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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Mechanical Properties of Fluids

BITSAT 2016 · Physics · Mechanical Properties of Fluids

The approximate depth of an ocean is 2700 m . The compressibility...

The approximate depth of an ocean is 2700m2700 \mathrm{ m}. The compressibility of water is 45.4×10−11Pa−145.4 \times 10^{-11} \mathrm{ Pa}^{-1} and density of water is 103kg/\mathrmm310^{3} \mathrm{ kg} / \mathrmm^{3}. What fractional compression of water will be obtained at the bottom of the ocean?

Options

  1. A1.0×10−21.0 \times 10^{-2}
  2. B1.2×10−21.2 \times 10^{-2}
  3. C1.4×10−21.4 \times 10^{-2}
  4. D0.8×10−20.8 \times 10^{-2}

Correct answer

B. 1.2×10−21.2 \times 10^{-2}

Step-by-step solution

Compressibility of water, K=45.4×10−11Pa−1\mathrm{K}=45.4 \times 10^{-11} \mathrm{ Pa}^{-1} density of water P=103kg/\mathrmm3\mathrm{P}=10^{3} \mathrm{ kg} / \mathrmm^{3} depth of ocean, h=2700m\mathrm{h}=2700 \mathrm{ m} We have to find ΔVV=?\frac{\Delta \mathrm{V}}{\mathrm{V}}=? As we know, compressibility, K=1B=(ΔV/V)P(\mathrm{K}=\frac{1}{\mathrm{ B}}=\frac{(\Delta \mathrm{V} / \mathrm{V})}{\mathrm{P}}(\mathrm{P}=\mathrm{Pgh})) So, (ΔV/V)=KPgh=45.4×10−11×103×10×2700=1.2258×10−2(\Delta \mathrm{V} / \mathrm{V})=\mathrm{KPgh}=45.4 \times 10^{-11} \times 10^{3} \times 10 \times 2700=1.2258 \times 10^{-2}
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