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The pressure acting on a submarine is 3 × 10⁵ 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 × 10⁵ Pa. The density of water is 10³ kg m⁻³, g = 10 ms⁻²)
- 200/3 %
- 200/5 %
- 5/200 %
- 3/200 %
Correct answer: 200/3 %
Solution
At first depth P = Patm + rho*g*h = 3e5 with Patm=1e5, so rho*g*h = 2e5. Doubling depth makes that term 4e5, giving P = 1e5 + 4e5 = 5e5. Percentage increase = (5e5-3e5)/3e5 = 2/3 = 200/3 %.
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