Question

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 $10^{3} \,kg\, m ^{-3}, g = 10\, m\,s ^{-2}$ )

• A. $\frac{200}{3} \%$
• B. $\frac{200}{5} \%$
• C. $\frac{5}{200} \%$
• D. $\frac{3}{200} \%$

Answer 1

Answer: (A)

$P_{1}=\rho g d+P_{0}=3 \times 10^{5} P a$
$\therefore \rho g d=2 \times 10^{5} P a$
$P_{2}=2 \rho g d+P_{0}$
$=4 \times 10^{5}+10^{5}=5 \times 10^{5} Pa$
$\%$ increase $=\frac{P_{2}-P_{1}}{P_{1}} \times 100$
$=\frac{5 \times 10^{5}-3 \times 10^{5}}{3 \times 10^{5}} \times 100=\frac{200}{3} \%$