When an air bubble rises from the bottom of lake to the surface, its radius is doubled. The atmospheric…

When an air bubble rises from the bottom of lake to the surface, its radius is doubled. The atmospheric pressure is equal to that of a column of water of height ' $H$ '. The depth of the lake is
  1. H
  2. 2 H
  3. 7 H
  4. 8 H

Solution

$\begin{aligned} & P_1 V_1=P_2 V_2 \\ & \Rightarrow\left(P_0+h \rho g\right) \times \frac{4}{3} \pi r^3=P_0 \times \frac{4}{3} \pi(2 r)^3 \end{aligned}$
Where, $\mathrm{h}=$ depth of lake $\Rightarrow \mathrm{h} \rho \mathrm{~g}=7 \mathrm{P}_0 \Rightarrow \mathrm{~h}=7 \times \frac{\mathrm{H} \mathrm{\rho g}}{\rho g}=7 \mathrm{H}$

Asked in: MHT CET 2024 (09 May Shift 1)

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