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If a disc of mass $M$ and radius $R$ rotates with an angular acceleration $a_{\text {, }}$ of the torque acting on the disc is
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Verified Answer
The correct answer is:
$\frac{M R^2 a}{2}$
$\therefore$ Moment of inertia of disc, $Y=\frac{1}{2} M R^2$
Angular acceleration, $\alpha=a$ (given)
So, torque $\tau=Y \alpha$,
$$
\tau=\left(\frac{1}{2} M R^2\right)\langle a\rangle \Rightarrow \tau=\frac{M R^2 a}{2}
$$
Angular acceleration, $\alpha=a$ (given)
So, torque $\tau=Y \alpha$,
$$
\tau=\left(\frac{1}{2} M R^2\right)\langle a\rangle \Rightarrow \tau=\frac{M R^2 a}{2}
$$
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