Question

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

Options

• 300 m

• 400 m

• 600 m

• 500 m

Answer 1

A submarine experiences a pressure of 5.05 × 10⁶ Pa at depth of d₁ in the sea. When it goes further to a depth of d₂, it experiences a pressure of 8.08 × 10⁶ Pa. Then d₁ - d₂ is approximately (density of water = 10³ kg/m³ and acceleration due to gravity = 10 ms^-2) 300 m.

Explanation:

P₁ = P₀ + ρgd₁

P₂ = P₀ + ρgd₂

ΔP = P₂ - P₁ = ρgΔd

`3.03 xx 10^6 = 10^3 xx 10 xx Deltad`

⇒ Δd = 300 m