# 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 =$13600Kg/m^3$

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

$P_{top}$=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.