Q96 / 125 · BITSAT 2016 · Physics
The approximate depth of an ocean is $2700 \mathrm{~m}$. The compressibility of water is $45.4 \times 10^{-11} \mathrm{~Pa}^{-1}$ and density of water is $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. What fractional compression of water will be obtained at the bottom of the ocean?
- A$1.0 \times 10^{-2}$
- Correct answerB$1.2 \times 10^{-2}$
- C$1.4 \times 10^{-2}$
- D$0.8 \times 10^{-2}$
Solution
Compressibility of water, $\mathrm{K}=45.4 \times 10^{-11} \mathrm{~Pa}^{-1}$
density of water $\mathrm{P}=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
depth of ocean, $\mathrm{h}=2700 \mathrm{~m}$ We have to find $\frac{\Delta \mathrm{V}}{\mathrm{V}}=?$
As we know, compressibility, $\mathrm{K}=\frac{1}{\mathrm{~B}}=\frac{(\Delta \mathrm{V} / \mathrm{V})}{\mathrm{P}}(\mathrm{P}=\mathrm{Pgh})$
So, $(\Delta \mathrm{V} / \mathrm{V})=\mathrm{KPgh}$
$=45.4 \times 10^{-11} \times 10^{3} \times 10 \times 2700$
$=1.2258 \times 10^{-2}$
Hence, the correct answer is option: 'B'.