Uniform rod AB is hinged at end A in horizontal position as shown in the fig. The other end is connected to a block through a massless string ‘m’ as shown. The pulley is smooth and massless. Masses of block and rod are same and equal to ‘m’. If acceleration of block just after release from this position is xg/8 then the value of ‘x’ will be

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Detailed Solution

FBD for block
mg – T = ma
torque equation
$Tl - mg \frac{l}{2} = \frac{{ml}^{2}}{3} α$
$a = αl$
Solve above equation

$a = \frac{3 g}{8}$

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