GEOMETRICAL OPTICS
EXERCISES- Q1 to Q10
1. A concave mirror having a radius of curvature 40 cm is placed in front of an illuminated point source at a distance of 30 cm from it. Find the location of the image.
ANSWER: Taking pole as origin and the right from pole as positive X-axis, u = 30 cm, R = 40 cm, v = ?
The mirror formula
1/v+1/u = 2/R
→1/v + 1/30 = 2/40 = 1/20
→1/v = 1/20 - 1/30 =(30-20)/600
→1/v = 10/600 = 1/60
→v = 60 cm
Since u and f are along the same side of the mirror and taken positive and v has the positive sign so the image formed is 60 cm from the mirror on the side of the object. See the diagram below,
Diagram for Q-1
Two rays PO and PS meet at Q after reflection from the mirror thus forming a real image at Q.
2. A concave mirror forms an image of 20 cm high object on a screen placed 5.0 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between the mirror and the object.
ANSWER: Since the image is formed on a screen it is a real and inverted image and in front of the mirror. So, v = 5 m. (Taking the right side from the pole as positive X-axis and upward as positive Y-axis). Object height h₁ = 20 cm and the image height h₂ = -50 cm. Since for a mirror, h₂/h₁ = -v/u
→-50/20 = - 5/u
→u = 2*5/5 = 2 m
So the object is 2.0 m in front of the mirror.
The mirror formula is also written as
1/f = 1/u + 1/v
→1/f = 1/2 + 1/5 =7/10
→f = 10/7 m = 1.43 m
So the focal length of the mirror is 1.43 m.
3. A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object size.
ANSWER: Given that m = h₂/h₁ = 2 or -2
If the image is real it will be inverted and m = -2 but if it is imaginary, it will be erect and m = 2.
but m = h₂/h₁ = -v/u
→v = -mu
Given f = 20 cm. From the mirror formula,
1/v + 1/u = 1/f
→-1/(mu) +1/u = 1/f
→(1-1/m)/u =1/f
→(m-1)/mu =1/f
→u = f(m-1)/m
If m = -2
u = 20(-2-1)/(-2) =20*3/2 = 30 cm
If m = 2,
u = 20(2-1)/2 =20/2 = 10 cm
So for the given condition the object positions will be 30 cm or 10 cm in front of the mirror.
4. A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Find its distance from the mirror if the image formed is 0.6 cm in size.
ANSWER: For the convex mirror given,
m = h₂/h₁ = 0.6/1.0 =0.6
But h₂/h₁ = -v/u =0.6
→v = -0.6 u
f = -7.5 cm
From the mirror formula,
1/v+1/u = 1/f
→-1/0.6u +1/u =-1/7.5
→(1-10/6)/u =-1/7.5
→-4/6u = -1/7.5
→u = 4*7.5/6 = 30/6 = 5 cm
So the object is placed 5 cm in front of the mirror.
5. A candle flame 1.6 cm high is imaged in a ball bearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image.
ANSWER: u = 20 cm
The ball bearing will act as a convex mirror with R = -0.4/2 =-0.2 cm
From the mirror formula
1/v+1/u =2/R
→1/v +1/20 = -2/0.2 = -10
→1/v = -10-0.05 =-10.05
→v =-1/10.05 =-0.0995 cm ≈-1.0 mm
The negative sign shows that the image will form inside the ball bearing at 1.0 mm distance.
m = h₂/h₁=-v/u ≈ 0.1/20
→h₂/1.6 =0.1/20
→h₂ = 0.16/20 cm =1.6/20 mm =0.08 mm
6. A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size, and nature of the image.
ANSWER: Taking the usual sign conventions,
u = 7.5 cm,
f = -6.0 cm
from the mirror formula
1/v + 1/u = 1/f
→1/v + 1/7.5 =-1/6
→1/v = -1/6-1/7.5 =-(7.5+6)/45 =-13.6/45
→v = -45/13.6 =-3.30 cm
The image is formed 3.30 cm behind the convex mirror.
h₁ = 3 cm
h₂/h₁ = -v/u
→h₂/3 = -(-3.30)/7.5
→h₂ = 3*3.30/7.5 ≈1.32 cm
So the height of the image is 1.32 cm and it is erect and virtual.
7. A U-shaped wire is placed before a concave mirror having a radius of curvature 20 cm as shown in the figure (18-E1). Find the total length of the image.
