

9. तस्य विना तृणमपि न चलति।
तेन विना तृणमपि न चलति।

10. यथा गुरुः रोचते।
यथा गुरवे रोचते।

2nd PUC Sanskrit Workbook Answers Chapter 10 कृष्णशास्त्रीमहोदयः

11. मध्याहः आरोहति दिवाकरः।
मध्याह्नम् आरोहति दिवाकरः।

12. इन्द्रः दुष्यन्तस्य प्रियसखा आसीत्।
इन्द्रः दुष्यन्तस्य प्रियसखः आसीत्।

2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः

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1. सन्धिं विभजत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 1
उत्तरम्
शशवच्च = शशवत् + च ।
विषयामिषम् = विषय + आमिषम्
हीनश्च = हीनः + च ।

सन्धिं योजयत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 2
उत्तरम्
हि + अकाले = ह्यकाले
देवाः + ते = देवास्ते
सिंहवत् + च = सिंहवच्च

2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः

2. विग्रहवाक्यं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 3
उत्तरम्
धेनुसहस्त्रम् = धेनुना सहस्रम्
जितेन्द्रियः = जितानि इन्द्रियाणि येन सः।
कामक्रोधौ = कामः च क्रोधः च ।

समस्तपदं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 4
उत्तरम्
कामधेन्वोः गुणाः = कामधेनुगुणाः।
मात्रा सदृशी = मातृसदृशी।
नराः एव विग्रहाः = नरविग्रहाः।

3. लिङ्ग-विभक्ति-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 5
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 13

4. लकार-पुरुष-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 6
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 14

2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः

5. पदपरिचयं कुरुत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 7
उत्तरम्
हृष्टः – क्तप्रत्ययः भूतकृदन्तः।
चिन्तितम् – क्तप्रत्ययः भूतकृदन्तः।
संरक्ष्यः – ल्यबन्ताव्ययम् ।

6. प्रयोगं परिवर्तयत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 8
उत्तरम्
1. स्त्रिणा गृहं रक्ष्यते।
स्त्री गृहं रक्षति।

2. वत्सः मातरं गच्छति।
वत्सेन माता गम्यते।

3. वित्तेन धर्मः रक्ष्यते।
वित्तं धर्मं रक्षति।

7. सलक्षणम् अलंकारं निर्दिशत।
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 9
उत्तरम्
अ) विद्या प्रवासे मातृसदृशी।
अत्र उपमालङ्कारः।
तल्लक्षणम् – ‘उपमा यत्र सादृश्यलक्ष्मीरुल्लसति द्वयोः।’

2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः

आ) बकवत् अर्थान् चिन्तयेत्।
अत्र उपमालङ्कारः।
तल्लक्षणम् – ‘उपमा यत्र सादृश्यलक्ष्मीरुल्लसति द्वयोः।’

इ) विद्यारत्नेन यो हीनः स हीनः सर्ववस्तुषु ।
अत्र रुपकालङ्कारः।
तल्लक्षणम् – ‘उपमानोपमेययोरभेदः रुपकम्।’

योग्यताविस्तारः-

1. सन्धिं विभजत –
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 10
उत्तरम्
नवाम्रः – नव + आम्र।
एकेनापि – एकेन + अपि।
विषयामिषम् – विषय + अमिषम् ।
वृतवच्च – कृतवत् + च

2. सन्धिं योजयत –
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 11
उत्तरम्
निद्रा + अतिसेवा = निद्रातिसेवा
तत् + निरुन्ध्यात् = तन्निरुद्यात् ।
न + एव = नैव
देवाः + ते = देवास्तेः

2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः

3. विग्रहवाक्यं लिखत –
2nd PUC Sanskrit Workbook Answers Chapter 9 नीतिसारः 12
उत्तरम्
अतिनिद्रातिसेवे – अति निद्रा अति सेवा च ।
साधुहितम् – साधवे हितम्।
धर्महतोः – धर्मस्य हेतोः।
गुप्तधनम् – गुप्तं च तत् धनं च।

2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

Students can Download 2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्, Notes Pdf, 2nd PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

1. सन्धिं विभजत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 1
उत्तरम्
गण्योऽस्मि = गण्यो + अस्मि
यदीह = यत् + इह
तावच = तावत् + च

सन्धिं योजयत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 2
उत्तरम्
अस्त्र + अभ्यासः = अस्त्राभ्यासः
मम + ऊरुद्वयम् = ममोरुद्वयम्
शक्तो + असि = शक्तोऽसि

2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

2. विग्रहवाक्यं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 3
उत्तरम्
उत्कटेच्छा = उत्कटा च असौ इच्छा च ।
ज्ञानभण्डारः = ज्ञानस्य भण्डारः।
निद्राभङ्गः = निद्रायाः भङ्गः।

समस्तपदं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 4
उत्तरम्
भार्गवस्य अस्त्रम् = भार्गवास्त्रम्
धवला कीर्तिः यस्यः सः = धवलकीर्तिः
गुरोः प्रसादम् = गुरुप्रसादम्

3. लिङ्ग-विभक्ति-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 5
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 14

2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

4. लकार-पुरुष-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 6
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 15
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 16

