Question

The force acting on a window of area $50\text{cm}\times 50\text{cm}$ of a submarine at a depth of $2000\text{m}$ in an ocean, interior of which is maintained at sea level atmospheric pressure is
(Density of sea water $={10}^3{\text{kg/m}}^3,g=10{\text{m/s}}^2$ )

• A. ${10}^6\text{N}$
• B. $5\times {10}^5\text{N}$
• C. $25\times {10}^6\text{N}$
• D. $5\times {10}^6\text{N}$

Answer 1

The correct option is D $5\times {10}^6\text{N}$

The pressure outside the submarine is $P=P_a+\rho gh$
Pressure inside the submarine is $P_a.$
$\therefore$ Net pressure acting on the window is $P_g=P-P_a=\rho gh={10}^3\times 10\times 2000$
$\Rightarrow P_g=2\times {10}^7\text{Pa}$
Area of window is
$A=50\text{cm}\times 50\text{cm}=2500\times {10}^{-4}{\text{m}}^2$
So, force acting on the window is
$F=P_{g}A=2\times {10}^7\times 2500\times {10}^{-4}$
$\Rightarrow F=5\times {10}^6\text{N}$