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BITSAT 2011

Exam: BITSAT
Year: 2011
Subject: Physics
Chapter: Rotational Motion

BITSAT 2011 · Physics · Rotational Motion

A constant torque of 31.4 N-m is exerted on a pivoted wheel. If...

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} / \mathrms^{2},

then the moment of inertia of the wheel is

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 $5.5 \mathrm{ kg} \mathrm{ m}^{2}$

Correct answer

A. $2.5 \mathrm{ kg} \mathrm{ m}^{2}$

Step-by-step solution

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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