Question

The approximate depth of an ocean is $2700m$. The compressibility of water is $45.4\times {10}^{-11}P{a}^{-1}$ and density of water is ${10}^{3}kg/m^3$. What fractional compression of water will be obtained at the bottom of the ocean?

• A. $1.0\times {10}^{-2}$
• B. $1.2\times {10}^{-2}$
• C. $1.4\times {10}^{-2}$
• D. $0.8\times {10}^{-2}$

Answer 1

The correct option is B $1.2\times {10}^{-2}$

Compressibility of water,
$K=45.4\times {10}^{-11}P{a}^{-1}$
density of water $P={10}^{3}kg/m^3$
depth of ocean, $h=2700m$
We have to find $\frac{△V}{V}=?$
As we know, compressibility,
$K=\frac{1}{B}=\frac{\left(△V/V\right)}{P}(P=Pgh)$
So, $\left(△V/V\right)=KPgh$
$=45.4\times {10}^{-11}\times {10}^3\times 10\times 2700$
$=1.2258\times {10}^{-2}$.