A couple of 10⁵ (Nm) is applied to a disc of mass 10 kg and 50 m radius of gyration. The value of angular acceleration (in rad / s²) will be

We know that

$\tau =\mathrm{I}\alpha \dots \left(\mathrm{i}\right)$

here $\tau ={10}^5 \mathrm{Nm}$

$\mathrm{I}={\mathrm{mK}}^2\dots \left(\mathrm{ii}\right)$

m = 10 kg and K = radius of

gyration = 50 m

So $\mathrm{I}=\left(10\right)(50{)}^2$  (from eq. (ii))

So $\mathrm{I}=2.5\times {10}^4{\mathrm{kgm}}^2$

Now from eq. $\text{ (i) }{10}^5=2.5\times {10}^4\left(\alpha \right)$

So $\alpha =4{\mathrm{rads}}^{-2}$