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A rigid massless beam is balanced by a particle of mass 4m in left hand side and a pulley particle system in right hand side. The value of $\frac{x}{y}$ is:

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$\frac{7}{6}$

$\frac{5}{6}$

$\frac{7}{9}$

$\frac{11}{12}$

answer is D.

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Detailed Solution

$\begin{array}{l} a = \frac{(2 m - m) g}{3 m} = \frac{g}{3} for T in lower string, \\ T = 2 m g - 2 m \times \frac{g}{3} = 4 m g / 3 \\ & T^{1} = 2 T = \frac{8 m g}{3} \\ T^{1} + m g = \frac{8 m g}{3} + m g = \frac{11 m g}{3} \end{array}$

Let us take moment about hinge

$4 m g \times x = \frac{11 m g}{3} \times y \Rightarrow \frac{x}{y} = \frac{11}{12}$

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