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(D) Let the acceleration of the blocks be $a$. The forces acting along the incline for the block on the $60^{\circ}$ slope is $mg \sin 60^{\circ} - T

Vedclass · Accepted Answer

$\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) g$

Vedclass · Answer

(D) Let the acceleration of the blocks be $a$. The forces acting along the incline for the block on the $60^{\circ}$ slope is $mg \sin 60^{\circ} - T = ma$ and for the block on the $30^{\circ}$ slope is $T - mg \sin 30^{\circ} = ma$.Adding these two equations: $mg(\sin 60^{\circ} - \sin 30^{\circ}) = 2ma$.$a = \frac{g}{2} (\frac{\sqrt{3}}{2} - \frac{1}{2}) = \frac{g(\sqrt{3}-1)}{4}$.The acceleration vectors for the two blocks are $\vec{a}_1 = a(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})$ and $\vec{a}_2 = a(-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j})$.The acceleration of the centre of mass is $\vec{a}_{CM} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.$\vec{a}_{CM} = \frac{a}{2} [(\cos 60^{\circ} - \cos 30^{\circ}) \hat{i} - (\sin 60^{\circ} + \sin 30^{\circ}) \hat{j}]$.Substituting $a = \frac{g(\sqrt{3}-1)}{4}$,$\cos 60^{\circ} = 1/2$,$\cos 30^{\circ} = \sqrt{3}/2$,$\sin 60^{\circ} = \sqrt{3}/2$,$\sin 30^{\circ} = 1/2$:$|\vec{a}_{CM}| = \frac{a}{2} \sqrt{(\frac{1-\sqrt{3}}{2})^2 + (\frac{\sqrt{3}+1}{2})^2} = \frac{a}{2} \sqrt{\frac{1+3-2\sqrt{3} + 3+1+2\sqrt{3}}{4}} = \frac{a}{2} \sqrt{\frac{8}{4}} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.$|\vec{a}_{CM}| = \frac{g(\sqrt{3}-1)}{4\sqrt{2}}$.Thus,option $(d)$ is correct.

In the figure shown,the blocks have equal masses. Friction,mass of the string,and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity $= g$).

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