A uniform disc with mass M = 4kg and radius R = 10cm is mounted on...

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 $τ = Ia$

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

Putting the value of tension in equation \left(1\right)$, we get</p><p>$mg - 2R\alpha = ma ⇒ mg - 2R\dfrac{a}{R} = ma ⇒ mg - 2a = ma ⇒ ma + 2a = mg ⇒ a = \dfrac{mg}{m + 2} = \dfrac{2 \times 10}{2 + 2} = 5m{s}^{- 2}$</p><p>Now,$\alpha = \dfrac{5}{0 . 1} = 50rad{s}^{- 2}$</p><p>Tension in the cord is$T = 2 \times 0 . 1 \times 50 =$$10N.