75 mcq of physics

REGENTS PHYSICS

1. Two forces A and B act on a point P and their resultant force has the

magnitude 2 2A B . What is the angle between the forces A and B?

a. 0o

b. 90o

c. 180o

d. 270o

Sol: Correct option is b. 90o

a. If angle is 0o Resultant = A+B. Therefore (a) is incorrect.

b. For angle = 90o , Resultant = 2 2 2 22 (0)A B AB A B . Therefore (b) is correct.

c. For angle =180o , Resultant = A-B. Therefore (c) is incorrect.

d. For angle = 270o , Resultant 2 2A B . Therefore (d) is incorrect.

2. If there are two vectors X and Y then which of the following is true.

a. X+Y=Y+Xb. X+Y=X(X)Yc. X-Y=Y-Xd. X(X)Y=Y(X)Xe. X-Y=X(X)Y

Sol: Correct option is a. X+Y=Y+X

a. Addition of vectors is commutative.

b. X+Y and X(X) Y have different directions and magnitude.

c. X-Y and Y-X have different magnitude and direction.

d. X(X) Y and Y(X) X have different directions.

e. X-Y and X(X) Y have different directions and magnitude.

3. The ratio of displacement to distance is

a. <1b. =1c. >1d. 1e. 1

Sol: Correct option is d. 1

a. There may be a case when displacement = distance.

b. Displacement can be less than distance.

c. Displacement is never greater than distance.

d. Displacement can be less than or equal to distance.

e. Displacement is never greater than distance.

4. A body B has velocity of 0.50 miles/sec at point P and it de-accelerates at the rate of 0.1 miles/ 2sec . Find the time in which the body comes to rest.

a. 1 sb. 3 sc. 5 sd. 8 se. 10s

Sol: Correct option is c. 5 s

a. In 1 s final velocity = 0.4 miles/sec

b. In 3 s final velocity = 0.2 miles/sec

c. In 5 s final velocity = 0 and hence body comes to rest.

d. In 8 s body changes its direction of motion.

e. In 10 s body changes its direction of motion.

5. The initial velocity of a body is given by 3 7i j

and its acceleration

is 1 2i j

. What is its speed after 3 s?

a. 140

b. 89

c. 296

d. 103

e. 205

Sol: Correct option is e. 205

3 1 3 6xV and 7 2 3 13yV hence 2 26 13 205V

a. Does not match the above calculation.

b. 89 205

c. 296 205

d. 103 205

e. 3 1 3 6xV and 7 2 3 13yV hence 2 26 13 205V

6. A is particle P is acted upon by two forces 12 N and 4 N. What is the net force on the P?

a. 3 Nb. 6 Nc. 4 Nd. 18 Ne. Between 8 N and 16 N

Sol: Correct option is e. between 8 N and 16 N

a. We are not provided the angle between the forces.

b. We are not provided the angle between the forces.

c. We are not provided the angle between the forces.

d. The maximum resultant force can be 16 N.

e. The minimum resultant can be equal to 12-4 = 8 N and maximum resultant can be equal to 12+4 = 16 N. Hence it lies between 8 N and 16 N.

7. The apparent weight of a body in a lift is less than the real weight when the lift is going

a. Up with uniform speedb. Down with uniform speedc. Down with an accelerationd. Up with an acceleratione. At rest

Sol: The correct option is (c)

a. Apparent Weight = Real Weight

b. Apparent Weight = Real Weight

c. Here normal reacting or apparent weight is less.

d. Apparent Weight > Real Weight

e. Apparent Weight = Real Weight

8. A body is thrown in the downward direction with an initial velocity of 2 miles/sec and the other body is dropped simultaneously from a height h>100 miles. The distance between two bodies is 20 miles after

a. 2 secb. 3 secc. 9 secd. 10 sece. 20 sec

Sol: The correct option is (d)

a. After 2 sec distance = 4 miles

b. After 3 sec distance = 6 miles

c. After 9 sec distance = 18 miles

d. For first body 21

12

2S t gt

For second body 22

1

2S gt

1 2 2S S t

1 2 20S S

2 20t or 10t sec

e. After 20 sec distance = 40 miles

9. A body is thrown in the upward direction with velocity of 10 miles/sec. What is the time taken by the body to attain the maximum possible height? (Take g= 10 2/ secmiles )

a. 0.2 secb. 0.8 secc. 1 secd. 2 sece. 4 sec


a. After 0.2 sec V= 8 miles/sec

b. After 0.8 sec V= 2 miles/sec

c. After 1 sec V =0 that the body has attained the maximum possible height.

d. After 2 sec body is on ground

e. After 4 sec body has reached the ground.

