Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App