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Question: Answered & Verified by Expert
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
PhysicsRotational MotionBITSATBITSAT 2011
Options:
  • A $2.5 \mathrm{~kg} \mathrm{~m}^{2}$
  • B $3.5 \mathrm{~kg} \mathrm{~m}^{2}$
  • C $4.5 \mathrm{~kg} \mathrm{~m}^{2}$
  • D $5.5 \mathrm{~kg} \mathrm{~m}^{2}$
Solution:
1193 Upvotes Verified Answer
The correct answer is: $2.5 \mathrm{~kg} \mathrm{~m}^{2}$
$I=\frac{\tau}{\alpha}=\frac{31.4}{4 \pi}=\frac{31.4}{4 \times 3.14}=2.5 \mathrm{~kg} \mathrm{~m}^{2}$.

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