A uniform disc with mass M = 4   kg and radius R = 10   cm is mounted on a fixed horizontal axle…

Let $a$ be the acceleration of block and $\alpha$ be the angular acceleration of disc.

Now, for unwrapping without slipping, $R\alpha =a$.

Forces acting on block are shown in FDB.

Using Newton's second law

$mg-T=ma ...\left(1\right)$

Torque about centre of disc is $\tau =Ia$

$TR=\frac{M{R}^2}{2}\alpha =\frac{4R^2}{2}\alpha =\Rightarrow T=2R\alpha ...\left(2\right)$

Putting the value of tension in equation $\left(1\right)$, we get

$$\begin{gathered} mg-2R\alpha =ma \\ \Rightarrow mg-2R\frac{a}{R}=ma \\ \Rightarrow mg-2a=ma \\ \Rightarrow ma+2a=mg \\ \Rightarrow a=\frac{mg}{m+2}=\frac{2\times 10}{2+2}=5 \mathrm{m} {\mathrm{s}}^{-2} \end{gathered}$$

Now, $\alpha =\frac{5}{0.1}=50 \mathrm{rad} {\mathrm{s}}^{-2}$

Tension in the cord is $T=2\times 0.1\times 50=$$10 \mathrm{N}$.