10. A body B is dropped from the top of a roof of height H. It takes t seconds to reach the ground. Where is the ball at time t/2 sec?

a. H/4 meters from the groundb. H/9 meters from the ground.c. 3H/4 meters from the ground.d. H/11 meters from the ground.e. 2H/3 meters from the ground.


a. 3

4 4

H H

b. 3

9 4

H H

c. 3

4

H is the correct answer.

U=0, a=g, t=t, S=H

21 2 1 2

2 2 2

H t HH gt t

g g

U=0, a=g, 1 2

2

Ht

g

1 1 2

2 4 4

H HS g

g from roof

3

4

H meters from ground.

d. 3

11 4

H H

e. 2 3

3 4

H H

11. The angular acceleration of a particle moving along a circular path with uniform speed is

a. Zerob. Infinityc. Non-zerod. Variablee. None of the above.

Sol: The correct option is (a)

a. As speed is uniform angular acceleration is zero.

b. Acceleration infinity as speed is uniform.

c. Acceleration cannot be non-zero.

d. Acceleration is constant and zero.

e. Option a is correct.

12. A body is projected upwards with a velocity of 98 m/s. It will strike the ground in

a. 50 secb. 80 secc. 2 secd. 20 sece. 6 sec

Sol: Correct option is (d).

U=98 m/s, V=0 (at highest point)

A= -9.8 2/ secm

Time to reach highest point = 't = 10 sec

Time to reach ground = 2 't = 20 sec

a. 50 sec > 20 sec

b. 80 sec > 20 sec

c. The body doesn’t reach highest point in 2 sec.

d. 20 sec

e. The body doesn’t reach highest point in 6 sec.

13. The distance traveled by a particle falling from rest in first, second and third second is in the ratio

a. 1:3:6b. 1:8:7c. 1:3:5d. 1:9:11e. 1:2:5

Sol: Correct option is (c)

U=0

Also distance traveled in the nth second (2 1)2n

aS U n

1 2

aS 2

3

2

aS 3

5

2

aS

Ratio 1:3:5

a. 1:3:6 is not same as 1:3:5

b. 1:8:7 is not same as 1:3:5

c. Correct option

d. 1 2: 1: 9S S

e. 1 2: 1: 2S S

14. A particle is projected horizontally from the top of roof at height l with a speed of 2gl . The radius of curvature of the trajectory at the instant at projection will be

a.2

l

b.

c. 3

l

d. 2l

e. 2

3

l


2 22 , 2A

A RR

V glV gl a g R l

a g

a. 22

ll

b. 0Ra also aV

c. 23

ll

d. 2l is the correct option

e. 2

23

ll

15. A stone is released from an elevator going up with acceleration 8 2/m s . The acceleration of the stone after the release is

a. 7 2/m sb. 0 2/m sc. 3 2/m sd. 8 2/m se. 9.8 2/m s

Sol: Correct option is (e)

a. 27 /g m s

b. 20 /g m s

c. 23 /g m s

d. 28 /g m s

e. The force on the stone is only due to gravity, hence its acceleration is 29.8 /g m s

16. If a body is dropped from the window of a moving vehicle, then what will the path of the body?

a. Circular pathb. Parabolic pathc. Elliptical pathd. Straight line pathe. None of the above.

Sol: Correct option is (b)

a. No source of centripetal acceleration.

b. The body is having the initial velocity equal to that of the vehicle hence it follows parabolic path.

c. Since it is a projectile elliptical path is not possible.

d. Gravity is acting on the body.

e. Option (b) is correct.

17. Choose the wrong statement

a. Zero velocity of a particle does not necessarily mean that its acceleration is zero.b. Zero acceleration of a particle does not necessarily mean that its velocity is zero.c. A body in circular path having uniform speed has zero acceleration.d. If speed of particle is constant, its acceleration is zero.e. None of the above.

Sol: Correct option is (d)

a. Zero velocity implies that acceleration is zero.

b. Body may have velocity if acceleration is zero.

c. Uniform speed implies zero acceleration.

d. This statement is false since the direction of body may be changing.

e. Option (d) is correct.

