 # The pressure acting on a submarine is 3 × 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : Question:

The pressure acting on a submarine is $3 \times 10^{5}$ $\mathrm{Pa}$ 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 $1 \times 10^{5} \mathrm{~Pa}$ density of water is $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )

1. $\frac{200}{3} \%$

2. $\frac{200}{5} \%$

3. $\frac{5}{200} \%$

4. $\frac{3}{200} \%$

Correct Option: 1

Solution:

$\mathrm{P}_{1}=\rho \mathrm{gd}+\mathrm{P}_{0}=3 \times 10^{5} \mathrm{~Pa}$

$\therefore \rho g d=2 \times 10^{5} \mathrm{~Pa}$

$P_{2}=2 \rho g d+P_{0}$

$=4 \times 10^{5}+10^{5}=5 \times 10^{5} \mathrm{~Pa}$

$\%$ increase $=\frac{P_{2}-P_{1}}{P_{1}} \times 100$

$=\frac{5 \times 10^{5}-3 \times 10^{5}}{3 \times 10^{5}} \times 100=\frac{200}{3} \%$