The pressure acting on a submarine is $3 \times 10^5 Pa $ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:

(Assume that atmospheric pressure is $1 \times 10^5 Pa$, density of water is $10^3 kgm^{-3}$, $g=10 ms^{-2}$)

(A) $\frac {3}{200} $%
(B) $\frac {5}{200} $%
(C) $\frac {200}{3} $%
(D) $\frac {200}{5} $%

Solution

We have, $P_1 = P_a + h_1 dg$ & $P_2 = P_a + h_2 dg$

% increase in pressure = $\frac {P_2 – P_1 }{P_1} \times 100 = \frac {(h_2 – h_1 ) dg }{P_1 } \times 100$

$\therefore$ % increase = $\frac {(2h – h) \times 10^3 \times 10 }{3 \times 10^5 } \times 100 = \frac {h}{30} \times 100$

Also, $P_1 = 3 \times 10^5 = 10^5 + h \times 10^3 \times 10 $

$\therefore h = 20 m$

Now, % increase in pressure = $ \frac {h}{30} \times 100 = \frac {20}{30} \times 100 = \frac {200}{3}% $

Answer: (C)