Diagram for Q-7
ANSWER: With proper sign convention, for the inner arm of the wire, u = -30 cm,
R = -20 cm. To get the image distance v, we use the mirror formula
1/v + 1/u = 2/R
→1/v - 1/30 = -2/20
→1/v =-1/10 + 1/30 =-2/30 =-1/15
→v =-15 cm
Height of the image
h₂ = h₁*(-v/u) = -10*(15/30) =-5 cm
Negative sign means the image is inverted and its height is 5 cm.
Diagram for Q-7
For the outer arm, u = -40 cm
1/v = 2/R - 1/u =-2/20 + 1/40 =-1/10 + 1/40
=-3/40
→v = -40/3 cm ≈-13.33 cm
Hence the base of the image of wire =-15 + 40/3
=-5/3 cm ≈-1.67 cm
Height of the image =-10*(40/3)/40 =-10/3 cm ≈3.33 cm
The negative values of the image distances show that image is on the same side that of object. Hence total image length =5 cm +5/3 cm + 10/3 cm
=(15 + 5 + 10)/3 cm
=30/3 cm = 10 cm
8. A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.
ANSWER: h₂/h₁ = 1.4 = -v/u =-v/25
→v = -1.4*25 = -35 cm
FRom the miror formula
1/v + 1/u = 1/f
→-1/35+1/25 = 1/f
→1/f = (35-25)/(35*25) =10/875
→f = 87.5 cm
9. Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 7.6 m. The diameter of the moon is 3450 km and the distance between the earth and the moon is 3.8x10⁵ km.
ANSWER: Given, f = 7.6 m, u =3.8x10⁵ km ≈ ∞ (In comparison to the focal length the distance of the moon is practically infinity). Hence from the mirror formula
1/v + 1/u = 1/f
→1/v + 1/∞ = 1/7.6
→v = 7.6 m
It is also true that rays coming from infinity are parallel and they converge at the focal length.
The diameter of the moon = object height, h₁= 3450 km. Image height h₂ =?
h₂/h₁ = -v/u
→h₂ = -vh₁/u = -7.6*3450/3.8x10⁵ m
=-6900*100/10⁵ cm = -6.9 cm
The ngative sign shows that the image is inverted and diameter of the moon's image is 6.9 cm.
10. A particle goes in a circle of radius of 2.0 cm. A concave mirror of focal length 20 cm is placed with its principal axis passing through the center of the circle and perpendicular to its plane. The distance between the pole of the mirror and the center of the circle is 30 cm. Calculate the radius of the circle formed by the image.
1. A concave mirror having a radius of curvature 40 cm is placed in front of an illuminated point source at a distance of 30 cm from it. Find the location of the image.
ANSWER: Taking pole as origin and the right from pole as positive X-axis, u = 30 cm, R = 40 cm, v = ?
The mirror formula
1/v+1/u = 2/R
→1/v + 1/30 = 2/40 = 1/20
→1/v = 1/20 - 1/30 =(30-20)/600
→1/v = 10/600 = 1/60
→v = 60 cm
Since u and f are along the same side of the mirror and taken positive and v has the positive sign so the image formed is 60 cm from the mirror on the side of the object. See the diagram below,
Two rays PO and PS meet at Q after reflection from the mirror thus forming a real image at Q.
2. A concave mirror forms an image of 20 cm high object on a screen placed 5.0 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between the mirror and the object.
ANSWER: Since the image is formed on a screen it is a real and inverted image and in front of the mirror. So, v = 5 m. (Taking the right side from the pole as positive X-axis and upward as positive Y-axis). Object height h₁ = 20 cm and the image height h₂ = -50 cm. Since for a mirror, h₂/h₁ = -v/u
→-50/20 = - 5/u
→u = 2*5/5 = 2 m
So the object is 2.0 m in front of the mirror.
The mirror formula is also written as
1/f = 1/u + 1/v
→1/f = 1/2 + 1/5 =7/10
→f = 10/7 m = 1.43 m
So the focal length of the mirror is 1.43 m.
