Question

The volume of an air bubble is doubled as it rises from the bottom of the lake to its surface. The atmospheric pressure is 75cm of mercury. The ratio of density of mercury to that of lake water is 40/3. The depth of the lake in meter is:
(A) 10m
(B) 15m
(C) 20m
(D) 25m

Answer 1

Hint: Fluid pressure is the pressure at a point within a fluid rising due to the weight of the fluid.
Using following formula,
 $P_{fluid}=P+\rho gh$
P is pressure at the reference point, Here, P= ${\text{P}}_0$ , ${\text{P}}_0$ is atmosphere pressure , $\rho$ is density of fluid, g is acceleration due to gravity, h is height from reference point.

Complete step by step solution:
We have given, (V’) volume of air bubble $=2\times$ volume of air bubble at bottom (v).
i.e. $\text{{V}'=2V}$ ---------(1)
Also,
 $\begin{array}{rl}& {\displaystyle \frac{({\rho }^{\prime })\text{density of mercury}}{(\rho )\text{density of lake water}}}={\displaystyle \frac{40}{3}}\\ & {\displaystyle \frac{{\rho }^{\prime }}{\rho }}={\displaystyle \frac{40}{3}}\\ & \Rightarrow {\rho }^{\prime }={\displaystyle \frac{40}{3}}\rho \end{array}$ ------(2)
Now, the pressure exerted at the surface is equal to 75cm of mercury.
From Idle gas equation,
 $$ $\begin{array}{rl}& P=P^{\prime }\left({\displaystyle \frac{V^{\prime }}{V}}\right)\\ & P=P^{\prime }\left({\displaystyle \frac{2V}{V}}\right)\\ & P=2P^{\prime }\\ & \sin ce,P^{\prime }=P_0\end{array}$
 $\therefore P=2P_0$ --------(3)
Pressure at h depth From eq. (3)
 $\begin{array}{rl}& P_0+\rho gh\\ & 2P_0=P_0+\rho gh\\ & P_0=\rho gh\\ & \therefore {\rho }^{\prime }g{h}^{\prime }=\rho gh\\ & h=\left({\displaystyle \frac{{\rho }^{\prime }}{\rho }}\right)\times h^{\prime }\end{array}$ ----(4)
Here, h is depth of lake h’ is depth = 75 cm
                                                               = 0.75m
Now, Put all the values in eq. (4)
 $\begin{array}{rl}& h=\left({\displaystyle \frac{40}{3}}\right)\times 0.75\\ & h={\displaystyle \frac{40}{3}}\times {\displaystyle \frac{75}{100}}\\ & h=10m\end{array}$
Hence, option A is correct.

Note:
Factors that affect fluid Pressure:
There are two factors that affect fluid pressure. These two factors are the depth of the fluid and its density.
- Depth of the fluid: As the depth increases, the pressure exerted by the fluid becomes more.
- Density of the fluid: Denser fluids such as water exert more pressure than lighter fluids such as air. The molecules in the denser fluid are close to each other resulting in more collisions in the given area.