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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 $10^{3} {kgm}^{- 3},$ $g = 10 {ms}^{- 2})$

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$\frac{5}{200} %$

$\frac{200}{5} %$

$\frac{3}{200} %$

$\frac{200}{3} %$

answer is D.

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Detailed Solution

$P_{o} + ρ {gh=3×10}^{5} P a \Rightarrow ρ gh = 2 \times 10^{5} P a, \frac{Δ P}{P} \times 100 = \frac{ρ g h}{(P_{o} + ρ g h)} \times 100 = \frac{200}{3} %$

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