3. A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object size.
ANSWER: Given that m = h₂/h₁ = 2 or -2
If the image is real it will be inverted and m = -2 but if it is imaginary, it will be erect and m = 2.
but m = h₂/h₁ = -v/u
→v = -mu
Given f = 20 cm. From the mirror formula,
1/v + 1/u = 1/f
→-1/(mu) +1/u = 1/f
→(1-1/m)/u =1/f
→(m-1)/mu =1/f
→u = f(m-1)/m
If m = -2
u = 20(-2-1)/(-2) =20*3/2 = 30 cm
If m = 2,
u = 20(2-1)/2 =20/2 = 10 cm
So for the given condition the object positions will be 30 cm or 10 cm in front of the mirror.
4. A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Find its distance from the mirror if the image formed is 0.6 cm in size.
ANSWER: For the convex mirror given,
m = h₂/h₁ = 0.6/1.0 =0.6
But h₂/h₁ = -v/u =0.6
→v = -0.6 u
f = -7.5 cm
From the mirror formula,
1/v+1/u = 1/f
→-1/0.6u +1/u =-1/7.5
→(1-10/6)/u =-1/7.5
→-4/6u = -1/7.5
→u = 4*7.5/6 = 30/6 = 5 cm
So the object is placed 5 cm in front of the mirror.
ANSWER: u = 20 cm
The ball bearing will act as a convex mirror with R = -0.4/2 =-0.2 cm
From the mirror formula
1/v+1/u =2/R
→1/v +1/20 = -2/0.2 = -10
→1/v = -10-0.05 =-10.05
→v =-1/10.05 =-0.0995 cm ≈-1.0 mm
The negative sign shows that the image will form inside the ball bearing at 1.0 mm distance.
m = h₂/h₁=-v/u ≈ 0.1/20
→h₂/1.6 =0.1/20
→h₂ = 0.16/20 cm =1.6/20 mm =0.08 mm
6. A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size, and nature of the image.
ANSWER: Taking the usual sign conventions,
u = 7.5 cm,
f = -6.0 cm
from the mirror formula
1/v + 1/u = 1/f
→1/v + 1/7.5 =-1/6
→1/v = -1/6-1/7.5 =-(7.5+6)/45 =-13.6/45
→v = -45/13.6 =-3.30 cm
The image is formed 3.30 cm behind the convex mirror.
h₁ = 3 cm
h₂/h₁ = -v/u
→h₂/3 = -(-3.30)/7.5
→h₂ = 3*3.30/7.5 ≈1.32 cm
So the height of the image is 1.32 cm and it is erect and virtual.
7. A U-shaped wire is placed before a concave mirror having a radius of curvature 20 cm as shown in the figure (18-E1). Find the total length of the image.
ANSWER: With proper sign convention, for the inner arm of the wire, u = -30 cm,
R = -20 cm. To get the image distance v, we use the mirror formula
1/v + 1/u = 2/R
→1/v - 1/30 = -2/20
→1/v =-1/10 + 1/30 =-2/30 =-1/15
→v =-15 cm
Height of the image
h₂ = h₁*(-v/u) = -10*(15/30) =-5 cm
Negative sign means the image is inverted and its height is 5 cm.
For the outer arm, u = -40 cm
1/v = 2/R - 1/u =-2/20 + 1/40 =-1/10 + 1/40
=-3/40
→v = -40/3 cm ≈-13.33 cm
Hence the base of the image of wire =-15 + 40/3
=-5/3 cm ≈-1.67 cm
Height of the image =-10*(40/3)/40 =-10/3 cm ≈3.33 cm
The negative values of the image distances show that image is on the same side that of object. Hence total image length =5 cm +5/3 cm + 10/3 cm
=(15 + 5 + 10)/3 cm
=30/3 cm = 10 cm
8. A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.
ANSWER: h₂/h₁ = 1.4 = -v/u =-v/25
→v = -1.4*25 = -35 cm
FRom the miror formula
1/v + 1/u = 1/f
→-1/35+1/25 = 1/f
→1/f = (35-25)/(35*25) =10/875
→f = 87.5 cm
9. Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 7.6 m. The diameter of the moon is 3450 km and the distance between the earth and the moon is 3.8x10⁵ km.
ANSWER: Given, f = 7.6 m, u =3.8x10⁵ km ≈ ∞ (In comparison to the focal length the distance of the moon is practically infinity). Hence from the mirror formula
1/v + 1/u = 1/f
→1/v + 1/∞ = 1/7.6
→v = 7.6 m
It is also true that rays coming from infinity are parallel and they converge at the focal length.
