Answer :

Given data,

Pressure at the given depth, p = 80.0 atm

⇒ Pressure = 80× 1.01 × 10^{5} Pa.

Density of water at the surface, ρ_{1} = 1.03 × 10^{3} kgm^{-3}

Let the given depth be h.

Let ρ_{2} be the density of water at the depth h.

Let V_{1} be the volume of water of mass m at the surface.

Let V_{2} be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = V_{1}-V_{2}

⇒ V = m[(1/ρ_{1})-(1/ρ_{2})]

∴ Volumetric strain = ΔV/V_{1}

⇒ Volumteric Strain = m [(1/ρ_{1})-(1/ρ_{2})] × (ρ_{1}/m)

⇒ ΔV/V_{1} = 1-(ρ_{1}/ρ_{2}) ………… (1)

Bulk modulus, B = pV_{1}/ ΔV

ΔV/V_{1} = p/B

Compressibility of water = (1/B) = 45.8× 10^{-11} Pa^{-1}

__Compressibility is the fractional change in volume per unit increase in pressure. For each atmosphere increase in pressure, the volume of water would decrease 46.4 parts per million.__

∴ ΔV/V_{1} = 80× 1.013× 10^{5}× 45.8× 10^{-11} = 3.71 × 10^{-3} ……. (2)

From equations 1 & 2, we get:

1-(ρ_{1}/ρ_{2}) = 3.71 × 10^{-3}

⇒ ρ_{2} = 1.03× 10^{3} / [1-(3.71 × 10^{-3})]

⇒ ρ_{2} = 1.034× 10^{3} kgm^{-3}

Therefore, the density of water at the given depth (h) is 1.034× 10^{3} kgm^{-3}.

Rate this question :

Water is flowing HC Verma - Concepts of Physics Part 1

Water flows throuHC Verma - Concepts of Physics Part 1

A cubical block oPhysics - Exemplar

A vessel filled wPhysics - Exemplar

Iceberg floats inPhysics - Exemplar

Water flows throuHC Verma - Concepts of Physics Part 1

Bernoulli theoremHC Verma - Concepts of Physics Part 1

While watering a HC Verma - Concepts of Physics Part 1

Water is slowly cHC Verma - Concepts of Physics Part 1

Suppose the tube HC Verma - Concepts of Physics Part 1