A uniform disc of mass $2 \mathrm{kg}$ and radius $1 \mathrm{m}$ is mounted on an axle supported on fixed frictionless bearings. A light chord is wrapped around the rim of the disc and mass of $1 \mathrm{kg}$ is tied to the free end. If it is released from rest -

By FBD of particle

$$\begin{gathered} \mathrm{mg}-\mathrm{T}=\mathrm{ma} \\ 10-\mathrm{T}=\mathrm{a} ...\left(\mathrm{i}\right) \end{gathered}$$

By FBD disc

$$\begin{gathered} \mathrm{TR}=\mathrm{I\alpha }=I\frac{a}{R}\Rightarrow \mathrm{T}=\frac{M{R}^2}{2}\times \frac{\mathrm{a}}{{\mathrm{R}}^2} \\ \mathrm{T}=\frac{Ma}{2}=a...\left(\mathrm{ii}\right) \end{gathered}$$

By equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$\left(1\right) \mathrm{a}=5 \mathrm{m}/{\mathrm{s}}^2$ and $\mathrm{T}=5\mathrm{N}$ and $\mathrm{\alpha }=\mathrm{a}/\mathrm{R}=5 \mathrm{rad}/{\mathrm{s}}^2$

$\left(2\right)$ For angular displacement of disc : $\mathrm{\theta }=\mathrm{\omega t}+1/2{\mathrm{\alpha t}}^2$ = 40 rad

$\left(3\right)$ Work done by torque $=\int \mathrm{\tau d\theta }=\mathrm{\tau }\int \mathrm{d\theta }=5\times 40=200 \mathrm{J}$

$\left(4\right) ∆\mathrm{KE}=∆\mathrm{w}=200 \mathrm{J} \Rightarrow {\mathrm{K}}_2-{\mathrm{K}}_1=200 \mathrm{J}$