The diameter of the moon = object height, h₁= 3450 km. Image height h₂ =?
h₂/h₁ = -v/u
→h₂ = -vh₁/u = -7.6*3450/3.8x10⁵ m
=-6900*100/10⁵ cm = -6.9 cm
The ngative sign shows that the image is inverted and diameter of the moon's image is 6.9 cm.
10. A particle goes in a circle of radius of 2.0 cm. A concave mirror of focal length 20 cm is placed with its principal axis passing through the center of the circle and perpendicular to its plane. The distance between the pole of the mirror and the center of the circle is 30 cm. Calculate the radius of the circle formed by the image.
ANSWER: Taking pole as origin and the right from pole as positive X-axis, u = 30 cm, R = 40 cm, v = ?
The mirror formula
1/v+1/u = 2/R
→1/v + 1/30 = 2/40 = 1/20
→1/v = 1/20 - 1/30 =(30-20)/600
→1/v = 10/600 = 1/60
→v = 60 cm
Since u and f are along the same side of the mirror and taken positive and v has the positive sign so the image formed is 60 cm from the mirror on the side of the object. See the diagram below,
 |
| Diagram for Q-1 |
Two rays PO and PS meet at Q after reflection from the mirror thus forming a real image at Q.
2. A concave mirror forms an image of 20 cm high object on a screen placed 5.0 m away from the mirror. The height of the image is 50 cm. Find the focal length of the mirror and the distance between the mirror and the object.
ANSWER: Since the image is formed on a screen it is a real and inverted image and in front of the mirror. So, v = 5 m. (Taking the right side from the pole as positive X-axis and upward as positive Y-axis). Object height h₁ = 20 cm and the image height h₂ = -50 cm. Since for a mirror, h₂/h₁ = -v/u
→-50/20 = - 5/u
→u = 2*5/5 = 2 m
So the object is 2.0 m in front of the mirror.
The mirror formula is also written as
1/f = 1/u + 1/v
→1/f = 1/2 + 1/5 =7/10
→f = 10/7 m = 1.43 m
So the focal length of the mirror is 1.43 m.
3. A concave mirror has a focal length of 20 cm. Find the position or positions of an object for which the image-size is double of the object size.
ANSWER: Given that m = h₂/h₁ = 2 or -2
If the image is real it will be inverted and m = -2 but if it is imaginary, it will be erect and m = 2.
but m = h₂/h₁ = -v/u
→v = -mu
Given f = 20 cm. From the mirror formula,
1/v + 1/u = 1/f
→-1/(mu) +1/u = 1/f
→(1-1/m)/u =1/f
→(m-1)/mu =1/f
→u = f(m-1)/m
If m = -2
u = 20(-2-1)/(-2) =20*3/2 = 30 cm
If m = 2,
u = 20(2-1)/2 =20/2 = 10 cm
So for the given condition the object positions will be 30 cm or 10 cm in front of the mirror.
4. A 1 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Find its distance from the mirror if the image formed is 0.6 cm in size.
ANSWER: For the convex mirror given,
m = h₂/h₁ = 0.6/1.0 =0.6
But h₂/h₁ = -v/u =0.6
→v = -0.6 u
f = -7.5 cm
From the mirror formula,
1/v+1/u = 1/f
→-1/0.6u +1/u =-1/7.5
→(1-10/6)/u =-1/7.5
→-4/6u = -1/7.5
→u = 4*7.5/6 = 30/6 = 5 cm
So the object is placed 5 cm in front of the mirror.
5. A candle flame 1.6 cm high is imaged in a ball bearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image.
ANSWER: u = 20 cm
The ball bearing will act as a convex mirror with R = -0.4/2 =-0.2 cm
From the mirror formula
1/v+1/u =2/R
→1/v +1/20 = -2/0.2 = -10
→1/v = -10-0.05 =-10.05
→v =-1/10.05 =-0.0995 cm ≈-1.0 mm
The negative sign shows that the image will form inside the ball bearing at 1.0 mm distance.
m = h₂/h₁=-v/u ≈ 0.1/20
→h₂/1.6 =0.1/20
→h₂ = 0.16/20 cm =1.6/20 mm =0.08 mm
6. A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size, and nature of the image.
ANSWER: Taking the usual sign conventions,
u = 7.5 cm,
f = -6.0 cm
from the mirror formula
1/v + 1/u = 1/f
→1/v + 1/7.5 =-1/6
→1/v = -1/6-1/7.5 =-(7.5+6)/45 =-13.6/45
→v = -45/13.6 =-3.30 cm
The image is formed 3.30 cm behind the convex mirror.
h₁ = 3 cm
h₂/h₁ = -v/u
→h₂/3 = -(-3.30)/7.5
→h₂ = 3*3.30/7.5 ≈1.32 cm
So the height of the image is 1.32 cm and it is erect and virtual.
