GEOMETRICAL OPTICS
QUESTIONS FOR SHORT ANSWER
1. Is the formula "Real depth/Apparent depth =µ" valid if viewed from a position quite away from the normal?
ANSWER: No. Because in its derivation the formula
Sin i/sin r = µ
has been used and keeping the angles i and r very small, the approximation sin i ≈ tan i, and sin r ≈tan r have been taken. When viewed from a position quite away from the normal these approximations do not hold good. Hence the formula "Real depth/ Apparent depth=µ" is not valid.
2. Can you ever have a situation in which a light ray goes undeviated through a prism?
ANSWER: If the light ray falls perpendicularly on the surface and the angle of the prism is zero, i.e. i=i'=A=0
→ẟ = i+i'-A = 0
But in this case the sides of the prism are parallel and we may not call it a prism practically because it will look like a cuboid.
Another case is when the prism is submerged in a dense liquid which refractive index is the same as the prism. Since the speed of light will not change in both medium hence the light will travel undeviated through the prism.
3. Why does a diamond shine more than a glass piece cut to the same shape?
ANSWER: There are two reasons. First, due to the higher refractive index, the critical angle for diamond is less and the light goes a larger number of total internal reflections inside the diamond than the glass.
Second, there is more variation of µ between violet and red light for the diamond than the glass. Hence the dispersion is wider in diamond than the glass.
These two reasons combined make the diamond more shiny and sparkling.
4. A narrow beam of light passes through a slab obliquely and is then received by an eye (figure 18-Q1). The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation.
Figure for Q-4
ANSWER: When the index of refraction, µ changes the refracted rays in the slab change its direction. Thus they emerge out of the slab at different points. This results in the shift of the image of the source and the eye find the source disappearing/shifting a bit with time and the source appears twinkling like a star.
5. Can a plane mirror ever form a real image?
ANSWER: Generally a plane mirror does not form a real image because the light coming from real objects diverge after reflection from the plane mirror. But consider a situation of convex lens or concave mirror where a real image is forming, if the plane mirror is placed between the image and the lens/mirror, the incident rays after reflection form a real image at a different place.
6. If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?
ANSWER: No, at the position of the virtual image, light rays do not actually converge, hence the piece of paper will not be affected.
Yes, if the image is real the paper will burn after sufficient time. For example, a convex lens can be used to form a real image of the sun at its focal point. If a piece of paper is put at this point it will burn. This technique was used to make a fire long ago.
Yes, since the image is real, the light rays actually converge at the point.
7. Can a virtual image be photographed by a camera?
ANSWER: If the virtual image is taken as a virtual object, it can be photographed by a camera because of the camera photographs the real image formed at its plate or screen. The virtual objects can have real images.
But a virtual image can not make an impression on a photographic plate placed at this position due to the fact that virtual images cannot be projected on a screen.
8. In motor vehicles, a convex mirror is attached near the driver's seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror cannot do?
ANSWER: A convex mirror forms a virtual, erect and diminished image behind the mirror if the object is in front of it. So it covers a vast area behind the driver.
The plane mirror also forms virtual and erect image behind the mirror but it is of the same size. So the plane mirror does not cover a wide field behind the driver. Hence the convex mirror is used.
9. If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?
ANSWER: If the object is far away from a convex mirror the virtual image is near the focus. As the object moves towards the mirror the image moves towards the back of the mirror. So in an equal interval of time, the object moves a larger distance than the image. So the image moves slower than the object.
10. Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height?
ANSWER: Let the friend's height be AB. The ray from top AO enters the water at O and bends towards the normal at O. The refracted ray OC travels to the eye at C. The eye perceives the ray coming from A' which is the virtual image of A on the extended line CO. Thus the friend's height seems A'B instead of AB. A'B is taller than AB.
Diagram for Q-10
So the friend will appear taller than usual.
11. The equation of refraction at a spherical surface is
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Taking R = ∞, show that the equation leads to the equation
Real depth/Apparent depth = µ₁/µ₂
for refraction at a plane surface.
ANSWER: The equation of refraction at a spherical surface is
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Given that R = ∞, (Implies that the surface is plane)
Hence, µ₂/v - µ₁/u = 0
→µ₂/v = µ₁/u
→µ₁/µ₂ = u/v = h₁/h₂ = Real Depth/Apparent depth
12. A thin converging lens is formed with one surface convex and the other plane. Does the position of the image depend on whether the convex surface or plane surface faces the object?
ANSWER: From the lens makers formula
1/f =(µ-1){1/R₁ - 1/R₂}
Case-I, when the object is facing the convex surface.
