An object is located at 2 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 $1000 \mathrm{kgm}^{-3}$ and $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$.]

(D) $1.44 \times 10^{9} \mathrm{Nm}^{-2}$

(A) $1.44 \times 10^{7} \mathrm{Nm}^{-2}$

(B) $1.96 \times 10^{7} \mathrm{Nm}^{-2}$

(C) $2.26 \times 10^{9} \mathrm{Nm}^{-2}$

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Elasticity & SHM

An object is located at 2 km beneath the surface of the water. If the fractional compression

\frac{Δ V}{V}

1.36%

, the ratio of hydraulic stress to the corresponding hydraulic strain will be
[Given: density of water is

1000 kgm^{- 3}

and

g = 9.8 ms^{- 2}

Answer: (D)

The ratio of hydraulic stress to hydraulic strain is the bulk modulus. The stress is the pressure due to the column of water above the object,

p = ρ g h = 1000 (9.8) (2 \times 1 0^{3}) = 1.96 \times 1 0^{7} N m^{- 2}

and the strain is the fractional compression

1.36% = 1.36 \times 1 0^{- 2}

, so

B = \frac{p}{Δ V / V} = \frac{1.96 \times 1 0 ^{7}}{1.36 \times 1 0 ^{- 2}} = 1.44 \times 1 0^{9} N m^{- 2}

The ratio is

1.44 \times 1 0^{9} N m^{- 2}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option