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 $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )
Correct Option: 1
(1)
$\mathrm{P}_{1}=\rho \mathrm{gd}+\mathrm{P}_{0}=3 \times 10^{5} \mathrm{~Pa}$
$\therefore \rho \mathrm{gd}=2 \times 10^{5} \mathrm{~Pa}$
$\mathrm{P}_{2}=2 \rho \mathrm{gd}+\mathrm{P}_{0}$
$=4 \times 10^{5}+10^{5}=5 \times 10^{5} \mathrm{~Pa}$
\\%increase $=\frac{\mathrm{P}_{2}-\mathrm{P}_{1}}{\mathrm{P}_{1}} \times 100$
$=\frac{5 \times 10^{5}-3 \times 10^{5}}{3 \times 10^{5}} \times 100=\frac{200}{3} \%$
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