Question

The force acting on a window of area 50 cm x 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 x 10${}^5$ N
• B. 25 x 10${}^5$ N
• C. 5 x 10${}^6$ N
• D. 25 x 10${}^6$ N

Answer 1

Answer: (C)

5 x 10${}^6$ N

Here, $h=2000m,$

$\rho ={10}^{3}kg{m}^{-3},$

$g=10m{s}^{-2}$

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,

$P_g=P-P_a=\rho gh$

$={10}^{3}kg{m}^{-3}\times 10m{s}^{-2}\times 2000m$

$=2\times {10}^{7}Pa$
Area of a window is $A=50cm\times 50cm$

$=2500\times {10}^{-4}{m}^2$
Force acting on the window is
$F=P_{g}A$

$=2\times {10}^{7}Pa\times 2500\times {10}^{-4}{m}^2$

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