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A wheel of radius $0.4 \mathrm{~m}$ can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of $4 \mathrm{~kg}$ is hung. An angular acceleration of $8 \mathrm{rad}-\mathrm{s}^{-2}$ is produced in it due to the torque. Then, moment of inertia of the wheel is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$2 \mathrm{~kg}-\mathrm{m}^{2}$
$$
\begin{array}{l}
\text { Given, } r=0.4 \mathrm{~m}, \\
\alpha=8 \mathrm{rad} \mathrm{s}^{-1}, \\
m=4 \mathrm{~kg}, I=? \\
\text { Torque, } \tau=I \alpha \\
m g r=I \cdot \alpha \\
4 \times 10 \times 0.4=I \times 8 \text { or } I=\frac{16}{8}=2 \mathrm{~kg} \cdot \mathrm{m}^{2}
\end{array}
$$
\begin{array}{l}
\text { Given, } r=0.4 \mathrm{~m}, \\
\alpha=8 \mathrm{rad} \mathrm{s}^{-1}, \\
m=4 \mathrm{~kg}, I=? \\
\text { Torque, } \tau=I \alpha \\
m g r=I \cdot \alpha \\
4 \times 10 \times 0.4=I \times 8 \text { or } I=\frac{16}{8}=2 \mathrm{~kg} \cdot \mathrm{m}^{2}
\end{array}
$$
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