The pressure acting on a submarine is 3 × 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 × 10 5 Pa density of water is 10 3 kg m - 3 , g = 10 ms - 2 )

Step-by-step solution

P 1 = ρ g d + P 0 = 3 × 10 5   Pa ∴   ρ g d = 2 × 10 5   Pa P 2 = 2 ρ g d + P 0 = 4 × 10 5 + 10 5 = 5 × 10 5   Pa % increase = P 2 - P 1 P 1 × 100 = 5 × 10 5 - 3 × 10 5 3 × 10 5 × 100 = 200 3 %