logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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}$$\times 1000$
$\Rightarrow 156cm$ of Hg
$\large\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$
$\large\frac{V_2}{V_1}$$=1.3$
Hence (B) is the correct answer.
answered May 12, 2014 by sreemathi.v
 

Related questions

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