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Moment of inertia of a body about an axis is $4 \mathrm{~kg}-\mathrm{m}^2$. The body is initially at rest and a torque of $8 \mathrm{~N}$-m starts acting on it along the same axis. Work done by the torque in $20 \mathrm{~s}$, in joules, is
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Verified Answer
The correct answer is:
$3200$
Given, $I=4 \mathrm{~kg}-\mathrm{m}^2, \tau=8 \mathrm{~N}-\mathrm{m}$ and $t=20 \mathrm{~s}$
$$
\begin{gathered}
\tau=l \alpha \\
\alpha=\frac{\tau}{l}=\frac{8}{4}=2 \\
\theta=\frac{1}{2} \alpha t^2 \\
\Rightarrow \quad \theta=\frac{1}{2} \times 2 \times 20 \times 20=400 \\
\omega=\tau \theta=8 \times 400=3200 \mathrm{~J}
\end{gathered}
$$
$$
\begin{gathered}
\tau=l \alpha \\
\alpha=\frac{\tau}{l}=\frac{8}{4}=2 \\
\theta=\frac{1}{2} \alpha t^2 \\
\Rightarrow \quad \theta=\frac{1}{2} \times 2 \times 20 \times 20=400 \\
\omega=\tau \theta=8 \times 400=3200 \mathrm{~J}
\end{gathered}
$$
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