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 density of water is 10³ kg m⁻³, g = 10 ms⁻²)

Solution

Physics sir · 3 steps

Only the water part of the pressure doubles with depth; the atmosphere on top stays the same.

1. Split the pressure. Total pressure at depth h is P = P₀ + ρ g h. With P = 3 × 10⁵ Pa and P₀ = 1 × 10⁵ Pa, the water's share is ρ g h = 2 × 10⁵ Pa.
2. Double the depth. Doubling h doubles only ρ g h, to 4 × 10⁵ Pa. The new total is 1 × 10⁵ + 4 × 10⁵ = 5 × 10⁵ Pa.

   % increase = (5 × 10⁵ - 3 × 10⁵)/(3 × 10⁵) × 100 = 200/3 %

3. So. The pressure rises by 200/3 % ≈ 66.7%, which is option (b).

The idea to keep. Depth controls only the gauge part ρ g h. Pull out the atmospheric pressure first, scale what is left, then add the atmosphere back.

Answer from the NTA final answer key. Solution by PrepWiser.