A torque of `10^(3) N-m` acting on a rigid body, turns it throughr `30^(@)` in `0.2` second. Calculate work done by the body and power of torque.

Explanation

Here, `tau = 10^(3) N-m`
`theta = 30^(@) = (pi)/(6) radian, t = 0.2 s`.
`W = tau theta = 10^(3) xx (tau)/(6) = (22)/(7 xx 6) xx 10^(3)J`
`= 5.23 xx 10^(2)J`
`p = tau omega = (W)/(t) = (5.23 xx 10^(2))/(0.2) = 2.615 xx 10^(3) watt`