Question

What is the pressure at a point $10m$ below the surface of a lake(see fig)?
Take $g=10m{s}^{-2};1atm=1.01\times {10}^{5}Pa$;
Density of water$=1000kg/m^3$

• A. $3.25\times {10}^{5}Pa$
• B. $2.01\times {10}^{7}Pa$
• C. $3.25\times {10}^{7}Pa$
• D. $2.01\times {10}^{5}Pa$

Answer 1

The correct option is D $2.01\times {10}^{5}Pa$

Increase in pressure from the top surface to the given depth=$\frac{V\rho g}{A}=\frac{Ah\rho g}{A}=\rho gh$

$=1000\times 10\times 10={10}^{5}Pa$

Hence to total pressure at the given depth=$P_0+\Delta P$

$=(1.01\times {10}^{5}Pa)+\left({10}^{5}Pa\right)$

$=2.01\times {10}^{5}Pa$