At what depth from surface of water, the pressure will be equal to three times the atmosphere pressure? Given atmospheric pressure = 10 $$\text{N/cm}^2$$​ and g = 9.8 $$\text{m/s}^2$$​.

Correct option is C

​$$\textbf{Given:} \\\quad \bullet\ \text{Atmospheric pressure, } P_{\text{atm}} = 10\, \text{N/cm}^2 = 10 \times 10^4\, \text{N/m}^2 = 10^5\, \text{Pa} \\\quad \bullet\ \text{Density of water, } \rho = 1000\, \text{kg/m}^3 \\\quad \bullet\ \text{Acceleration due to gravity, } g = 9.8\, \text{m/s}^2 \\\\\textbf{Concept } \\\text{The total (absolute) pressure at a depth in a liquid is the sum of atmospheric pressure and the pressure due} \\\text{to the water column (hydrostatic pressure):} \\\quad P_{\text{abs}} = P_{\text{atm}} + \rho g h \\\\\text{According to the question, the absolute pressure is \textbf{three times} the atmospheric pressure:} \\\quad 3P_{\text{atm}} = P_{\text{atm}} + \rho g h$$​

$$3P_{\text{atm}} = P_{\text{atm}} + \rho g h \\\Rightarrow 2P_{\text{atm}} = \rho g h \\\Rightarrow h = \frac{2 \times 10^5}{1000 \times 9.8} \\\Rightarrow h = \frac{2 \times 10^5}{9800} = 20.4\, \text{m}$$​​