Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
0 votes

An air bubble is released from the bottom of a lake of depth 11m.The temperature at the bottom is $4^{\large\circ}C$ and that at the surface is $12^{\large\circ}C$.What is the ratio of bubble's radius at the surface and its radius at bottom?Assume density of heater is $100Kg/m^3$.Atmospheric pressure =$75$cm of Hg & density of mercury =$13600Kg/m^3$

$\begin{array}{1 1}(A)\;1.2\\(B)\;1.3\\(C)\;1.4\\(D)\;1.5\end{array} $

Can you answer this question?

1 Answer

0 votes
$P_{top}$=75cm of Hg
$P_{bottom}=75+\text{pressure of heater}=75+\large\frac{110}{13600}$$\times 1000$
$\Rightarrow 156cm$ of Hg
Hence (B) is the correct answer.
answered May 12, 2014 by sreemathi.v

Related questions

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