Ask Questions, Get Answers

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

A body floats on mercury with $\large\frac{1}{3}$ of its volume below the surface. Another liquid of density $\large\frac{1}{10}$ that of mercury is poured on top to cover the body completely. What is the fraction of cube now immersed.

\[\begin {array} {1 1} (a)\;0.33 \\ (b)\;0.25 \\ (c)\;0.15 \\ (d)\;0.44 \end {array}\]

Can you answer this question?

1 Answer

0 votes
Let V be volume of body using law of floatation.
In the first case :
The upthrust =weight of body
$W= \large\frac{1}{3} $$V \rho _{Hg}g$-------(1)
In the second case :
Let x fraction of the volume be immersed in mercury.
Then $W=x V \rho _{Hg}+(1-x) V \rho _{liquid}g$--------(2)
equating (1) and (2)
$\large\frac{\rho _{Hg}}{3}$$=x \rho _{Hg}+(1-x) \rho _{liquid}$
$\large\frac{\rho _{Hg}}{3}$$=x \rho _{Hg}+(1-x) \large\frac{\rho _{Hg}}{10}$
$\large\frac{1}{3}$$=x +\bigg(\large\frac{1-x}{10}\bigg)$
$\large\frac{1}{3}$$=x -\large\frac{x}{10}+\frac{1}{10}$
$x= \large\frac{7}{27}$$=0.25$
Hence b is the correct answer.


answered Oct 8, 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