Question

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

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

Answer 1

Answer: (C)

$5 \times 10^{6} \, N$

Answer 2

Here, $h - 2000 \,m$, $\rho = 10^{3} \,kg\, m^{-3}$, $g = 10 \,m \,s^{-2}$
The pressure outside the submarine is
$P = P_{a} + \rho\, gh$
where $P_{a}$ is the atmospheric pressure
Pressure inside the submarine is $P_{a}$
Hence, net pressure acting on the window is gauge pressure
Gauge pressure, $P_{g}=P-P_{a}= \rho\,gh$
$=10^{3} \, Kg\, m^{-3} \times 10 \, m\, s^{-2} \times 2000 \, m =2 \times 10^{7}\,pa$
Area of a window is
$A = 50\, cm \times 50 \, cm = 2500 \times 10^{-4} \, m^{2}$
Force acting on the window is
$F = P_{g} \, A$
$=2 \times 10^{7}\,pa \times 2500 \times 10^{-4} \, m^{2} =5 \times 10^{6} \, N$