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:

$\mathbf{β}\mathbf{=}\frac{\mathbf{∆}\mathbf{p}}{{\displaystyle \frac{\mathbf{∆}\mathbf{V}}{{\mathbf{V}}_{\mathbf{0}}}}}$

Where, $\mathbf{β}$is Bulk modulus

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

$\mathbf{∆}\mathbf{V}$is Change of the volume of the material due to the compression

${\mathbf{V}}_{\mathbf{0}}$ is Initial volume of the material

(a)

$∆V=\frac{V_0∆p}{\mathrm{β}}$

As pressure is increases $∆p$is positive

So, Volume is increases slightly.

(b)

As given pressure difference is proportional to depth,

So, $∆p$becomes twice when ship been twice as deep.

$∆V=\frac{V_0∆p}{\mathrm{β}}$

From above equation,

The volume change would be twice as great.

(c)

Given $∆p$is same and volume is equal

${\mathrm{β}}_{Load}=\frac{1}{4}{\mathrm{β}}_{Gold}$

$\mathrm{β}=\frac{∆\mathrm{p}}{{\displaystyle \frac{∆\mathrm{V}}{{\mathrm{V}}_0}}}$

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.