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 \(10^{3}\, kg \,m ^{-3}, g =10 \,ms ^{-2}\) )

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Correct Answer

(A) \(\frac{200}{3} \%\)

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Detailed explanation

\(P _{1}=\rho gd + P _{0}=3 \times 10^{5} Pa\)

\(\therefore \rho gd =2 \times 10^{5} Pa\)

\(P _{2}=2 \rho gd + P _{0}\)

\(=4 \times 10^{5}+10^{5}=5 \times 10^{5} Pa\)

\(\%\)increase \(=\frac{ P _{2}- P _{1}}{ P _{1}} \times 100\)

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

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