18. Two particles of mass 1M and 2M are dropped from height 1H

and 2H respectively. They land on the ground after time 1t and 2t . Which of the following relation is correct?

a. 1 1

2 2

t M

t M

b. 1 1

2 2

t H

t H

c. 1 1

2 2

2t H

t H

d. 1 2

2 1

t H

t H

e. 1 1 1

2 2 2

t M H

t M H


a. Time of flight doesn’t depend on mass.

b. 11

2Ht

g and 2

2

2Ht

g 1 1

2 2

t H

t H

c. This relation does not follow kinematics equations.

d. Relation does not follow kinematics equations.

e. Time of flight doesn’t depend on mass.

19. The velocity-time graph of a linear motion is shown in figure. The displacement from the origin after 8 sec is

a. 5 mb. 0 mc. 10md. 9 me. None of the above

Sol: The correct option is (b)

a. The lower part of the graph is also to be taken into account

b. Displacement = 0 m because the area of positive part = area of negative part.

c. Invalid

e. Invalid

f. Option (b) is correct

20. A man projects balls into air one after the other at an interval of 2seconds. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. To what height the ball rises.

a. 19.6 mb. 80.6 mc. 31.9 md. 8 me. 6.5 m

Sol: The correct option is (a)

t = 2 sec, a = 9.8 2/m s , V=09.8 2 / secU m

19.8 2 2 9.8 4

29.8 2 19.6

H

H m

a. Correct option

b. 80.6 19.6

c. 31.9 19.6

d. 8 19.6

e. 6.5 19.6

21. A body is dropped from a height of 125 m. What will be the distance traveled by the body during last one second of motion? Take g= 10

2/m s

a. 10 mb. 20 mc. 45 md. 80 me. 100 m


2

5

110 125 5

210

(2 5 1) 452

t t

S m

a. 10 45

b. 20 45

c. Correct option

d. 80 45

e. 100 45

22. A body moving with uniform speed along a straight line takes a bend in a curve with the same speed. Due to this

a. its velocity is not changed.b. its speed is not changed.c. its velocity is changedd. its speed and velocity is not changed.e. none of the above.


a. As direction is changed, therefore velocity cannot remain same.

b. Speed doesn’t changes.

c. Due to change in direction its velocity changes.

d. Speed is not changed but velocity changes.

e. Option (c) is correct.

23. Two bodies of mass 8 kg and 11 kg are dropped from a height H simultaneously. The time taken by both bodies to reach the ground are 1t and 2t respectively. What is the relation between 1t and 2t .

a. 1 28t tb. 1 211t t

c. 1 2

11

8t t

d. 1 2t t

e. 1 2

8

11t t


a. Time of flight doesn’t depend upon mass.

b. Time of flight doesn’t depend upon mass.

c. Time of flight doesn’t depend upon mass.

d. Bodies dropped from same height reach the ground simultaneously.

e. Time of flight doesn’t depend upon mass.

24. A body is thrown with the velocity of U making an angle of with the horizontal axis. Determine the maximum height up to which it goes

a. 2 sinU

g

b. 2

sinU

g

c. 2 2sin

2

U

g

d. sinU

g

e. U

g


Dealing in vertical direction sin ,U U a g and for maximum

height V=0 2 2sin

2

UH

g

a. 2 2 2sin sin

2

U U

g g

b. 2 2

2

sin sin

2

U U

g g

c. 2 2sin

2

U

g

is the answer according to above calculations

d. 2 2sin sin

2

U U

g g

e. U

gis the time of flight when body is thrown vertically upwards.

25. A body is thrown with the velocity of U making an angle of with the horizontal axis. Determine the horizontal range of the body

a. 2 2sinU

g

b. sinU

g

c. 2 sinU

g

d. 2 sin 2U

g

e. None of the above


Time of flight = 2 sinU

g

Horizontal range

=22 sin sin 2

cosU U

Ug g

a. 2 2 2sin sin 2U U

g g

b. It’s the time of going upwards.

c. It’s the total time of flight

d. Correct option according to calculations


26. The angle of projection at which the projectile has the maximum possible range with a given velocity V is

a. 90o

b. 30o

c. 60o

d. 45o

e. 0o


a. At 90o Range = 0

b. At 30o Range = 23

2

U

g

c. At 60o Range = 23

2

U

g

d. At 45o Range= 2U

g which is maximum because maximum value at

sin 2 1 which is at 45o

e. At 0o the body is not in a projectile motion.

