A solid sphere of mass 2 kg and radius 5 cm is rotating at the rate of 300 rpm. The torque required to stop it in `2pi` revolutions is

Options

A. `2.5 xx 10^(4) `dyne cm
B. `2.5xx10^(-4)` dyne cm
C. `2.5xx 10^(6)` dyne cm
D. `2.5xx10^(5)` dyne cm

Answer

D

Explanation

In one revolution angle discribed is `2 pi`
`therefore 2pi` revolution corresponds to `4 pi^(2)`.
`tau = I alpha = ((2)/(5)MR^(2))((omega_(2)^(2)-omega_(1)^(2))/(2theta))`
`= (2)/(5)xx2xx25xx10^(-4)xx(4pi^(2)(0-25))/(2xx4pi^(2))`
`= 2.5xx10^(-2)Nm`.
`= 2.5xx10^(-2)xx10^(7)` dyne cm.