$a)\;\large\frac{mg}{l}$$R^2$

$b)\; \large\frac{mg}{2}$$R$

$c) \;mgR$

$d)\; \large\frac{mg}{2l}$$R^2$

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Gravitation potential energy

$=\int \limits_{\theta=0} ^{\theta=\pi/2} (r d\theta) \rho g (r \cos \theta)$

When $\rho=\large\frac{m}{l}$ $'Rd \theta'$ small segement of the chain and $'R \cos \theta'$ is the height of the chain from base.

$\quad= \rho gr^2 \sin \theta\bigg ]_0^{\pi/2}= \rho g R^2$

$\quad=\large\frac{mg}{l}$$R^2$

Hence a is the correct answer.

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