A constant torque of $31.4 \mathrm{~N}-\mathrm{m}$ is exerted on a pivoted wheel. If angular acceleration of…

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
  1. $2.5 \mathrm{~kg} \mathrm{~m}^{2}$
  2. $3.5 \mathrm{~kg} \mathrm{~m}^{2}$
  3. $4.5 \mathrm{~kg} \mathrm{~m}^{2}$
  4. $5.5 \mathrm{~kg} \mathrm{~m}^{2}$

Solution

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

Asked in: BITSAT 2011

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