Question

A submarine experiences a pressure of 5.05 × 10⁶ Pa at 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:
1. 300 m
2. 400 m
3. 600 m
4. 500 m

Answer 1

Correct Answer - Option 1 : 300 m

Concept:

Pressure difference:

ΔP stands for the difference between two measured pressure values. This can be measured either at different times/dates or at different positions in a system.

The equation for the pressure difference is: ΔP = P₁ – P₂

Calculation:

Given,

At depth \({{\rm{d}}_1}\), submarine’s pressure,

P₁ = 5.05 × 10⁶ Pa

At depth \({\rm{\;}}{{\rm{d}}_2}\), submarine’s pressure,

P₂ = 8.08 × 10⁶ Pa

Density of water, ρ = 10³ kg/m³

Acceleration due to gravity, g = 10 ms^-2

P₁ = P₀ + ρgd₁

P₂ = P₀ + ρgd₂

ΔP = P₂ – P₁

ΔP = P₀ + ρgd₂ – P₀ – ρgd₁

ΔP = ρgd₂ – ρgd₁

ΔP = ρg(d₂ – d₁)

ΔP = ρgΔd

P₂ - P₁ = ρgΔd

8.08 × 10^{6 – 5.05} × 10⁶ = 10³ × 10 × Δd

3.03 × 10⁶ = 10³ × 10 × Δd

\({\rm{\Delta d}} = \frac{{3.03 \times {{10}^6}}}{{{{10}^4}}}\) 

Δd = 3.03 × 100 = 303

Δd ≈ 300 m

Therefore, the value of d₂ – d₁ is approximately 300 m.