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A uniform rod AB is hinged at end A in the horizontal position as shown in Fig. The other end is connected to a block through a mass less string as shown. The pulley is smooth and mass less. Masses of block and rod are same and equal to m. Then acceleration of the block just after release from this position is (g is acceleration due to gravity)

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$\frac{3 g}{8}$

$\frac{g}{4}$

$\frac{3 g}{16}$

$\frac{6 g}{13}$

answer is C.

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Detailed Solution

$(C) m g - T = m (2 a)$ ……(i)
$T l - m g \frac{1}{2} = \frac{m l^{2}}{3} α$ ……..(ii)
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$α = \frac{2 α}{l}$
After solving we get $α = \frac{3 g}{8 l}; T = \frac{5 m g}{3}; α = \frac{3 g}{16}$

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