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12.Kinetic Theory of Gases
medium
When an air bubble of radius $‘r’$ rises from the bottom to the surface of a lake, its radius becomes $5r/4$ (the pressure of the atmosphere is equal to the $10 \,m$ height of water column). If the temperature is constant and the surface tension is neglected, the depth of the lake is .... $m$
A
$3.53$
B
$6.53$
C
$9.53$
D
$12.53$
Solution
In this case, $(4 / 3) \pi r^{3}\left(p_{0}+p_{1}\right)=(4 / 3) \pi r^{3}\left(5^{3} / 4^{3}\right) P _{0}$
where, $P_{0}=$ atm pressure, $P_{1}=$ water pressure
Rearranging, we get, $P_{1}=\left[\left(5^{3} / 4^{3}\right)-1\right] P_{0}$
Or, $\rho gH =(\frac{61}{64}) P _{0}$
Or, $H =\frac{(\frac{61}{64}) P _{0}}{\rho g}$
$P_{0}=10 g$ and $\rho$ of water $=1$
Putting these values we get, $H =9.53 \; m$
Standard 11
Physics