R₂ =∞, and R₁ = R₁
→1/f = (µ-1){1/R₁ - 0} =(µ-1)/R₁
Diagram for Q - 12
Case-II, when the object is facing the plane surface of the lens.
R₁ =∞, and R₂ = -R₁ {since the center of curvature is towards left of origin at P}
→1/f = (µ-1){0 - 1/(-R₁)} =(µ-1)/R₁
Thus in both cases, the focal length does not change and the position of the image does not change whether the convex surface or the plane surface faces the object.
13. A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?
ANSWER: Assuming that the tube is in the air and the lens is made of a material having the refractive index µ > 1, it can not be said with certainty that there is a diverging lens (concave lens) inside the tube. A converging lens (convex lens) having its focal point well inside the tube will also let this beam emerge out as a diverging beam.
14. An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?
ANSWER: Though the air bubble inside the water has convex surfaces, it acts as a diverging lens because in this case the refractive index of the lens (air bubble) is less than the surrounding. Hence it will act just opposite than the usual glass lens in the air which acts as a converging lens.
15. Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without losing the energy of light. This increases the intensity. Describe how the converging lenses should be placed to do this.
ANSWER: It can be achieved by placing the larger focal length lens towards the incoming beam and then the smaller focal length lens at a distance f+f' from the first lens. See the diagram below.
Diagram for Q-15
In the above diagram, the focal length of the lens AB = f and the focal length of the lens CD = f', such that
f > f'.
The aperture of the incoming parallel beam is AB. The first lens converges it at the point F, PF = f. Since F is also the focal point of the lens CD, the rays coming from F become a parallel beam of aperture CD after passing through the second lens. The energy of the beam remains the same after passing through the two lenses. The triangle APF and FP'D are similar triangles hence
PF/P'F =AP/DP' =2AP/2DP' =AB/CD
→AB/CD =PF/P'F = f/f'
Since f > f', AB > CD
So the aperture is reduced by this arrangement. The intensity is the energy per unit area. The energy of the light beam is constant but aperture (hence the area) reduces, thus intensity increases in this process.
16. If a spherical mirror is dipped in water, does its focal length change?
ANSWER: The mirrors reflect rays and incident and reflected rays remain in the same medium, its focal length f = R/2 does not depend on the refractive index of the medium in which it is placed. So if a spherical mirror is dipped in water its focal length will not change.
17. If a thin lens is dipped in water, does its focal length change?
ANSWER: The focal length of a lens is given as
1/f = (µ₂/µ₁ - 1){1/R₁-1/R₂}
When dipped in water, the factor µ₂/µ₁ changes. Hence the focal length will also change.
18. Can mirrors give rise to chromatic aberration?
ANSWER: In mirrors laws of reflection are followed. The angle of incidence and the angle of reflection are the same for each wavelength of the light. Hence the focal length of a mirror is the same for each wavelength of the light. Thus mirrors can not give rise to chromatic aberration.
19. Laser light is focused by a converging lens. Will there be a significant chromatic aberration?
ANSWER: Laser lights are monochromatic i.e. its wavelengths are nearly the same. Thus they will focus at the same point and no significant chromatic aberration will be there.
1. Is the formula "Real depth/Apparent depth =µ" valid if viewed from a position quite away from the normal?
ANSWER: No. Because in its derivation the formula
Sin i/sin r = µ
has been used and keeping the angles i and r very small, the approximation sin i ≈ tan i, and sin r ≈tan r have been taken. When viewed from a position quite away from the normal these approximations do not hold good. Hence the formula "Real depth/ Apparent depth=µ" is not valid.
2. Can you ever have a situation in which a light ray goes undeviated through a prism?
ANSWER: If the light ray falls perpendicularly on the surface and the angle of the prism is zero, i.e. i=i'=A=0
→ẟ = i+i'-A = 0
But in this case the sides of the prism are parallel and we may not call it a prism practically because it will look like a cuboid.
Another case is when the prism is submerged in a dense liquid which refractive index is the same as the prism. Since the speed of light will not change in both medium hence the light will travel undeviated through the prism.
3. Why does a diamond shine more than a glass piece cut to the same shape?
ANSWER: There are two reasons. First, due to the higher refractive index, the critical angle for diamond is less and the light goes a larger number of total internal reflections inside the diamond than the glass.
Second, there is more variation of µ between violet and red light for the diamond than the glass. Hence the dispersion is wider in diamond than the glass.
These two reasons combined make the diamond more shiny and sparkling.
4. A narrow beam of light passes through a slab obliquely and is then received by an eye (figure 18-Q1). The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation.