7. A U-shaped wire is placed before a concave mirror having a radius of curvature 20 cm as shown in the figure (18-E1). Find the total length of the image.
 |
| Diagram for Q-7 |
ANSWER: With proper sign convention, for the inner arm of the wire, u = -30 cm,
R = -20 cm. To get the image distance v, we use the mirror formula
1/v + 1/u = 2/R
→1/v - 1/30 = -2/20
→1/v =-1/10 + 1/30 =-2/30 =-1/15
→v =-15 cm
Height of the image
h₂ = h₁*(-v/u) = -10*(15/30) =-5 cm
Negative sign means the image is inverted and its height is 5 cm.
 |
| Diagram for Q-7 |
For the outer arm, u = -40 cm
1/v = 2/R - 1/u =-2/20 + 1/40 =-1/10 + 1/40
=-3/40
→v = -40/3 cm ≈-13.33 cm
Hence the base of the image of wire =-15 + 40/3
=-5/3 cm ≈-1.67 cm
Height of the image =-10*(40/3)/40 =-10/3 cm ≈3.33 cm
The negative values of the image distances show that image is on the same side that of object. Hence total image length =5 cm +5/3 cm + 10/3 cm
=(15 + 5 + 10)/3 cm
=30/3 cm = 10 cm
8. A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.
ANSWER: h₂/h₁ = 1.4 = -v/u =-v/25
→v = -1.4*25 = -35 cm
FRom the miror formula
1/v + 1/u = 1/f
→-1/35+1/25 = 1/f
→1/f = (35-25)/(35*25) =10/875
→f = 87.5 cm
9. Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 7.6 m. The diameter of the moon is 3450 km and the distance between the earth and the moon is 3.8x10⁵ km.
ANSWER: Given, f = 7.6 m, u =3.8x10⁵ km ≈ ∞ (In comparison to the focal length the distance of the moon is practically infinity). Hence from the mirror formula
1/v + 1/u = 1/f
→1/v + 1/∞ = 1/7.6
→v = 7.6 m
It is also true that rays coming from infinity are parallel and they converge at the focal length.
The diameter of the moon = object height, h₁= 3450 km. Image height h₂ =?
h₂/h₁ = -v/u
→h₂ = -vh₁/u = -7.6*3450/3.8x10⁵ m
=-6900*100/10⁵ cm = -6.9 cm
The ngative sign shows that the image is inverted and diameter of the moon's image is 6.9 cm.
10. A particle goes in a circle of radius of 2.0 cm. A concave mirror of focal length 20 cm is placed with its principal axis passing through the center of the circle and perpendicular to its plane. The distance between the pole of the mirror and the center of the circle is 30 cm. Calculate the radius of the circle formed by the image.
ANSWER: Given, u = 30 cm, f = 20 cm. From the mirror formula
1/v + 1/u = 1/f
→1/v =1/20 -1/30 =1/60
→v = 60 cm.
Object height h₁ = 2*2.0 cm = 4.0 cm
Let us find the image height h₂
h₂/h₁ = -v/u
→h₂ =-60*4/30 cm =-8 cm
The negative sign is for the inverted image. The diameter of the image circle = 8 cm, hence the radius of the image circle = 4.0 cm.
ANSWER: Given, u = 30 cm, f = 20 cm. From the mirror formula
1/v + 1/u = 1/f
→1/v =1/20 -1/30 =1/60
→v = 60 cm.
Object height h₁ = 2*2.0 cm = 4.0 cm
Let us find the image height h₂
h₂/h₁ = -v/u
→h₂ =-60*4/30 cm =-8 cm
The negative sign is for the inverted image. The diameter of the image circle = 8 cm, hence the radius of the image circle = 4.0 cm.
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CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
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- 18 - Geometrical Optics
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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