Solve the following :

Here,

$I=0.5 \mathrm{~kg}_{-} \mathrm{m}^{2}$

$\mathrm{R}=0.1 \mathrm{~m}$

Translatory Motion Equation

$4 g^{\sin 45-T_{2}=4 a}-(\mathrm{i})$

$T_{1}-2 g \sin 45=2 a$-(ii)

Rotational Motion Equation $\tau=\mathrm{I} \alpha$

$T_{2} R-T_{1} R=I\left(\frac{a}{R}\right)_{-\text {(iii) }}$

Using (i),(ii) and (iii)

$\mathrm{a}=0.25 \mathrm{~m} / \mathrm{s}^{2}$