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)

$P=h \rho g$

$\beta=\frac{p}{\frac{\Delta V}{V}}=\frac{2 \times 10^{3} \times 10^{3} \times 9.8}{1.36 \times 10^{-2}}$

$=1.44 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$

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