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Question:

A submarine experiences a pressure of 5.05 × 10⁶ Pa at a depth of d₁ in a sea. When it goes further to a depth of d₂, it experiences a pressure of 8.08 × 10⁶ Pa. Then d₂ − d₁ is approximately (density of water = 10³ kg/m³ and acceleration due to gravity = 10 m s⁻²)

300m

400m

600m

500m

Solution:

The correct option is C
300m
P₀ + ρgd₁ = P₁
P₀ + ρgd₂ = P₂
ρg(d₂ − d₁) = P₂ − P₁
10³ × 10 (d₂ − d₁) = 3.03 × 10⁶
d₂ − d₁ = 303m ≈ 300m