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 Multiple Choice QuestionsMultiple Choice Questions

61.

A stone of mass 2 kg is tied to a string of length 0.5 m. If the breaking tension of the string is 900 N, then the maximum angular velocity, the stone can have in uniform circular motion is

  • 30 rad s-1

  • 20 rad s-1

  • 10 rad s-1

  • 25 rad s-1


62.

The resultant of two vectors P and Q is R. If the magnitude of Q is doubled, the new resultant becomes perpendicular to P. Then the magnitude of R is 

  • P + Q

  • Q

  • P

  • P + Q2


63.

A motor car is moving with a speed of 20 ms-1 on a circular track of radius 100 m. If its speed is increasing at the, rate of 3 ms-1, its resultant acceleration is

  • 3 ms-2

  • 5 ms-2

  • 2.5 ms-2

  • 3.5 ms-2


64.

An object is dropped from rest. Its v-t graph is


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65.

Two particles A and B are projected with same speed so that the ratio of their maximum heights reached is 3 : 1. If the speed of A is doubled without altering other parameters, the ratio of the horizontal ranges attained by A and B is

  • 1 : 1

  • 2 : 1

  • 4 : 1

  • 3 : 2


66.

An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of 0.25 ms-2 is (g = 10 ms-2)

  • 51.25 N

  • 48.75 N

  • 52.75 N

  • 47.25 N


67.

A ball is projected from the ground at a speed of 10 ms-1 making an angle of 30° with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. The initial height of the second ball is (g = 10 ms-2)

  • 6.25 m

  • 2.5 m

  • 1.25 m

  • 5 m


68.

If the two vectors A = 2i^ + 3j^ + 4k^ and B = i^ + 2j^ - nk^ are perpendicular then the value of n is

  • 1

  • 2

  • 3

  • 4


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69.

A monkey climbs up and another monkey climbs down a rope hanging from a tree with same uniform acceleration separately. If the respective masses of monkeys are in the ratio 2 : 3, the common acceleration must be

  • g/5

  • 6g

  • g/2

  • g


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70.

From the top of tower, a stone is thrown up. It reaches the ground in t1 second. A second stone thrown down with the same speed reaches the ground in t2 second. A third stone released from rest reaches the ground in tsecond. Then

  • t3 = t1 + t22

  • t3 = t1t2

  • 1t3 = 1t1 - 1t2

  • t32 = t22 - t12


B.

t3 = t1t2

t3 = t1t2


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