The torque exerted on the disc is given, by
$\hspace15mm \tau=TR \hspace15mm ...(i)$
Also $\hspace15mm \tau=1 \alpha \hspace15mm ...(ii)$
From Eqs. (i) and (ii), we get
$\hspace15mm I\alpha=TR$
$\hspace15mm \alpha=\frac {TR}{I}= \frac {2TR}{MR^2}$
or $\hspace15mm \alpha=\frac {2T}{MR}$