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A rod of mass ‘ $m$ ’ lengths ‘ $l$ ’ is in equilibrium in the position shown. The ground is rough enough to prevent slipping.

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The normal force $N_{1}$ at B is $\frac{m g}{2}$ (Before cutting the string)

The normal force $N_{2}$ at B immediately after cutting the string is $\frac{7 m g}{16}$

The friction force at point B is $\frac{\sqrt{3} m g}{16}$ after cutting

The friction force at point B is zero before cutting the string.

answer is A, D, B.

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Detailed Solution

$T + N_{1} = m g$

$m g \frac{l}{2} \sin θ - T l \sin θ = 0 \Rightarrow T = m g / 2, N_{1} = m g / 2 m g \frac{l}{2} \sin θ = \frac{m l^{2}}{3} α$

$\Rightarrow α = \frac{3 g \sin θ}{2 l} a_{c m} = \frac{l}{2} α = \frac{3 g \sin θ}{4} m g - N_{2} = m {(a_{c m})}_{y} = m \frac{3 g}{4} \sin^{2} θ N_{2} = m g [1 - \frac{3}{4} \sin^{2} θ] = \frac{7 m g}{16}$

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