ANSWER: When the index of refraction, µ changes the refracted rays in the slab change its direction. Thus they emerge out of the slab at different points. This results in the shift of the image of the source and the eye find the source disappearing/shifting a bit with time and the source appears twinkling like a star.
5. Can a plane mirror ever form a real image?
ANSWER: Generally a plane mirror does not form a real image because the light coming from real objects diverge after reflection from the plane mirror. But consider a situation of convex lens or concave mirror where a real image is forming, if the plane mirror is placed between the image and the lens/mirror, the incident rays after reflection form a real image at a different place.
6. If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?
ANSWER: No, at the position of the virtual image, light rays do not actually converge, hence the piece of paper will not be affected.
Yes, if the image is real the paper will burn after sufficient time. For example, a convex lens can be used to form a real image of the sun at its focal point. If a piece of paper is put at this point it will burn. This technique was used to make a fire long ago.
Yes, since the image is real, the light rays actually converge at the point.
7. Can a virtual image be photographed by a camera?
ANSWER: If the virtual image is taken as a virtual object, it can be photographed by a camera because of the camera photographs the real image formed at its plate or screen. The virtual objects can have real images.
But a virtual image can not make an impression on a photographic plate placed at this position due to the fact that virtual images cannot be projected on a screen.
8. In motor vehicles, a convex mirror is attached near the driver's seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror cannot do?
ANSWER: A convex mirror forms a virtual, erect and diminished image behind the mirror if the object is in front of it. So it covers a vast area behind the driver.
The plane mirror also forms virtual and erect image behind the mirror but it is of the same size. So the plane mirror does not cover a wide field behind the driver. Hence the convex mirror is used.
9. If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?
ANSWER: If the object is far away from a convex mirror the virtual image is near the focus. As the object moves towards the mirror the image moves towards the back of the mirror. So in an equal interval of time, the object moves a larger distance than the image. So the image moves slower than the object.
10. Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height?
ANSWER: Let the friend's height be AB. The ray from top AO enters the water at O and bends towards the normal at O. The refracted ray OC travels to the eye at C. The eye perceives the ray coming from A' which is the virtual image of A on the extended line CO. Thus the friend's height seems A'B instead of AB. A'B is taller than AB.
So the friend will appear taller than usual.
11. The equation of refraction at a spherical surface is
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Taking R = ∞, show that the equation leads to the equation
Real depth/Apparent depth = µ₁/µ₂
for refraction at a plane surface.
ANSWER: The equation of refraction at a spherical surface is
µ₂/v - µ₁/u =(µ₂-µ₁)/R
12. A thin converging lens is formed with one surface convex and the other plane. Does the position of the image depend on whether the convex surface or plane surface faces the object?
ANSWER: From the lens makers formula
1/f =(µ-1){1/R₁ - 1/R₂}
Case-I, when the object is facing the convex surface.
R₂ =∞, and R₁ = R₁
→1/f = (µ-1){1/R₁ - 0} =(µ-1)/R₁
Case-II, when the object is facing the plane surface of the lens.
R₁ =∞, and R₂ = -R₁ {since the center of curvature is towards left of origin at P}
→1/f = (µ-1){0 - 1/(-R₁)} =(µ-1)/R₁
Thus in both cases, the focal length does not change and the position of the image does not change whether the convex surface or the plane surface faces the object.
13. A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?
ANSWER: Assuming that the tube is in the air and the lens is made of a material having the refractive index µ > 1, it can not be said with certainty that there is a diverging lens (concave lens) inside the tube. A converging lens (convex lens) having its focal point well inside the tube will also let this beam emerge out as a diverging beam.
14. An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?
ANSWER: Though the air bubble inside the water has convex surfaces, it acts as a diverging lens because in this case the refractive index of the lens (air bubble) is less than the surrounding. Hence it will act just opposite than the usual glass lens in the air which acts as a converging lens.
15. Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without losing the energy of light. This increases the intensity. Describe how the converging lenses should be placed to do this.
ANSWER: It can be achieved by placing the larger focal length lens towards the incoming beam and then the smaller focal length lens at a distance f+f' from the first lens. See the diagram below.
In the above diagram, the focal length of the lens AB = f and the focal length of the lens CD = f', such that
f > f'.
The aperture of the incoming parallel beam is AB. The first lens converges it at the point F, PF = f. Since F is also the focal point of the lens CD, the rays coming from F become a parallel beam of aperture CD after passing through the second lens. The energy of the beam remains the same after passing through the two lenses. The triangle APF and FP'D are similar triangles hence
PF/P'F =AP/DP' =2AP/2DP' =AB/CD
→AB/CD =PF/P'F = f/f'
Since f > f', AB > CD
So the aperture is reduced by this arrangement. The intensity is the energy per unit area. The energy of the light beam is constant but aperture (hence the area) reduces, thus intensity increases in this process.
