A wheel has moment of inertia `5 xx 10^(-3) kg m^(2)` and is making `20 "rev" s^(-1)`. The torque needed to stop it in `10 s` is….. `xx 10^(-2) N-m`

Options

A. `2pixx10^(-2)`
B. `2.5 pixx10^(-3)`
C. `4pi xx10^(-2)`
D. `4.5 pixx10^(-4)`

Answer

A

Explanation

Given, moment of inertia
`l=5xx10^(-3)kgm^(2)`
`omega-_(0)=40pi` rad/sec
`t=10s`
Applying `omega=0` at
`implies 0=40pi+alpha+10`
`:.alpha=(-40pi)/10=-4pi rad//s^(2)`
The torque needed to stop the wheel `=l alpha=5xx10^(-3)xx4pi`
`=2pixx10^(-2)N//m`