5. पदपरिचयं कुरुत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 7
उत्तरम्
प्रतिसंहर्तुम् – तुमन्नन्ताव्ययम्
पतित्वा – क्त्वान्ताव्ययम्
पतन् – शतृ प्रत्ययः वर्तमानकृदन्त

2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

6. प्रयोगं परिवर्तयत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 8
उत्तरम्
त्वया किमेतत् आचरितम्।
त्वं किमेतत् आचरितवान् ।

मया धर्मविरुद्धं न आचरितम् ।
अहं धर्मविरुद्धं न आचरितान् ।

7. सलक्षणम् अलंकारं निर्दिशत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 9
उत्तरम्
त्वम् असत्यवादीव प्रतिभासि ।
अत्र उत्प्रेक्षालङ्कारः।
तल्लक्षणम् – ‘संभावना स्यादुत्प्रेक्षा’ इति।

2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

योग्यताविस्तारः-

1. सन्धिं विबजत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 10
उत्तरम्
उन्मील्याद्य – उन्मील्य + अद्य
अत्रैव – अत्र + एव
ममाप्यत्र – ममपि + अत्र
यद्योवम् – यदि + एवम्

2. सन्धिं योर ।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 11
उत्तरम्
भूमौ + उपविश्य – भूमावुपविश्य
स्व + उत्सङ्गे – स्वोत्सङ्गे
शिशुः + इव – शिशुरिव
कदाचित् + अपि – कदाचिदपि

2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम्

3. विग्रहवाक्यं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 12
उत्तरम्
गुरुप्रसादम् – गुरोः प्रसादः
असत्यवादी – न सत्यवादी

4. समस्तपदं लिखत –
2nd PUC Sanskrit Workbook Answers Chapter 8 विधिविलसितम् 13
उत्तरम्
उन्मीलिते अक्षिणी यस्य सः – उन्मीलिताक्षिः
न समर्थः – असमर्थः

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

Students can Download 2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः, Notes Pdf, 2nd PUC Sanskrit Textbook Answers, helps you to revise complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

1. सन्धिं विभजत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 1
उत्तरम्
वहन्तीव = वहन्ती + इव
मयैवम् = मया + एवम्
आननारविन्दे = आनन + अरविन्दे

सन्धिं योजयत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 2
उत्तरम्
सर्वो + अपि = सर्वोऽपि
यत् + अहम् = यदहम्
ग्रामस्य + एव = ग्रामस्यैव

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

2. विग्रहवाक्यं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 3
उत्तरम्
वस्त्रापणः = वस्त्रस्य आपणः
मुखकमलम् = मुखम् एव कमलम्
क्रोधाग्निः = क्रोधः एव अग्निः

समस्तपदं लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 4
उत्तरम्
मनुजः पशुः इव। = मनुजपशुः
कमनीयं रूपं यस्याः सा = कमनीयरूपा
अहनि अहनि = प्रत्यहम्

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

3. लिङ्ग-विभक्ति-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 5
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 12

4. लकार-पुरुष-वचनानि लिखत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 6
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 13

5. पदपरिचयं कुरुत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 7
उत्तरम्
उन्नमय्य – ल्यबन्ताव्ययम्
त्यक्तवान् – क्तवतु प्रत्ययः, भूतकृदन्तः
भवितुम् – तुमुन्नन्ताव्ययम्

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

6. प्रयोगं परिवर्तयत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 8
उत्तरम्
मया सा दृश्यते।
अहं तां पश्यामि।

अहं दशमकक्ष्यां पठितवान् ।
मया दशमकक्ष्या पठिता।

7. सलक्षणम् अलंकारं निर्दिशत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 9
उत्तरम्
अ) अस्याः आननारविन्दे अपूर्वाकर्षणम् आसीत् ।
अत्र रूपकालङ्कारः।
तल्लक्षणम् – ‘उपमानोपमेययोः अभेदः रूपकम्।’

आ) पुत्थलिका तरुण्याः सादृश्यं वहन्तीव सन्दृश्यते।
अत्र उपमालङ्कारः वर्तते।
तल्लक्षणम् – ‘उपमायत्रसादृश्यलक्ष्मीरुल्लसति द्वयोः।’

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

योग्यताविस्तारः-

1. कन्नडभाषया – आङ्ग्लभाषया वा अनुवदत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 10
उत्तरम्
अ) परेद्यवि सुन्दरः तत्र प्रेषितः असीत् ।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 14
Sundar was sent there the next day.

आ) मां द्रष्टं सा मदीयं कार्यालयम् आगच्छत् ।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 15
She came to my office to see me.

इ) किमर्थं स्वग्रामं त्यक्तवान्?
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 16
Why did he leave his village?

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

2. संस्कृतभाषया अनुवदत।
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 11
उत्तरम्
2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 17
Her lotus face is attractive.
तस्याः मुखारविन्दम् आकर्षणीयमस्ति।

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 18
Shanthi did not accept the money.
शान्ती धनं न स्वीकृतवती।

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः

2nd PUC Sanskrit Workbook Answers Chapter 7 सा शान्तिः 19
She washed off evil thoughts from my mind.
मम मनसः मलिनं निर्मूलनं कृतवती।