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Properties of Matter question

2024 · 31 Jan · Shift 1 · Q84
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Properties of Matter question

2024 · 31 Jan · Shift 1 · Q84

JEE MainPhysicsProperties of MatterNumerical+4 / −1
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02%0.02 \%0.02% is ‾\underline{\hspace{2cm}}​mmm. (Take density of sea water =103kgm−3=10^3 \mathrm{kgm}^{-3}=103kgm−3, Bulk modulus of rubber =9×108 Nm−2=9 \times 10^8 \mathrm{~Nm}^{-2}=9×108 Nm−2, and g=10 ms−2g=10 \mathrm{~ms}^{-2}g=10 ms−2)
Numerical answer
View written solutionFree

Correct answer: 18

  1. Use bulk modulus relation

For a small change in volume,

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

Taking magnitudes,

ΔP=B(ΔVV)\Delta P = B\left(\frac{\Delta V}{V}\right)ΔP=B(VΔV​)

Given:

  • Bulk modulus of rubber: B=9×108 N m−2B = 9\times 10^8\ \text{N m}^{-2}B=9×108 N m−2
  • Decrease in volume = 0.02%0.02\%0.02%

Convert percentage to fraction:

0.02%=0.02100=2×10−40.02\% = \frac{0.02}{100} = 2\times 10^{-4}0.02%=1000.02​=2×10−4

So,

ΔVV=2×10−4\frac{\Delta V}{V} = 2\times 10^{-4}VΔV​=2×10−4

  1. Find required pressure increase

ΔP=9×108×2×10−4\Delta P = 9\times 10^8 \times 2\times 10^{-4}ΔP=9×108×2×10−4

ΔP=18×104=1.8×105 Pa\Delta P = 18\times 10^4 = 1.8\times 10^5\ \text{Pa}ΔP=18×104=1.8×105 Pa

  1. Relate pressure increase to depth in sea water

Hydrostatic pressure at depth hhh is

ΔP=ρgh\Delta P = \rho g hΔP=ρgh

Given:

  • ρ=103 kg m−3\rho = 10^3\ \text{kg m}^{-3}ρ=103 kg m−3
  • g=10 m s−2g=10\ \text{m s}^{-2}g=10 m s−2

Thus,

1.8×105=103×10×h1.8\times 10^5 = 10^3\times 10 \times h1.8×105=103×10×h

1.8×105=104h1.8\times 10^5 = 10^4 h1.8×105=104h

h=1.8×105104=18 mh = \frac{1.8\times 10^5}{10^4} = 18\ \text{m}h=1041.8×105​=18 m

  1. Final answer

The required depth is

18\boxed{18}18​

This matches the stored correct answer.

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