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)Nm`
B. `2pixx10^(2)Nm`
C. `4pixx10^(-2)Nm`
D. `4pixx10^(2)Nm`

Answer

A

Explanation

`alpha =(2pi(n_(2)-n_(1)))/(t)`
`= (2pi(0-20))/(10)=-4 pi rad//s^(2)`
Negative sign means retardation
Now, `tau = I alpha`
`= 5xx10^(-3)xx4 pi = 2 pi xx10^(-2)Nm`.