Question

If pressure at the half depth of a lake is equal to $\frac{3}{4}$ times the pressure at its bottom, then find the depth of the lake . [Take g=$10m/s^2]$

• A. $\frac{P_0}{\rho g}$
• B. $\frac{2P_0}{\rho g}$
• C. $\frac{P_0}{2\rho g}$
• D. $\frac{3P_0}{\rho g}$

Answer 1

The correct option is A $\frac{P_0}{\rho g}$

Let depth of the lake be $h$ and pressure at bottom $=P$

Then $P=P_0+\rho gh\to \left(1\right)$ $(P_0=$ atmospheric pressure, $\rho$ = density of water)

At half depth $\left(h/2\right)$ pressure is $\frac{3P}{4}$ then :

$\frac{3P}{4}=P_0+\rho g\frac{h}{2}\to \left(2\right)$

On subtracting equation 2 from 1 we get :

$\frac{P}{4}=\rho g\frac{h}{2}$

$\Rightarrow P=2\rho gh$, substituting this value of $P$ in equation 1:

$2\rho gh=P_0+\rho gh$

$\Rightarrow h=\frac{P_0}{\rho g}\to$ Depth of the lake