Correct Option A
Solution:
As we know, Torque $\tau=\mathrm{I} \alpha$
Given, $\alpha=2 \mathrm{rad} / \mathrm{s}^{2}$
As we know, Tangential acceleration $\mathrm{a}=\mathrm{r} \alpha$
$\mathrm{a}_{t}=\frac{1}{2} \times 2=1 \mathrm{~ms}^{-2}$
$\mathrm{a}_{t}=1 \mathrm{~ms}^{-2}$
$\mathrm{v}=\mathrm{u}+\mathrm{at}$
$=0+2$
$\mathrm{v}=2 \mathrm{~m} / \mathrm{s}$
Also, Radial acceleration
$\mathrm{a}_{\mathrm{r}}=\frac{\mathrm{v}^{2}}{\mathrm{r}}=\frac{4}{0.5}=8 \mathrm{~ms}^{-2}$
Thus, calculating net acceleration
$\mathrm{a}=\sqrt{\mathrm{a}_{\mathrm{r}}^{2}+\mathrm{a}_{\mathrm{t}}^{2}}$
$=\sqrt{64+1}$
$=8 \mathrm{~ms}^{-2}$ (approx)