Question

If the atmospheric pressure is ${10}^5\frac{N}{m^2}$ and the density of water ${10}^3\frac{kg}{m^3}$. What will be the depth below the surface of water if pressure is equal to thrice the atmospheric pressure?

• A. 10² m
• B. 10 m
• C. 20 m
• D. 30 m

Answer 1

The correct option is C

20 m

Given, atmospheric pressure $P_0={10}^5\frac{N}{m^2}$
Density of water, $\rho ={\frac{1}{0}}^{3}kg{m}^3$
Acceleration due to gravity = 10 $\frac{m}{s^2}$
Pressure at a depth h below the surface of water = atmospheric pressure + pressure due to water column of height h.
Or $3P_0=P_0+h\rho +h\rho g$
Or $2P_0$ = $h\rho g$
Hence, h = $\frac{2P_0}{\rho g}$ = $\frac{2\times {10}^5}{{10}^3\times 10}$ = 20 m.