A particle of mass 5 kg is moving along a circular path of radius 1 m. It is making 300 rp., uniformly. Its kinetic energy would be:

Options

A. 100 `pi^2` J
B. 50 `pi^2` J
C. 250 `pi^2` J
D. zero

Answer

c

Explanation

Here the `E_k=1//2 m v^2`
or `E_k=1/2mr^2omega^2=(mr^2(2pin)^2)/(2)`
=`1/2xxmr^2xx4 pi^2.n^2`
=`2m r^2 pi^2.n^2`
=`2xx5xx(1)^2pi^2xx(300/60)^2`
=`10pi^2xx25=250 pi^2J`