A constant torque of 3.14 N mis exerted on a pivoted wheel. If the angular acceleration of the wheel is `4 pi "rad/" s^(2)` then the moment of inertia of the wheel is

Options

A. `0.25 kg m^(2)`
B. `2.5 kg m^(2)`
C. `4.5 kg m^(2)`
D. `25 kg m^(2)`

Answer

A

Explanation

Here , `tau = 3.14 N m , alpha = 4 pi "rad/"s^(2)`
As `tau = I alpha`
`therefore I = (tau)/(alpha) = (3.14)/(4pi) = (3.14)/(4 xx 3.14) = 0.25 kg m^(2)`