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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 rad- $\mathrm{s}^{-2}$ is produced in it due to the torque. Then, moment of inertia of the wheel is $\left(g=10 \mathrm{~ms}^{-2}\right)$
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The correct answer is:
$1 \mathrm{~kg}-\mathrm{m}^2$
Given, $r=0.4 \mathrm{~m}$,
$\begin{aligned} \alpha & =8 \mathrm{rad} \mathrm{s}^{-1}, \\ m & =4 \mathrm{~kg}, \quad I=? \\ \text { Torque, } \quad \tau & =I \alpha \\ m g r & =I \cdot \alpha\end{aligned}$
$\begin{aligned} \text { or } & & 4 \times 10 \times 0.4 & =I \times 8 \\ \Rightarrow & & I & =\frac{16}{8}=2 \mathrm{~kg}-\mathrm{m}^2 \\ \text { or } & & I & =2 \mathrm{~kg}-\mathrm{m}^2\end{aligned}$
$\begin{aligned} \alpha & =8 \mathrm{rad} \mathrm{s}^{-1}, \\ m & =4 \mathrm{~kg}, \quad I=? \\ \text { Torque, } \quad \tau & =I \alpha \\ m g r & =I \cdot \alpha\end{aligned}$
$\begin{aligned} \text { or } & & 4 \times 10 \times 0.4 & =I \times 8 \\ \Rightarrow & & I & =\frac{16}{8}=2 \mathrm{~kg}-\mathrm{m}^2 \\ \text { or } & & I & =2 \mathrm{~kg}-\mathrm{m}^2\end{aligned}$
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