JEE Main 2019 — Physics solution

JEE Main2019Physics

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 ms^–2) :

A
600 m
B
500 m
C
300 m
D
400 m

Solution

$Δ P = P_{2} - P_{1} = ρ g Δ H 3.03 \times 1 0^{6} = 1 0^{3} \times 10 \times Δ H \Rightarrow Δ H ≃ 300 m$

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