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# A cylinder of mass m and density $\rho$ hanging from a string is lowered into a vessel of cross. sectional area s containing a liquid of density $\sigma( < \rho)$ until it is fully immersed. The increase in pressure at the bottom of vessel is

$\begin {array} {1 1} (a)\;\frac{m \rho g}{\sigma s} \\ (b)\;\frac{mg}{s}\\ (c)\;\frac{m \sigma g}{s} \\ (d)\;zero \end {array}$

Upthrust in the cylinder
$\qquad= \bigg(\large\frac{m}{\rho}\bigg)$$\sigma g [when \large\frac{m}{\rho}$$=V$] Volume of cylinder
From's Newton's third law force. exerted by the cylinder on the liquid is also $\bigg(\large\frac{m}{\rho}\bigg)$$\sigma g$
$\therefore$ Increase in pressure $=\large\frac{Force}{area}$
$\qquad= \large\frac{\Large\frac{m}{\rho}\sigma g}{s}$
$\qquad= \large\frac{m \sigma g}{\rho s}$
Hence c is the correct answer.

edited Feb 18, 2014 by meena.p