27. The angle of projection at which the maximum height is equal to the horizontal range of the projectile is

a. 45o b. 1tan (1) c. 90o d. 1tan (4) e. None of the above.


a. At 45o horizontal range is maximum

b. At 1tan (1) implies 45o which is explained above.

c. At 90o Horizontal range = 0

d. For maximum height = horizontal range

2 21sin 2 sin cos

tan (4)2

u u

g g

e. Option (d) is correct

28. A body is dropped from the same height when another body is fired horizontally. They will hit the ground.

a. One after the otherb. Depends on the air resistancec. Simultaneouslyd. The body dropped hits the ground first.e. None of the above.

Sol: The correct option is (c)a. Force is vertical direction is same for both.

b. No air resistance is stated.

c. As force in vertical direction is same for both, therefore they will land simultaneously.

d. Force on both bodies is vertical direction is same.


29. A ball is thrown upwards and it returns to ground following a parabola. Which of the following remains constant.

a. Velocity of ball.b. Vertical components of velocity.c. Momentum of balld. Horizontal components of velocity.e. None of the above.


a. Velocity changes due to de-acceleration.

b. Vertical component has the acceleration.

c. As velocity changes momentum changes.

d. As no force is acting in horizontal direction hence velocity in horizontal direction remains constant.


30. At the top of the trajectory of a projectile the velocity vector and acceleration vector are.

a. Parallel to each otherb. Perpendicular to each otherc. Inclined to each other at angle of 60o

d. Inclined to each other at angle of 30o

e. None of the above.


a. Velocity is in horizontal direction and acceleration is in vertical direction.

b. At top vertical component at velocity is zero and horizontal component is perpendicular to acceleration.

c. Invalid

d. Invalid


31. In projectile motion the horizontal motion takes place with

a. Constant acceleration.b. Variable acceleration.c. Constant velocity.d. Variable velocity.e. Constant retardation.


a. No acceleration in horizontal direction.

b. No acceleration in horizontal direction.

c. As there is no acceleration hence velocity is constant.

d. Velocity remains constant.

e. Retardation is in vertical direction.

32. A projectile is thrown with an initial velocity of x i y j

m/s. If range is equal to the maximum height obtained then

a. y= 2xb. y=4xc. y=xd. y=3xe. y=9x


Range =2 2x xy

yg g . Therefore if range = maximum height

Maximum height = 2

2

y

gThen 4

y

x

a. 2y

x

b. 4 4y

y xx

c. 1y

x

d. 3y

x

e. 9y

x

33. A projectile is fired so as to have maximum possible range equal to 800 m. Taking origin as point of projection the coordinate where the velocity of projectile is minimum are.

a. (50, 100)b. (100, 300)c. (200, 200)d. (400, 200)e. (100, 200)


When projectile attains maximum range then maximum height = 800

200.4 4

R Therefore the coordinates are (400, 200)

a. (50, 100) (400, 200)

b. (100, 300) (400, 200)

c. (200, 200) (400, 200)

d. Correct option

e. (100, 200) (400, 200)

34. Two bodies with speeds U, 3

Ucover the same horizontal distance. If

the angle of projection of ball one is 15o with horizontal, then angle of projection is

a. 1 2sin

7

b. 1 1tan

3

c. 11 1sin

2 8

d. 11 1sin

2 18

e. 1 2sin

13


2

2 sin 2(15 )sin 2 13

sin 218

ouu

g g

11 1sin

2 18

a. 1 12 1 1sin sin

7 2 18

b. 1 11 1 1tan sin

3 2 18

c. 1 11 1 1 1sin sin

2 8 2 18

d. 11 1sin

2 18

is correct according the calculation

e. 1 12 1 1sin sin

13 2 18

35. The velocity at the maximum height of a projectile is one third of its velocity of projection U. Its range on horizontal plane is

a. 24 2

27

u

g

b. 28 3

19

u

g

c. 217 2

31

u

g

d. 218 5u

g

e. 24 2

9

u

g


2 222 sin cos 2 2 1 4 2

cos 23 3 9

u uu u g

g g

a. 2 24 2 4 2

27 9

u u

g g

b. 2 28 3 4 2

19 9

u u

g g

c. 2 217 2 4 2

31 9

u u

g g

d. 2 218 5 4 2

9

u u

g g

e. 24 2

9

u

g is the correct option as per calculation.

