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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Aicharya · Accepted Answer

For a pulley and block system on a smooth double inclined plane as shown below

Force equation for both the blocks, \(\Rightarrow \quad m g \cos 30^{\circ}-T=m a\) ...(i) \(\Rightarrow \quad T-m g \cos 60^{\circ}=m a\) ...(ii) From above equation we get, \(a=\frac{(\sqrt{3}-1)}{4} g\) \(\because\) Magnitude of the acceleration of centre of mass, \(\mathbf{a}_{C M}=\left|\frac{m \mathbf{a}_1+m \mathbf{a}_2}{m+m}\right|\) Here, \(\mathbf{a}_{C M}=\left|\frac{\left(\frac{\sqrt{3}-1}{4}\right) g \hat{\mathbf{i}}+\left(\frac{\sqrt{3}-1}{4}\right) g \hat{\mathbf{j}}}{2}\right|\) \(\begin{aligned} & =\frac{g}{2} \sqrt{\left(\frac{\sqrt{3}-1}{4}\right)^2+\left(\frac{\sqrt{3}-1}{4}\right)^2} \\ & =\frac{g}{2} \times \sqrt{2} \times \frac{\sqrt{3}-1}{4} \\ & =\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) g \end{aligned}\) Hence, option (d) is correct.

In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley…

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