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Properties of Matter question

2015 · Q150
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Properties of Matter question

2015 · Q150

NEETPhysicsProperties of MatterMCQ+4 / −1
The approximate depth of an ocean is 2700 m. The compressiblity of water is 45.4 ×\times× 10−-−11 Pa−-−1 and density of water is 103 kg/m3. What fractional compression of water will be obtained at the bottom of the ocean?
  1. A
    1.2 ×\times× 10−-−2
  2. B
    1.4 ×\times× 10−-−2
  3. C
    0.8 ×\times× 10−-−2
  4. D
    1.0 ×\times× 10−-−2
View written solutionFree

Correct answer: A

Compressibility of water, K = 45.4 × 10–11 Pa–1

density of water P = 103 kg/m3

depth of ocean, h = 2700 m

We have to find ΔVV=?{{\Delta V} \over V} = ?VΔV​=?

As we know, compressibility,

K=1B=(ΔV/V)P(P=ρgh)K = {1 \over B} = {{\left( {\Delta V/V} \right)} \over P}\left( {P = \rho gh} \right)K=B1​=P(ΔV/V)​(P=ρgh)

So, (ΔVV)=Kρgh\left( {{{\Delta V} \over V}} \right) = K\rho gh(VΔV​)=Kρgh

= 45.4 × 10–11 × 103 × 10 × 2700 = 1.2258 × 10–2

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