Mechanical Properties Of Fluids Ques 14
14. 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?
[2015]
(a) $1.0 \times 10^{-2}$
(b) $1.2 \times 10^{-2}$
(c) $1.4 \times 10^{-2}$
(d) $0.8 \times 10^{-2}$
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Answer:
Correct Answer: 14.(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, $(\Delta V / V)=K \rho gh$
$=45.4 \times 10^{-11} \times 10^{3} \times 10 \times 2700=1.2258 \times 10^{-2}$
BETA
