# A uniform chain of length l and mass m is placed as shown . Find the gravitational potential of chain

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

$c) \;mgR$

$d)\; \large\frac{mg}{2l}$$R^2 ## 1 Answer Gravitation potential energy =\int \limits_{\theta=0} ^{\theta=\pi/2} (r d\theta) \rho g (r \cos \theta) When \rho=\large\frac{m}{l} 'Rd \theta' small segement of the chain and 'R \cos \theta' is the height of the chain from base. \quad= \rho gr^2 \sin \theta\bigg ]_0^{\pi/2}= \rho g R^2 \quad=\large\frac{mg}{l}$$R^2$
Hence a is the correct answer.

edited Feb 17, 2014 by meena.p