The depth of an ocean trench is $4000 \mathrm{~m}$. The compressibility of water is $5 imes 10^{-1

Question

The depth of an ocean trench is $4000 \mathrm{~m}$. The compressibility of water is $5 	imes 10^{-10} \mathrm{~Pa}^{-1}$ and the density of water is $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. Assuming $g = 10 \mathrm{~m/s}^2$, the fractional compression of water at the bottom of the trench is

Professor PC Thomas Classes · Answer

(a) : The excess pressure $\Delta P$ at a depth $h = 4000 \mathrm{~m}$ is $\Delta P = ho g h = 10^3 	imes 10 	imes 4000 = 4 	imes 10^7 \mathrm{~Pa}$. Compressibility $K = \frac{1}{B} = \frac{\Delta V / V}{\Delta P}$. The fractional compression is $\frac{\Delta V}{V} = K \cdot \Delta P$. Substituting the values, we get $\frac{\Delta V}{V} = (5 	imes 10^{-10} \mathrm{~Pa}^{-1}) 	imes (4 	imes 10^7 \mathrm{~Pa}) = 20 	imes 10^{-3} = 2.0 	imes 10^{-2}$.