An object is located at 2 km beneath the surface of the water. If the fractional compression (ΔV)/V is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ______[Given: density of water is 1000 kg m^(-3) and g=9.8 m s^(-2).

Correct answer: (D) 1.44×10^9 N m^(-2). P=hρg β=p/(ΔVV)=(2×10^3×10^3×9.8)/(1.36×10^(-2)) =1.44×10^9 N m^(-2)

Q65 / 90 · JEE Main 2021 (17 Mar Shift 2) · Physics

An object is located at $2 km$ beneath the surface of the water. If the fractional compression $\frac{Δ V}{V}$ is $1.36 %$ , the ratio of hydraulic stress to the corresponding hydraulic strain will be ______[Given: density of water is $1000 kg m^{- 3}$ and $g = 9.8 {m s}^{- 2} .$

$P = h ρ g$

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

$= 1.44 \times 10^{9} N m^{- 2}$

Hence, the correct answer is option: 'D'.