Question

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 the vessel is :

• A. $\frac{m\rho g}{\sigma s}$
• B. $\frac{mg}{s}$
• C. $\frac{m\sigma g}{\rho s}$
• D. Zero

Answer 1

The correct option is C $\frac{m\sigma g}{\rho s}$

$\rho$ = density of cylinder
$\sigma$= density of the liquid which is in vessel
$F_u$ = upthrust force
$s=$ cross sectional area of vessel
From Newton's third law,
excess force acting on the liquid is equal force of buoyancy acting on the cylinder.
Hence,
$\Delta P=\frac{F}{s}=\frac{V\sigma g}{s}=\frac{m\sigma g}{s\rho }$