Question

If pressure at half depth of a lake is equal to $\frac{2}{3}rd$ pressure at the bottom of the lake, then what is the depth of the lake?
(Take atmospheric pressure as ${10}^5\text{Pa}$, density of water as $1000{\text{kg/m}}^3$ and $g=10{\text{m/s}}^2$)

• A. $6.67\text{m}$
• B. $60\text{m}$
• C. $10\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is D $20\text{m}$

Let $h$ be the depth of lake and $P_o$ be the atmospheric pressure.
According to the question,
Pressure at half depth of lake
$=\frac{2}{3}\times$ (Pressure at bottom of lake)
$P_o+\rho g\frac{h}{2}=\frac{2}{3}(\rho gh+P_o)$
$\Rightarrow 3(P_o+\rho g\frac{h}{2})=2(\rho gh+P_o)$
$[\because P_o={10}^5\text{Pa},\rho ={10}^3{\text{kg/m}}^3]$
$\Rightarrow 3P_o+\frac{3}{2}\rho gh=2\rho gh+2P_o$
$\Rightarrow P_o=(2-\frac{3}{2})\rho gh$
$\Rightarrow P_o=\frac{\rho gh}{2}$
$\Rightarrow h=\frac{2P_o}{\rho g}=\frac{2\times {10}^5}{{10}^3\times 10}=20\text{m}$