1. Pressure at depth in a liquid

The total pressure on the submarine at depth $h$ is

$P = P_{\text{atm}} + \rho g h$

Given initially,

$P_1 = 3 \times 10^5\,\text{Pa}$

Also,

$P_{\text{atm}} = 1 \times 10^5\,\text{Pa}$

So the hydrostatic pressure initially is

$\rho g h = P_1 - P_{\text{atm}}$

$\rho g h = 3 \times 10^5 - 1 \times 10^5 = 2 \times 10^5\,\text{Pa}$

2. Pressure when depth is doubled

If depth becomes $2h$, hydrostatic pressure becomes

$\rho g (2h) = 2\rho g h = 2(2 \times 10^5) = 4 \times 10^5\,\text{Pa}$

Hence new total pressure is

$P_2 = P_{\text{atm}} + \rho g (2h)$

$P_2 = 1 \times 10^5 + 4 \times 10^5 = 5 \times 10^5\,\text{Pa}$

3. Increase in pressure

$\Delta P = P_2 - P_1 = 5 \times 10^5 - 3 \times 10^5 = 2 \times 10^5\,\text{Pa}$

4. Percentage increase

$\%\text{ increase} = \frac{\Delta P}{P_1} \times 100$

$= \frac{2 \times 10^5}{3 \times 10^5} \times 100$

$= \frac{2}{3} \times 100 = \frac{200}{3}\%$

5. Option check

• A: $\frac{200}{5}\% = 40\%$
• B: $\frac{200}{3}\%$ ✅
• C: $\frac{3}{200}\%$
• D: $\frac{5}{200}\%$

Therefore, the correct answer is Option B.