Mechanical Properties of Fluids - Result Question 4
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4. The approximate depth of an ocean is $2700 m$. The compressibility of water is $45.4 \times 10^{-11} Pa^{-1}$ and density of water is $10^{3} kg / m^{3}$. What fractional compression of water will be obtained at the bottom of the ocean?
======= ####4. The approximate depth of an ocean is $2700 m$. The compressibility of water is $45.4 \times 10^{-11} Pa^{-1}$ and density of water is $10^{3} kg / m^{3}$. What fractional compression of water will be obtained at the bottom of the ocean?
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/mechanical-properties-of-fluids/mechanical-properties-of-fluids—result-question-4.md (a) $1.0 \times 10^{-2}$
(b) $1.2 \times 10^{-2}$
(c) $1.4 \times 10^{-2}$
(d) $0.8 \times 10^{-2}$
[2015]
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Answer:
Correct Answer: 4. (b)
Solution:
- (b) Compressibility of water,
$K=45.4 \times 10^{-11} Pa^{-1}$
density of water $P=10^{3} kg / m^{3}$
depth of ocean, $h=2700 m$
We have to find $\frac{\Delta V}{V}=$ ?
As we know, compressibility,
$K=\frac{1}{B}=\frac{(\Delta V / V)}{P}(P=\rho gh)$
So, $(\Delta V / V)=K \rho gh$
$=45.4 \times 10^{-11} \times 10^{3} \times 10 \times 2700=1.2258 \times 10^{-2}$