36. A train is moving with a speed S on a curved road of radius R. A spring balance loaded with a body m is suspended from the roof of the train. The reading of spring balance is

a. 2mS

mgR

b. 2mS

R

c. mg

d. 2

2( )mS

mgR

e. None of the above


a. Gravitational force and centripetal force are not in same direction.

b. Gravitation force also acts on body.

c. Centripetal force also acts on body.

d. Gravitational and centripetal force at perpendicular to each other.


37. The speed of revolution of a particle going around a circle is doubled and its angular speed is also doubled then centripetal acceleration becomes.

a. Halvedb. Four timesc. Remains samed. Doublede. Three Times


Centripetal acceleration = 2 2V V

VwVrw

a. Not appropriate according to above calculation.

b. Correct option according to above calculations.

c. Not appropriate

d. Not appropriate

e. Not appropriate

38. The road of a bride is in the form of a circle of radius 10 m. What is the maximum speed with which a car can cross the bride without leaving ground at highest point?

a. 20 m/sb. 30 m/sc. 9.8 m/sd. 98 m/se. 9.8 m/s


For maximum velocity normal reaction should be zero

2

max 9.8 10 98mV

mg V gr Vr

m/s

Where V is maximum velocity and r is the radius.

a. 98 m/s 20 m/s

b. 98 m/s 30 m/s

c. 98 m/s 9.8 m/s

d. Correct option according to above calculation

e. 98 m/s 9.8 m/s

39. A mass M is hanging by a rod at length l. The velocity U which must be given to it in order to just reach the top is

a. 3 gl

b. gl

c. 2 gl

d. 6 gl

e. 8 gl


For just reaching the top 212 2

2mV mgl V gl

a. 3 2gl gl

b. 2gl gl

c. 2 gl is the correct option according to above calculation.

d. 6 2gl gl

e. 8 2gl gl

40. If the over bridge is concave instead of being convex, the thrust on the road at the highest position will be (for mass m traveling with velocity V and radius r)

a. mg

b.2mV

r

c.2mV

mgr

d.2mV

mgr

e. 2mg


a. Centripetal force is not taken into account.

b. Gravitational force is not taken into account.

c.2 2mV mV

N mg N mgr r

where N is the thrust.

d. Option (c) is correct.

e. Not appropriate.

41. A block of mass 3 kg is suspended from a string 9 m long. The block is rotated with a constant speed of 6 m/s. The tension of the string when it is horizontal is

a. 6 Nb. 12 Nc. 18 Nd. 24 Ne. 30 N


At horizontal position T = 2 3 6 6

129

mVN

r

a. Not appropriate

b. Correct option according to above calculation

c. 18 N 12 N

d. Gravitational force acts in perpendicular direction.

e. 30 N 12 N

42. Swimming is possible on account of

a. Newton’s law of gravitationb. Second law of motionc. First law of motiond. Third law of motione. None of the above.


a. Not appropriate

b. Second law equates force to product of mass and acceleration.

c. Not appropriate

d. Each action has equal and opposite reaction.


43. We can derive Newton’s

a. Second law from first lawb. Third law from first lawc. All laws are independentd. Third and first law from second law.e. None of the above.


a. We can’t derive second law from first law.

b. We can’t derive third law from first law.

c. There exists a relation between the laws.

d. We can derive third and first law from second law.


44. A vessel containing water is provided an acceleration a towards right along a straight horizontal path which of the following diagram represents the surface of the liquid.


a. Pseudo Force is acting

b. As acceleration is towards right, therefore pseudo force is towards left.

c. Option (b) is correct.

d. Inappropriate.


45. A canon after firing recoils due to

a. Newton’s first law of motion.b. Newton’s second law of motion.c. Newton’s third law of motion.d. Newton’s fourth law of motion.e. None of the above.


a. First law only tells above motion.

b. Second law tells the relationship between force and acceleration.

c. Each action has equal and opposite reaction.

d. No fourth law exists.


