Q. A couple of $\left(10\right)^{5}\left(\right.Nm\left.\right)$ is applied to a disk of mass $10kg$ and $50m$ radius of gyration. The value of angular acceleration $\left(\right.$ in $rad/s^{2}$ ) will be

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A

$50$

B

$5$

C

$4$

D

$20$

Solution:

We know that
$\tau=Iα...\left(\right.i\left.\right)$
here $\tau=10^{5}Nm$
$I=\left(mK\right)^{2}...\left(\right.ii\left.\right)$
$m=10kgandK=radiusofgyration=50m$
So $I=\left(\right.10\left.\right)\left(\right.50\left(\left.\right)^{2}$ (from eq. $\left(\right.ii\left.\right)$ )
So $I=2.5\times 10^{4}kgm^{2}$
Now from eq. $\left(\right.i\left.\right)$ $\left(10\right)^{5}=2.5\times \left(10\right)^{4}\left(\right.\alpha \left.\right)$
So $\alpha =4rads^{- 2}$

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