A solid cylinder of mass `50 kg` and radius `0.5 m` is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other end hanging freely. Tension in the string required to produce an angular acceleration of `2` revolution `s^(-2)` is

Options

A. 25 N
B. 50 N
C. `78.5` N
D. 157 N

Answer

D

Explanation

` rArr m = 50 kg , r = 0.5 M rArr alpha = 2 rev /s^(2)`
`rArr ` Torque produce key by the tension in the string
`= T xx R = T xx 0.5 = T/2 Nm …..(i)`
we know `tau = //alpha .....(ii)`
From Eqs . i and ii , `T/2 = //alpha`
` = ((MR^(2))/(2)) xx (2 xx 2pi) ("rad")/(s^(2))`
`T/2 = (50 xx(0.5)^(2))/ xx 4pi`
` rArr T = 50 xx1/4 xx4pi = 50 pi = 157 N `