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 atmosphere pressure is $1\times {10}^5$ Pa; density of water is 10³kgm^-3 , g = 10ms^-2)

Given,

$\mathrm{P}=3\times {10}^5\mathrm{Pa}, {\mathrm{P}}_{\circ }=1\times {10}^5\mathrm{Pa}$

The following formula can be used to calculate the pressure at depth h:

$\mathrm{P}={\mathrm{P}}_{\circ }+\mathrm{h\rho g\_\_\_\_\_\_}\left(1\right)$

substituting the values,

$$\begin{gathered} 3\times {10}^5=1\times {10}^5+\mathrm{h\rho g} \\ \Rightarrow \mathrm{h\rho g}=2\times {10}^5\mathrm{\_\_\_\_\_\_\_}\left(2\right) \end{gathered}$$

from the question $\mathrm{h\rho g}$ value when the height gets doubled is,

$2\mathrm{h\rho g}=4\times {10}^5\mathrm{\_\_\_\_\_}\left(3\right)$

Thus, the pressure increased as the height increased,

$\mathrm{P}\text{'}={\mathrm{P}}_{\circ }+2\mathrm{h\rho g}=(1\times {10}^5)+(4\times {10}^5)=5\times {10}^5\mathrm{Pa}$

The percentage pressure increase is,

$\mathrm{Percentage} \mathrm{increase}=\frac{\mathrm{Final} \mathrm{pressure}(\mathrm{P}\text{'})-\mathrm{initial} \mathrm{pressure}\left(\mathrm{P}\right)}{\mathrm{initial} \mathrm{pressure}}\times 100$

$\mathrm{Percentage} \mathrm{increase}=\frac{(5\times {10}^5)-(3\times {10}^5)}{(3\times {10}^5)}\times 100$.

$\mathrm{Percentage} \mathrm{increase}=\frac{200}{3}\%$

Hence the correct answer is $\frac{200}{3}\%.$