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Properties of Matter question

2024 · 27 Jan · Shift 1 · Q90
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Properties of Matter question

2024 · 27 Jan · Shift 1 · Q90

JEE MainPhysicsProperties of MatterNumerical+4 / −1
If average depth of an ocean is 4000 m4000 \mathrm{~m}4000 m and the bulk modulus of water is 2×109 Nm−22 \times 10^9 \mathrm{~Nm}^{-2}2×109 Nm−2, then fractional compression ΔVV\frac{\Delta V}{V}VΔV​ of water at the bottom of ocean is α×10−2\alpha \times 10^{-2}α×10−2. The value of α\alphaα is ‾\underline{\hspace{2cm}}​ (Given, g=10 ms−2,ρ=1000 kg m−3\mathrm{g}=10 \mathrm{~ms}^{-2}, \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}g=10 ms−2,ρ=1000 kg m−3)
Numerical answer
View written solutionFree

Correct answer: 2

  1. Pressure at the bottom of the ocean

The excess pressure due to a water column of depth hhh is

P=ρghP = \rho g hP=ρgh

Given:

  • ρ=1000 kg m−3\rho = 1000\,\text{kg m}^{-3}ρ=1000kg m−3
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2
  • h=4000 mh = 4000\,\text{m}h=4000m

So,

P=1000×10×4000=4×107 PaP = 1000 \times 10 \times 4000 = 4 \times 10^7\,\text{Pa}P=1000×10×4000=4×107Pa
  1. Use bulk modulus relation

Bulk modulus is defined as

B=−ΔPΔV/VB = -\frac{\Delta P}{\Delta V/V}B=−ΔV/VΔP​

Taking magnitude for fractional compression,

ΔVV=ΔPB\frac{\Delta V}{V} = \frac{\Delta P}{B}VΔV​=BΔP​

Given:

  • B=2×109 PaB = 2 \times 10^9\,\text{Pa}B=2×109Pa
  • ΔP=4×107 Pa\Delta P = 4 \times 10^7\,\text{Pa}ΔP=4×107Pa

Thus,

ΔVV=4×1072×109\frac{\Delta V}{V} = \frac{4 \times 10^7}{2 \times 10^9}VΔV​=2×1094×107​ ΔVV=2×10−2\frac{\Delta V}{V} = 2 \times 10^{-2}VΔV​=2×10−2
  1. Compare with the given form

It is given that

ΔVV=α×10−2\frac{\Delta V}{V} = \alpha \times 10^{-2}VΔV​=α×10−2

So,

α=2\alpha = 2α=2

Final Answer: 222

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