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Question

A uniform chain of length $$l$$ is placed on a smooth horizontal table, such that half of its length hangsover one edge. It is released from rest, the velocity with which it leaves the table is:


A
3gl4
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B
3gl2
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C
2gl3
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D
gl3
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Solution

The correct option is C $$\displaystyle \sqrt { \frac { 3gl }{ 4 } } $$
Keeping the horizontal level of table as reference of gravitational potential, decrease in the gravitational potential of the chain=$$-\dfrac{m}{2}g\dfrac{l}{4}-(-mg\dfrac{l}{2})$$
$$=\dfrac{3}{8}mgl$$
This loss in gravitational potential increases the kinetic energy of chain.
$$\implies \dfrac{3}{8}mgl=\dfrac{1}{2}mv^2$$
$$\implies v=\sqrt{\dfrac{3gl}{4}}$$

Physics

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