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

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.

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