Chat with tutor

Ask Questions, Get Answers


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.

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

1 Answer

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.


Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.