N bullets each of mass m kg are fired with a velocity v ms^–1 at the rate of n bullets per second upon a wall. The reaction offered by the wall to the bullets is given by:

(1) nmv

(2) $\frac{Nmv}{n}$

(3) $n\frac{Nm}{v}$

(4) $n\frac{Nv}{m}$

With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is $\frac{2}{3}$ of his weight 

(1) $\frac{2}{3}g$

(2) g

(3) $\frac{1}{3}g$

(4) Zero

A ball of mass m moves with speed v and it strikes normally with a wall and reflected back normally, if its time of contact with wall is t then find force exerted by ball on wall 

(1) $\frac{2mv}{t}$

(2) $\frac{mv}{t}$

(3) mvt

(4) $\frac{mv}{2t}$

A body of mass 5 kg starts from the origin with an initial velocity $\overrightarrow{u}=30\hat{i}+40\hat{j}m{s}^{-1}$. If a constant force $\overrightarrow{F}=-(\hat{i}+5\hat{j})N$ acts on the body, the time in which the y–component of the velocity becomes zero is 

(1) 5 seconds

(2) 20 seconds

(3) 40 seconds

(4) 80 seconds

A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is:

(1) 2000 N

(2) 1000 N

(3) 5000 N

(4) 125 N

A particle moves in the \(XY\text-\)plane under the action of a force \(F\) such that the components of its linear momentum \(p\) at any time \(t\) are \(p_x = 2 \cos t\),  \(p_y = 2 \sin t\). The angle between \(F\) and \(p\) at time \(t\) will be:

Swimming is possible on account of 

(1) First law of motion

(2) Second law of motion

(3) Third law of motion

(4) Newton's law of gravitation

On a stationary sailboat, the air is blown at the sails from a fan attached to the boat. The boat will:

(1) remain stationary

(2) spin around

(3) move in a direction opposite to that in which air is blown

(4) move in the direction in which the air is blown

A man is standing at a spring platform. Reading of spring balance is 60 kg-wt. If a man jumps outside the platform, then reading of spring balance:

(1) first increases then decrease to zero.

(2) decreases.

(3) increases.

(4) remains same.