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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Some Mechanical Properties of Matter

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Some Mechanical Properties of Matter

Determine the estimated change in the density of water in the ocean...

Determine the estimated change in the density of water in the ocean at a depth of 400 m400 \text{ m} below the surface. The density of water at the surface is 1030 kg/m31030 \text{ kg/m}^3 and the bulk modulus of water is 2×109 N/m22 \times 10^9 \text{ N/m}^2.

Options

  1. A2 kg/m32 \text{ kg/m}^3
  2. B4 kg/m34 \text{ kg/m}^3
  3. C4×10−3 kg/m34 \times 10^{-3} \text{ kg/m}^3
  4. D2×10−3 kg/m32 \times 10^{-3} \text{ kg/m}^3

Correct answer

A. 2 kg/m32 \text{ kg/m}^3

Step-by-step solution

The change in pressure at a depth hh is given by:

ΔP=ρgh\Delta P = \rho g h

The bulk modulus BB relates the change in pressure to the fractional change in density:

B=ΔPΔρρB = \dfrac{\Delta P}{\dfrac{\Delta \rho}{\rho}}

Rearranging for the change in density Δρ\Delta \rho:

Δρ=ρΔPB=ρ2ghB\Delta \rho = \dfrac{\rho \Delta P}{B} = \dfrac{\rho^2 g h}{B}

Substituting the given values (using g=9.8 m/s2g = 9.8 \text{ m/s}^2):

Δρ=(1030)2×9.8×4002×109\Delta \rho = \dfrac{(1030)^2 \times 9.8 \times 400}{2 \times 10^9}

Δρ=1060900×39202×109\Delta \rho = \dfrac{1060900 \times 3920}{2 \times 10^9}

Δρ=4.158×1092×109\Delta \rho = \dfrac{4.158 \times 10^9}{2 \times 10^9}

Δρ≈2.08 kg/m3\Delta \rho \approx 2.08 \text{ kg/m}^3

Rounding to the nearest integer, the estimated change in density is 2 kg/m32 \text{ kg/m}^3.
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