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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

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

1 Answer

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.


answered Oct 4, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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