Question

At a depth of 1000 m in an ocean, what is the force acting on the window area $20 cm \times 20 \,cm$ of a submarine, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is $\left.1.03 \times 10^{3} kgm ^{-3}, g=10\, ms ^{-2}\right)$

• A. $2 \times 10^{5} N$
• B. $3 \times 10^{5} N$
• C. $4 \times 10^{5} N$
• D. $5 \times 10^{5} N$

Answer 1

Answer: (C)

The pressure outside the submarine is $P=P_{a}+$ $\rho g h$ and the pressure inside it is $P_{a}$. Hence, the net pressure acting on the window is the garage pressure $P_{g}=\rho g h$. Since the area of the window is $A=0.04 m ^{2},$ the force acting on it is
$F=P_{g} \cdot A=\rho g h . A$
$=\left(1.03 \times 10^{3} kg m ^{-3}\right) .\left(10 m s ^{-2}\right)(1000 m ) \times\left(0.04 m ^{2}\right)$
$=4.12 \times 10^{5} N =4 \times 10^{5} N$