# An object is located at 2 km beneath the surface of the water.

Question:

An object is located at $2 \mathrm{~km}$ beneath the surface of the water. If the fractional compression $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ is $1.36 \%$, the ratio of hydraulic stress to the corresponding hydraulic strain will be [Given : density of water is $1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and $\mathrm{g}=9.8 \mathrm{~ms}^{-2} .$ ] (1) $1.96 \backslash$ times $10^{\wedge}\{7\} \backslash$ mathrm $\{\mathrm{Nm}\}^{\wedge}\{-2\} \$$1. (1)$1.96 \backslash$times$10^{\wedge}\{7\} \backslash \operatorname{mathrm}\{\mathrm{Nm}\}^{\wedge}\{-2\} \mathrm{S}$2. (2)$1.44 \times 10^{7} \mathrm{Nm}^{-2}$3. (3)$2.26 \times 10^{9} \mathrm{Nm}^{-2}$4. (4)$1.44 \times 10^{9} \mathrm{Nm}^{-2}$Correct Option: , 4 Solution: (4)$P=h \rho g\beta=\frac{p}{\frac{\Delta V}{V}}=\frac{2 \times 10^{3} \times 10^{3} \times 9.8}{1.36 \times 10^{-2}}=1.44 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\$