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111.

A rocket of mass 100 kg burns 0.1 kg of fuel per sec. If velocity of exhaust gas is 1 km/sec, then it lifts with an acceleration of

1000 ms

^{-2}100 ms

^{-2}10 ms

^{-2}1 ms

^{-2}

112.

A bullet emerge from a barrel of length 1.2 m with a speed of 640 ms^{-1} . Assuming constant acceleration, the approximate time that it spends in the barrel after the gun is fired is

4 ms

40 ms

400 μs

1 s

113.

The acceleration a (in ms^{-2} ) of a body, starting from rest varies with time t (in s) following the equation a = 3 t + 4. The velocity of the body at time t = 2 s will be

10 ms

^{-1}18 ms

^{-1}14 ms

^{-1}26 ms

^{-1}

114.

Figure below shows the distance-time graph of the motion of a car. It follows from the graph that the car is

at rest

in uniform motion

in non-uniform acceleration

uniformly accelerated

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115.

A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The fragments having equal masses fly off along mutually perpendicular direction with speed v. What will be the speed of the third (lighter) fragment ?

116.

If a person can throw a stone to maximum height of h metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

$\frac{\mathrm{h}}{2}$

h

2h

3h

117.

A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

t

^{1/2}t

^{3/4}t

^{3/2}t

^{2}

118.

A particle is moving with a constant speed v in a circle. What is the magnitude of average velocity after half rotation ?

2v

$2\frac{\mathrm{v}}{\mathrm{\pi}}$

$\frac{\mathrm{v}}{2}$

$\frac{\mathrm{v}}{2\mathrm{\pi}}$

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119.

A box of mass 2 kg is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and g = 10 ms^{-2}. This maximum acceleration of the car, for the box to remain stationary, is

8 ms

^{-2}6 ms

^{-2}4 ms

^{-2}2 ms

^{-2}

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A particle is travelling along a straight line OX. The distance x (in metre) of the particle from O at a time t is given by x = 37 + 27t − t^{3}, where t is time in seconds. The distance of the particle from O when it comes to rest is

81 m

91 m

101 m

111 m

B.

91 m

$\mathrm{Given},\mathrm{x}=37+27\mathrm{t}-{\mathrm{t}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=27-3{\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{problem},\phantom{\rule{0ex}{0ex}}\mathrm{v}=0\Rightarrow 27-3{\mathrm{t}}^{2}=0\phantom{\rule{0ex}{0ex}}\mathrm{Here},\mathrm{t}=3\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{x}=37+27\times 3-(3{)}^{2}\phantom{\rule{0ex}{0ex}}=91\mathrm{m}$

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