Question

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Answer 1

Given P = `1.1 xx 10^8` Pa , V = 0.32 `m^3` , `K = 1.6 xx 10^11` `Nm^(-2)`

Bulk modulus for steel = `1.6 xx 10^11 Nm^(-2)`

Using relation, `K = P/((triangle V)/V) = (PV)/(triangle V)`

or `triangle V = (PV)/K`

`=> triangle V = (1.1xx10^8xx0.32)/(1.6xx10^11) m^3`

`= 2.2 xx 10^(-4) m^3`

Answer 2

Water pressure at the bottom, p = 1.1 × 10⁸ Pa

Initial volume of the steel ball, V = 0.32 m³

Bulk modulus of steel, B = 1.6 × 10¹¹ Nm^–2

The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.

Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.

`"Bulk modulus", B= P/((triangle V)/V)`

`triangle V = B/(pV)`

`= (1.1xx10^8xx0.32)/(1.6xx10^11)`

`= 2.2 xx 10^(-4) m^3`

Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10^–4 m³