A constant torque of 31.4 ~N-m is exerted on a pivoted wheel. If angular acceleration of wheel is 4 π rad / s^2, then the moment of inertia of the wheel is

Correct answer: (A) 2.5 ~kg ~m^2. I=(τ)/(α)=31.4/(4 π)=31.4/(4 × 3.14)=2.5 ~kg ~m^2 .

Q93 / 125 · BITSAT 2011 · Physics

A constant torque of $31.4 \mathrm{~N}-\mathrm{m}$ is exerted on a pivoted wheel. If angular acceleration of wheel is $4 \pi \mathrm{rad} / \mathrm{s}^{2},$ then the moment of inertia of the wheel is

$I=\frac{\tau}{\alpha}=\frac{31.4}{4 \pi}=\frac{31.4}{4 \times 3.14}=2.5 \mathrm{~kg} \mathrm{~m}^{2}$.

Hence, the correct answer is option: 'A'.