question_answer

A particle of mass m = 1 kg moves in a circle of radius R = 2m with uniform speed \[v=3\pi \,\text{m/s}\]. m/s. The magnitude of impulse given by centripetal force to the particle in one second is

A)  \[\sqrt{2}\pi \text{Ns}\]                             

B)  \[\sqrt{3}\pi \text{Ns}\]

C)  \[2\sqrt{3}\pi \text{Ns}\]                           

D)  \[3\sqrt{2}\pi \text{Ns}\]

Correct Answer: D

Solution :

 The angular displacement of the particle in t = 1 sec is \[\theta =\omega t=\frac{v}{R}t=\frac{3\pi }{2}\] \[\therefore \]The magnitude of impulse by centripetal force in t = 1 second = change in momentum \[=\sqrt{2}mv=3\sqrt{2}\pi Ns\]