46. A body of mass 75 10 kg initially at rest is pulled by a force of 65 10N through a distance of 5 , the speed of the body is

a. 0.1 m/sb. 1 m/sc. 10 m/sd. 100 m/se. .01 m/s


2 7 2 61 1tan 5 10 5 10 5

2 2mV Force dis ce V

1V m/s

a. 0.1 m/s 1 m/s

b. Correct option due to above calculation.

c. 10 m/s 1 m/s

d. 100 m/s 1 m/s

e. .01 m/s 1 m/s

47. A rocket is ejecting 0.8 kg of gases per second at a velocity of 80 m/s. The accelerating force on the rocket is

a. 60 Nb. 80 Nc. 64 Nd. 90 Ne. 10 N


Acceleration = 80

1 = 80 2/m s

Force = 0.8 kg X 80 2/m s = 64 N

a. 60 N 64 N

b. 80 N 64 N

c. Correct option due to above calculation

d. 90 N 64 N

e. 10 N 64 N

48. A force of 10 N acts on a body of mass 2 kg for an interval of time during which the body attains a velocity of 40 m/s. The time in seconds for which the force acts on the body is

a. 10 m/sb. 8 m/sc. 5 m/sd. 12 m/se. 20 m/s


Acceleration = 240 /5 /

2

m sm s

kg

Time = 2

40 /8 /

5 /

m sm s

m s

a. 10 m/s 8 m/s

b. 8 m/s is correct according to above calculation.

c. 5 m/s 8 m/s

d. 12 m/s 8 m/s

e. 20 m/s 8 m/s

49. A body of mass of 10 kg moving with a constant velocity of 80 m/s on a friction less road. The force required to keep the body moving with same velocity is

a. 20 Nb. 60 Nc. 7 Nd. Zero Newton’se. 79 N


a. No force is required

b. No force is required.

c. No force is required.

d. As there is no friction hence no force is required.

e. No force is required.

50. A player caught a ball of mass 1 kg moving at rate of 10 m/s. If the catching process is completed in 0.5 s, the force exerted by ball on hands of the player is

a. 2 Nb. 20 Nc. 40 Nd. 70 Ne. 90 N



0.5 /

m sm s

m s

Force = 1 Kg X 220 /m s = 20 N

a. 2 N 20 N

b. Correct option according to above calculation.

c. 40 N 20 N

d. 70 N 20 N

e. 90 N 20 N

51. A lift is moving up with acceleration equal to 1

3of that due to

gravity. The apparent weight of a 90 kg man standing in life is (Take g = 10 2/m s )

a. 40 kgb. 70 kgc. 50 kgd. 100 kge. 120 kg


4 490

3 3 3120 120

mg g gN mg N m N

N g kg

a. Apparent weight will be more than real weight.

b. Apparent weight will be more than real weight.

c. Apparent weight will be more than real weight.

d. 100 kg 120 kg

e. Correct option according to above calculation

52. A body of mass 1 kg moves at a constant speed of 10 m/s. A constant force acts for 0.2 seconds on the body and gives it a speed of 2 m/s in opposite direction. The force acting on a body is

a. 20 Nb. -30 Nc. 30 N

d. -60 Ne. 70 N


Acceleration = 210 ( 2)60 /

0.2m s

Force = 60 2/m s X 1 kg = -60 N

a. Force can’t be positive

b. -30 N -60 N

c. Force cant be positive

d. Correct due to above calculation.

e. 70 N -60 N

53. A monkey of mass 10 kg is holding a vertical rope. The rope can break when a mass of 15 kg is suspended from it. What is the maximum acceleration with which monkey can climb up along the rope? (Take g = 10 2/m s )

a. 2 2/m sb. 5 2/m sc. 3 2/m sd. -2 2/m se. – 8 2/m s


2

( ), 15

15 10( ) 5 10 5 /2

T mg ma T m g a T g

gg g a g a a a m s

a. 2 2/m s 5 2/m s

b. 5 2/m s is correct according to above calculation.

c. 3 2/m s 5 2/m s

d. Acceleration can’t be negative.

e. Acceleration can’t be negative.

