A wheel having moment of inertia `2 "kg-m"^(2)` about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be

Options

A. `(pi)/(15)` N-m
B. `(2pi)/(15)` N-m
C. `(pi)/(18)` N-m
D. `(pi)/(12)` N-m

Answer

A

Explanation

We have, `omega_(2)=omega_(1)+alphatrArr omega_(1)=omega_(2)-alphat`

Here, `omega_(1)=0 and omega_(2)=60 "rpm" =(60xx2pi)/(60)"rad s"^(-1)`
`rArr 0=2pixx(60)/(60)-alpha.60 or alpha=(2pi)/(60)`
`:.` Torque, `tau=Ialpha=2xx(2pi)/(60)=(pi)/(15) "N-m"`.