1.

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:

Answer»

100 `pi^2` J
50 `pi^2` J
250 `pi^2` J
zero

Solution :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`


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