16. If a spherical mirror is dipped in water, does its focal length change?
ANSWER: The mirrors reflect rays and incident and reflected rays remain in the same medium, its focal length f = R/2 does not depend on the refractive index of the medium in which it is placed. So if a spherical mirror is dipped in water its focal length will not change.
17. If a thin lens is dipped in water, does its focal length change?
ANSWER: The focal length of a lens is given as
1/f = (µ₂/µ₁ - 1){1/R₁-1/R₂}
When dipped in water, the factor µ₂/µ₁ changes. Hence the focal length will also change.
18. Can mirrors give rise to chromatic aberration?
ANSWER: In mirrors laws of reflection are followed. The angle of incidence and the angle of reflection are the same for each wavelength of the light. Hence the focal length of a mirror is the same for each wavelength of the light. Thus mirrors can not give rise to chromatic aberration.
19. Laser light is focused by a converging lens. Will there be a significant chromatic aberration?
ANSWER: Laser lights are monochromatic i.e. its wavelengths are nearly the same. Thus they will focus at the same point and no significant chromatic aberration will be there.
ANSWER: No. Because in its derivation the formula
Sin i/sin r = µ
has been used and keeping the angles i and r very small, the approximation sin i ≈ tan i, and sin r ≈tan r have been taken. When viewed from a position quite away from the normal these approximations do not hold good. Hence the formula "Real depth/ Apparent depth=µ" is not valid.
2. Can you ever have a situation in which a light ray goes undeviated through a prism?
ANSWER: If the light ray falls perpendicularly on the surface and the angle of the prism is zero, i.e. i=i'=A=0
→ẟ = i+i'-A = 0
But in this case the sides of the prism are parallel and we may not call it a prism practically because it will look like a cuboid.
Another case is when the prism is submerged in a dense liquid which refractive index is the same as the prism. Since the speed of light will not change in both medium hence the light will travel undeviated through the prism.
3. Why does a diamond shine more than a glass piece cut to the same shape?
ANSWER: There are two reasons. First, due to the higher refractive index, the critical angle for diamond is less and the light goes a larger number of total internal reflections inside the diamond than the glass.
Second, there is more variation of µ between violet and red light for the diamond than the glass. Hence the dispersion is wider in diamond than the glass.
These two reasons combined make the diamond more shiny and sparkling.
4. A narrow beam of light passes through a slab obliquely and is then received by an eye (figure 18-Q1). The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation.
 |
| Figure for Q-4 |
ANSWER: When the index of refraction, µ changes the refracted rays in the slab change its direction. Thus they emerge out of the slab at different points. This results in the shift of the image of the source and the eye find the source disappearing/shifting a bit with time and the source appears twinkling like a star.
5. Can a plane mirror ever form a real image?
ANSWER: Generally a plane mirror does not form a real image because the light coming from real objects diverge after reflection from the plane mirror. But consider a situation of convex lens or concave mirror where a real image is forming, if the plane mirror is placed between the image and the lens/mirror, the incident rays after reflection form a real image at a different place.
6. If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?
ANSWER: No, at the position of the virtual image, light rays do not actually converge, hence the piece of paper will not be affected.
Yes, if the image is real the paper will burn after sufficient time. For example, a convex lens can be used to form a real image of the sun at its focal point. If a piece of paper is put at this point it will burn. This technique was used to make a fire long ago.
Yes, since the image is real, the light rays actually converge at the point.
7. Can a virtual image be photographed by a camera?
ANSWER: If the virtual image is taken as a virtual object, it can be photographed by a camera because of the camera photographs the real image formed at its plate or screen. The virtual objects can have real images.
But a virtual image can not make an impression on a photographic plate placed at this position due to the fact that virtual images cannot be projected on a screen.
8. In motor vehicles, a convex mirror is attached near the driver's seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror cannot do?
ANSWER: A convex mirror forms a virtual, erect and diminished image behind the mirror if the object is in front of it. So it covers a vast area behind the driver.
The plane mirror also forms virtual and erect image behind the mirror but it is of the same size. So the plane mirror does not cover a wide field behind the driver. Hence the convex mirror is used.
9. If an object far away from a convex mirror moves towards the mirror, the image also moves. Does it move faster, slower or at the same speed as compared to the object?
