Q. A cylinder of mass $m$ and density $\rho$ hanging from a string is lowered into a vessel of cross-sectional area $A$ containing a liquid of density $\sigma( < \rho)$ until it is fully immersed. The increase in pressure at the bottom of the vessel is
A) Zero
B) $\frac{mg}{A}$
C) $\frac{mg\ \rho}{\sigma A}$
D) $\frac{m\ \sigma\ g}{\rho\ A}$
Explanation
(d) Volume of cylinder $= \frac{m}{\rho}$
Upthrust on cylinder $= \left( \frac{m}{\rho} \right)\sigma\ g$
From Newton’s third law, the downward force exerted by cylinder on the liquid is $= \left( \frac{m}{\rho} \right)\sigma\ g$
$\therefore\ $Increase in pressure $= \frac{m\ \sigma\ g}{\rho\ A}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.