1. A parachutist of weight ‘w’ strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3 g. Force exerted on him by ground during landing is

a) w

b) 2w

c) 3w

d) 4w

Explanation: Resultant force is w + 3w = 4w

2. At a place where the acceleration due to gravity is \[10m sec^{-2}\] a force of 5 kg-wt acts on a body of mass 10 kg initially at rest. The velocity of the
body after 4 second is

a) 5 \[m sec^{-1}\]

b) 10 \[m sec^{-1}\]

c) 20 \[m sec^{-1}\]

d) 50 \[m sec^{-1}\]

Explanation:

3. In a rocket of mass 1000 kg fuel is consumed at a rate of 40 kg/s. The velocity of the gases ejected from the rocket is \[ 5\times10^{4}m\diagup s\] . The thrust on the
rocket is

a) \[ 2\times10^{3}N\]

b) \[ 5\times10^{4}N\]

c) \[ 2\times10^{6}N\]

d) \[ 2\times10^{9}N\]

Explanation:

4. A man is standing on a weighing machine placed in a lift. When stationary his weight is recorded as 40 kg. If the lift is accelerated upwards with an
acceleration of \[2m\diagup sec^{2}\] , then the weight recorded in the machine will be \[\left(g=10m\diagup sec^{2}\right)\]

a) 32 kg

b) 40 kg

c) 42 kg

d) 48 kg

Explanation: In stationary lift man weighs 40 kg i.e. 400 N.

When lift accelerates upward it's apparent weight = m(g + a) = 40(10 + 2) = 480 N i.e. 48 kg

For the clarity of concepts in this problem kg-wt can be used in place of kg

5. A body of mass 4 kg weighs 4.8 kg when suspended in a moving lift. The acceleration of
the lift is

a) 9.8 \[m sec^{-2}\] downwards

b) 9.8 \[m sec^{-2}\] upwards

c) 1.96 \[m sec^{-2}\] downwards

d) 1.96 \[m sec^{-2}\] upwards

Explanation: As the apparent weight increase therefore we can say that acceleration of the lift is in upward direction.

R = m(g + a)

4.8 g = 4(g + a)

a = 0.2g = 1.96 m /s

^{2}

6. An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of \[5m s^{-2}\] . Taking g to be \[10m s^{-2}\] , then the tension in the cable is

a) 6000 N

b) 9000 N

c) 60000 N

d) 90000 N

Explanation: T = m(g + a) = 6000 (10 + 5) = 90000 N

7. A vehicle of 100 kg is moving with a velocity of 5 m/sec. To stop it in \[\frac{1}{10}sec\] , the required force in opposite direction is

a) 5000 N

b) 500 N

c) 50 N

d) 1000 N

Explanation:

8. A boy having a mass equal to 40 kilograms is standing in an elevator. The force felt by the feet of the boy will be greatest when the elevator
\[\left(g=9.8 metres\diagup sec^{2}\right)\]

a) Stands still

b) Moves downward at a constant velocity of 4 metres/sec

c) Accelerates downward with an acceleration equal to 4 \[ metres\diagup sec^{2}\]

d) Accelerates upward with an acceleration equal to 4 \[ metres\diagup sec^{2}\]

Explanation: Accelerates upward with an acceleration equal to 4 \[ metres\diagup sec^{2}\]

9. A rocket has an initial mass of \[20\times 10^{3}kg\] . If it is to blast off with an initial acceleration of \[4 ms^{-2}\] , the initial thrust needed is \[\left(g\cong10 ms^{-2}\right)\]

a) \[6\times 10^{4}N\]

b) \[28\times 10^{4}N\]

c) \[20\times 10^{4}N\]

d) \[12\times 10^{4}N\]

Explanation:

10. The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ is 3 : 2. The value of ‘a’ is
(g-Acceleration due to gravity of the earth)

a) \[\frac{3}{2}g\]

b) \[\frac{g}{3}\]

c) \[\frac{2}{3}g\]

d) g

Explanation: