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Properties of Matter question

2021 · 31 Aug · Shift 1 · Q64
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Properties of Matter question

2021 · 31 Aug · Shift 1 · Q64

JEE MainPhysicsProperties of MatterNumerical+4 / −1
When a rubber ball is taken to a depth of ‾\underline{\hspace{2cm}}​ m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = 9.8 ×\times× 108 Nm −-− 2, Density of sea water = 103 kgm −-− 3, g = 9.8 m/s2)
Numerical answer
View written solutionFree

Correct answer: 500

  1. Use the definition of bulk modulus

The bulk modulus BBB is

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

Taking magnitude,

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

Given:

  • B=9.8×108 N m−2B = 9.8 \times 10^8\ \text{N m}^{-2}B=9.8×108 N m−2
  • Volume decrease =0.5%=0.5100=0.005=0.5\% = \frac{0.5}{100} = 0.005=0.5%=1000.5​=0.005

So,

ΔP=9.8×108×0.005\Delta P = 9.8 \times 10^8 \times 0.005ΔP=9.8×108×0.005 ΔP=4.9×106 Pa\Delta P = 4.9 \times 10^6\ \text{Pa}ΔP=4.9×106 Pa
  1. Relate pressure increase to depth in sea water

At depth hhh, the increase in pressure 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=9.8 m s−2g = 9.8\ \text{m s}^{-2}g=9.8 m s−2

Thus,

4.9×106=103×9.8×h4.9 \times 10^6 = 10^3 \times 9.8 \times h4.9×106=103×9.8×h h=4.9×1069.8×103h = \frac{4.9 \times 10^6}{9.8 \times 10^3}h=9.8×1034.9×106​ h=500 mh = 500\ \text{m}h=500 m
  1. Final answer

The required depth is

500\boxed{500}500​
  1. Comparison with stored answer

Stored correct answer = 500500500

Our derived answer matches the stored answer.

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