54. A ball weighing 1 kg moving with a speed of 3 m/s hits a wall perpendicularly. It bounces back with same speed. What is the average force exerted by wall on the ball if the two were in contact for 110 s.

a. 50 Nb. 60 Nc. 70 Nd. 80 Ne. 90 N


Acceleration = 23 ( 3)60 /

0.1m s

Force = 1 kg x 260 /m s = 60 N

a. 50 N 60 Nb. Correct option according to above calculation c. 70 N 60 Nd. 80 N 60 Ne. 90 N 60 N

55. A body of mass 10 kg is moving horizontally at speed of 2 m/s A vertically upward force of 10 N acts on it for 2 seconds. What will be the distance of the block from the point where the force started acting?

a. 2 5m

b. 5 2m

c. 8 3m

d. 4 3m

e. 7 2m

Sol: Correct option is (a)

Acceleration = 2101 /

10

Nm s

kg

Vertical distance = 210 1 (2) 2

2m

Horizontal distance = 2 x 2 = 4 m

Total distance = 4 16 20 2 5

a. Correct option according to above calculation.

b. 5 2 2 5m m

c. 8 3 2 5m m

d. 4 3 2 5m m

e. 7 2 2 5m m

56. A block of mass m is placed on a smooth inclined plane of inclination . The inclined plane is itself placed in a lift which is accelerating upwards at the rate a in a direction making an angle with the horizontal. What will be the acceleration of the block down the inclined plane?

a. sing ab. sing a c. ( ) sina g d. sing e. a


a. Not appropriate

b. a is the pseudo acceleration.

c. Not appropriate

d. Pseudo acceleration must be considered.

e. Gravitational acceleration must be considered.

57. A rifle fires P bullets per seconds and mass of each bullet is M. If V is speed of each bullet, then force exerted on rifle is

a. MPb. MVc. MPV

d.MP

Ve. VP


a. Velocity is to be considered.

b. No. of bullets per second (p) is to be considered.

c. Force = Impulse x t = MVP

d. MP

V is not appropriate.

e. Mass of bullet is to be considered.

58. A ball weighing 1 kg is dropped from a certain height above the ground. It bounces back after an elastic collision and if speed with which ball strikes the ground is 5 m/s then the impulse imparted by ball to the floor is

a. 10 kg m/sb. 6 kg m/sc. 5 kg m/sd. 2 kg m/se. 0


Change in velocity = 5 – (-5) = 10 m/s

Impulse = 1 kg x 10 m/s = 10 kg m/s

a. Correct option according to above calculation.

b. 6 kg m/s 10 kg m/s

c. Reverse velocity is to be taken into account.

d. 2 kg m/s 10 kg m/s

e. Not appropriate

59. A spring balance fastened to the ceiling of a moving lift indicates 80 kg wt as the weight of 60 kg man. The acceleration of lift is

a. 2

g

b. 2gc. g

d. 3

g

e. 4

g


T- 60 g = 60 a

80 g- 60 g= 60 a

a = 3

g

a. 2

greading will be different.

b. at 2g reading will be different.

c. at g reading will be twice.

d. correct option according to above calculation.

e. at 4

g reading will be different.

60. Static friction is

a. less than dynamic frictionb. more than dynamic frictionc. equal to dynamic frictiond. no relatione. none of the above


a. Not appropriate

b. Correct option

c. Never equal

d. Not appropriate


61. The maximum static frictional force is

a. Thrice the surface area in contact.b. Equal to surface area in contactc. Independent of surface area in contact

d. Four times the surface area in contact.e. None of the above.


a. No relation

b. No relation

c. Independent of surface area in contact.

d. No relation


62. Maximum value of static friction is called

a. Dynamic frictionb. Angle of frictionc. Sliding frictiond. Limiting frictione. None of the above

Sol: Option (d) is correct.

a. Not appropriate

b. Angle of friction is not the maximum value.

c. Sliding friction is not appropriate

d. Correct according to definition


63. A block can slide on a smooth inclined plane of inclination kept on the floor of a lift. When the lift is ascending with acceleration a the acceleration of block relative to incline is

a. gb. ac. sing d. ( ) sing a e. ( )sing a


a. Pseudo acceleration is to be taken into account

b. Gravitational acceleration is to be taken into account.

c. Pseudo acceleration is to be taken into account.

d. Correct option is pseudo acceleration is included.

e. Not appropriate.

