91Ó°ÊÓ

Specimen: Solid gold bar

Bulk modulus is given by the ratio of pressure applied to the corresponding relative decrease in the volume of the material.

Mathematically, it is represented as follows:

$\mathit{β}\mathbf{=}\frac{\mathbf{Δ}\mathbf{p}}{\frac{\mathbf{Δ}\mathbf{V}}{{\mathbf{V}}_{\mathbf{0}}}}$

Where, $\mathit{β}$is Bulk modulus

$\mathit{Δ}\mathit{p}$Is change of the pressure or force applied per unit area on the material

$\mathit{Δ}\mathit{V}$is Change of the volume of the material due to the compression

(a)

\(\Delta V = - \frac{{{V_0}\Delta p}}{\beta }\)

As pressure is increases$\mathrm{Δ}p$is positive

So, Volume is increases slightly.

(b)

As given pressure difference is proportional to depth,

So,$\mathrm{Δ}p$becomes twice when ship been twice as deep.

$\mathrm{Δ}V=-\frac{V_0\mathrm{Δ}p}{\mathrm{β}}$

From above equation,

The volume change would be twice as great.

(c)

Given $\mathrm{Δ}p$is same and volume is equal

$$\begin{gathered} {\mathrm{β}}_{Lead}=\frac{1}{4}{\mathrm{β}}_{Gold} \\ \mathrm{β}=\frac{\mathrm{Δ}p}{\frac{\mathrm{Δ}V}{V_0}} \end{gathered}$$

So, so the volume change of the lead ingot would be four times that of the gold. Because the volume change is inversely proportional to the bulk modulus for a given pressure change.