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A thin uniform inelastic chain of mass ‘ $m$ ’ and length ‘ $l$ ’ is suspended so that its lower end just touches a smooth inelastic plane inclined at $45^{o}$ . The chain is released from rest.

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The force exerted by the chain on the plane at any subsequent instant is $\frac{m g^{2} t^{2}}{\sqrt{2} l}$

The force exerted by the chain on the plane at any subsequent instant is $\frac{3 m g^{2} t^{2}}{2 \sqrt{2} l}$

The total impulse on the plane while the chain falls is $\sqrt{g l}$

The total impulse on the plane while the chain falls is $\sqrt{\frac{g l}{2}}$

answer is B.

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Detailed Solution

$v = g t$

$F_{1} = λ v^{2} \cos 45^{0} = \frac{m}{t} . g^{2} t^{2} . \frac{1}{\sqrt{2}}$

$F_{2} = m g \sin 45^{0} = \frac{m}{l} . \frac{1}{2} g t . g \frac{1}{\sqrt{2}} = \frac{3 g^{2} t^{2}}{2 \sqrt{2} l}$

$F_{n e t} = 3 m g^{2} t^{2} / 2 \sqrt{2} l$

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