ANSWER: If the object is far away from a convex mirror the virtual image is near the focus. As the object moves towards the mirror the image moves towards the back of the mirror. So in an equal interval of time, the object moves a larger distance than the image. So the image moves slower than the object.
10. Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height?
ANSWER: Let the friend's height be AB. The ray from top AO enters the water at O and bends towards the normal at O. The refracted ray OC travels to the eye at C. The eye perceives the ray coming from A' which is the virtual image of A on the extended line CO. Thus the friend's height seems A'B instead of AB. A'B is taller than AB.
 |
| Diagram for Q-10 |
11. The equation of refraction at a spherical surface is
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Taking R = ∞, show that the equation leads to the equation
Real depth/Apparent depth = µ₁/µ₂
for refraction at a plane surface.
ANSWER: The equation of refraction at a spherical surface is
µ₂/v - µ₁/u =(µ₂-µ₁)/R
Given that R = ∞, (Implies that the surface is plane)
Hence, µ₂/v - µ₁/u = 0
→µ₂/v = µ₁/u
→µ₁/µ₂ = u/v = h₁/h₂ = Real Depth/Apparent depth
12. A thin converging lens is formed with one surface convex and the other plane. Does the position of the image depend on whether the convex surface or plane surface faces the object?
ANSWER: From the lens makers formula
1/f =(µ-1){1/R₁ - 1/R₂}
Case-I, when the object is facing the convex surface.
R₂ =∞, and R₁ = R₁
→1/f = (µ-1){1/R₁ - 0} =(µ-1)/R₁
 |
| Diagram for Q - 12 |
Case-II, when the object is facing the plane surface of the lens.
R₁ =∞, and R₂ = -R₁ {since the center of curvature is towards left of origin at P}
→1/f = (µ-1){0 - 1/(-R₁)} =(µ-1)/R₁
Thus in both cases, the focal length does not change and the position of the image does not change whether the convex surface or the plane surface faces the object.
13. A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?
ANSWER: Assuming that the tube is in the air and the lens is made of a material having the refractive index µ > 1, it can not be said with certainty that there is a diverging lens (concave lens) inside the tube. A converging lens (convex lens) having its focal point well inside the tube will also let this beam emerge out as a diverging beam.
14. An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?
ANSWER: Though the air bubble inside the water has convex surfaces, it acts as a diverging lens because in this case the refractive index of the lens (air bubble) is less than the surrounding. Hence it will act just opposite than the usual glass lens in the air which acts as a converging lens.
15. Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without losing the energy of light. This increases the intensity. Describe how the converging lenses should be placed to do this.
ANSWER: It can be achieved by placing the larger focal length lens towards the incoming beam and then the smaller focal length lens at a distance f+f' from the first lens. See the diagram below.
 |
| Diagram for Q-15 |
In the above diagram, the focal length of the lens AB = f and the focal length of the lens CD = f', such that
f > f'.
The aperture of the incoming parallel beam is AB. The first lens converges it at the point F, PF = f. Since F is also the focal point of the lens CD, the rays coming from F become a parallel beam of aperture CD after passing through the second lens. The energy of the beam remains the same after passing through the two lenses. The triangle APF and FP'D are similar triangles hence
PF/P'F =AP/DP' =2AP/2DP' =AB/CD
→AB/CD =PF/P'F = f/f'
Since f > f', AB > CD
So the aperture is reduced by this arrangement. The intensity is the energy per unit area. The energy of the light beam is constant but aperture (hence the area) reduces, thus intensity increases in this process.
16. If a spherical mirror is dipped in water, does its focal length change?
ANSWER: The mirrors reflect rays and incident and reflected rays remain in the same medium, its focal length f = R/2 does not depend on the refractive index of the medium in which it is placed. So if a spherical mirror is dipped in water its focal length will not change.
17. If a thin lens is dipped in water, does its focal length change?
ANSWER: The focal length of a lens is given as
1/f = (µ₂/µ₁ - 1){1/R₁-1/R₂}
When dipped in water, the factor µ₂/µ₁ changes. Hence the focal length will also change.
18. Can mirrors give rise to chromatic aberration?
ANSWER: In mirrors laws of reflection are followed. The angle of incidence and the angle of reflection are the same for each wavelength of the light. Hence the focal length of a mirror is the same for each wavelength of the light. Thus mirrors can not give rise to chromatic aberration.
19. Laser light is focused by a converging lens. Will there be a significant chromatic aberration?
ANSWER: Laser lights are monochromatic i.e. its wavelengths are nearly the same. Thus they will focus at the same point and no significant chromatic aberration will be there.
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Links to the Chapters
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Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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