If `2 xx 10^5` J of work is done in producing rotational kinetic energy of a wheel having moment of inertia of `10 kg.m^2` then the angular velocity (in `"rad s"^(-1)`) acquired by the wheel is equal to:

Answer

200

Explanation

From work-energy theorem,
Work done produces change in kinetic energy,
`therefore W = 1/2 Iomega^(2)`…………. `(therefore K.E. = 1.2 Iomega^(2))`
`therefore omega = sqrt((2W)/I) = sqrt((2 xx 2 xx 10^(5))/10) = 200` rad/s