64. Pushing force making an angle to the horizontal is applied on a body of weight w placed on ground. If the angle of friction is , the magnitude of force required to move the body is

a. cos

cos(

w

b. cos

sin(

w

c. sin

cos(

w

d. tanw e. None of the above


( sin )

cos ( sin )

( sin ) tan cos

sin

cos( )

friction

required

required

required

F p w

F p p w

F p w p

wF

a. Not appropriate

b. Not appropriate

c. Correct according to above calculation

d. Not appropriate

e. Option (c) is correct

65. A block of mass 4 kg rests on a rough inclined plane making angle of 30o with horizontal. The coefficient of static friction between the block and plane is 0.9. The frictional force on the block is

a. gb. 2 gc. 3 gd. 4 ge. 5 g


Frictional force = 4 kg x g x sin 30o = 2g

a. g 2g

b. correct option

c. 3g 2 g

d. 4 g 2 g

e. 5 g 2g

66. A body is moving across a rough floor with an initial speed of 8 m/s.it stopped after moving for 4 seconds. If resisting force of friction was on an average 20 N, the mass of box in kg is

a. 5 kgb. 7 kgc. 8 kgd. 4 kge. 10 kg



4

m sm s

s

Also force = 20 Nmass = 10 kg

a. 5 kg 10 kg

b. Wrong calculation

c. 8 kg 10 kg

d. Wrong calculation

e. Correct option according to above calculation.

67. A fireman weighing 40 kg slides down a pole. If the resisting force of friction is a constant 320 N, his acceleration in 2/m s is

a. 1 2/m sb. 2 2/m sc. 3 2/m sd. 4 2/m se. 5 2/m s


40 x a = 400 – 320 22 /a m s

a. This acceleration is for more friction.

b. Correct option according to above calculation

c. Not appropriate

d. Not appropriate

e. Not appropriate

68. A block of mass 7 kg and surface area 2 2m just begins to slide down

an inclined plane when the angle of inclination is 6

. Keeping the

mass same, the surface area of block is halved. At what angle the block will now start sliding down.

a. 6

b. 3

c. 2

d. 4

e. 8


a. Angle remains same as motion does not depend on area.

b. Motion is independent of area.

c. Motion is independent of area.

d. Motion is independent of area.

e. Not appropriate.

69. The work done in moving a body up a rough inclined plane is given by

a. sinmg d b. ( sin cos )mg mg d c. ( sin cos )mg mg d d. cosmg d e. None of the above


Required force = ( sin cos )mg mg

Work done = ( sin cos )mg mg d

a. Inappropriate

b. is not taken into account.

c. Correct option

d. sinmg is not considered.

e. option (c) is correct.

70. If , r and d represent coefficients of friction, normal reaction and distance moved, then expression for work against friction is

a. rdb. rc. dd. rde. None of the above


Force = r

Work done = rd

a. Correct option according to calculation

b. d is not taken into account.

c. r is not taken into account.

d. is not taken into account.

e. Option (a) is correct.

71. A block of mass 2 kg is at rest on a horizontal surface in a cart. The coefficient of static friction between the block and surface is 0.8. If the acceleration of cart is 0.10 2/m s , then the frictional force acting on the block is

a. 16 Nb. 0.2 Nc. 10 Nd. 0.4 Ne. 8 N


Pseudo force = 2 kg x 0.10 2/m s = 0.2 N

Force of friction = applied force = 0.2 N

a. Maximum value of friction

b. Correct option according to calculation.

c. Not appropriate

d. Not appropriate

e. Not appropriate

72. The coefficient of restitution e for a perfectly elastic collision is

a. 0b. -1c. 1d. 2e.


For elastic collision there is both conservation of energy and momentum.

a. For e=0 momentum is not conserved.

b. For e= -1 momentum is not conserved.

c. Correct Option.

d. Not appropriate

e. Conservation laws not obeyed.

73. The same retarding force is applied to stop a train. If speed is halved then distance will be

a. two timesb. samec. four times

d. one-fourthe. halved.


Acceleration = same

Let speed V 2

2

US

a

Now if U is halved then S becomes one-fourth

a. If speed is increased by 2 times, then this is correct option

b. If speed is doubled then this is the correct option.

c. Not appropriate

d. Correct option according to above calculation.

e. Not appropriate

74. If a force f is applied on a body and it moves with a velocity V, then the power will be

a. 2F

Vb. FV

c. 2

F

V

d. F

V

e. V

F


Power = W FS

FVt t

a. Not appropriate

b. Correct option as per calculation

c. Inappropriate

d. Dimensionally incorrect.

e. Dimensionally incorrect.

75 mcq of physics

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