Question

When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\frac{5r}{4}$. Taking the atmospheric pressure to be equal to $10\, m$ eight of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :

• A. 11.2 m
• B. 8.7 m
• C. 9.5 m
• D. 10.5 m

Answer 1

Answer: (C)

9.5 m

Answer 2

Given, bubble of radius $r$ rises from bottom of lake. The radius of bubble at top becomes $5 r / 4$. Therefore, pressure in bubble at bottom is
$P_{1}=P_{0}+\rho g h+\frac{4 T}{r}$
Pressure in bubble at top is $P_{2}=P_{0}+\frac{4 T}{5 r / 4}$
Now, we know $P_{1} V_{1}=P_{2} V_{2}$, therefore,
$\left(P_{0}+\rho g h+\frac{4 T}{4}\right) \frac{4 \pi^{\prime}}{3} r^{3}=\left(P_{0}+\frac{4 T}{5 r / 4}\right) \frac{4 \pi}{3}\left(\frac{5 r}{4}\right)^{3}$
$P_{0}+\rho h g+\frac{4 T}{r}=\left(P_{0}+\frac{4 T \times 4}{5 r}\right) \frac{125}{64}$
Now given $P_{0}=10\, \rho g$, therefore,
$10\, \rho g+\rho g h+\frac{4 T}{r}=\frac{125}{64} \times 10 \rho g+\frac{16 T}{5 r} \times \frac{125}{64}$
Neglecting effect of temperature, we get
$10 \,\rho g+\rho g h=\frac{125}{64} \times 10 \,\rho g$
$\Rightarrow 10+h=\frac{125}{64} \times 10$
$\Rightarrow h=\frac{1250}{64}-10=9.53125 \,m$
$\Rightarrow h \sim 9.5\, m$