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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Mechanical Properties of Solids

NEET 2015 · Physics · Mechanical Properties of Solids

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

The approximate depth of an ocean is 2700 m\text{2700 m} . The compressibility of water is 45.4×10−11Pa−145.4 \times {10}^{- 11}Pa^{- 1} and density of water is 103kg/m3{10}^{3} {kg/m}^{3}. What fractional compression of water will be obtained at the bottom of the ocean ?

Options

  1. A0.8×10−20.8 \times {10}^{- 2}
  2. B1.0×10−21.0 \times {10}^{- 2}
  3. C1.2×10−21.2 \times {10}^{- 2}
  4. D1.4×10−21.4 \times {10}^{- 2}

Correct answer

C. 1.2×10−21.2 \times {10}^{- 2}

Step-by-step solution

Pressure at the bottom of the ocean
P=ρgdP = ρgd
=103×10×2700= {10}^{3} \times 10 \times 2700
=27×106Pa= 27 \times {10}^{6}Pa
∴∴ Fractional compression = compressibility ×\times pressure
=45.4×10−11Pa−1×27×106Pa= 45.4 \times {10}^{- 11}P{a}^{- 1} \times 27 \times {10}^{6}Pa
=1.2×10−2= 1.2 \times {10}^{- 2}
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