Question

The normal density of a material is $\rho$ and its bulk modulus of elasticity is $K$. The magnitude of increase in density of material when a pressure $P$ is applied uniformly on all sides will be:

• A. $\frac{\rho K}{P}$
• B. $\frac{K}{\rho P}$
• C. $\frac{PK}{\rho }$
• D. $\frac{\rho P}{K}$

Answer 1

The correct option is D

$\frac{\rho P}{K}$

Step 1: Given data

The density of material = $\rho$

Bulk Modulus of elasticity = $K$

Pressure applied on material = $P$

Step 2: Formula used

The following formula is used to compute the density:

$$\begin{gathered} density\left(\rho \right)=\frac{mass\left(M\right)}{volume\left(V\right)}·····\left(1\right) \\ ∆\rho =\frac{m}{V^2}∆V(\mathrm{when}\mathrm{the}\mathrm{mass}\mathrm{is}\mathrm{constant}) \end{gathered}$$

The general formula of the bulk modulus of elasticity is:

$$\begin{gathered} K=\frac{-P}{\left({\displaystyle \raisebox{1ex}{$∆V$}\!\left/ \!\raisebox{-1ex}{$V$}\right.}\right)} \\ \frac{∆V}{V}=\frac{-P}{K}······\left(2\right) \end{gathered}$$

Where $K=$Bulk modulus, $P=$Pressure, $∆V=$Change in volume and $V=$Initial volume

Step 3: Compute increase in density

If the mass remains constant in equation $\left(1\right)$ then it gives,

$$\begin{gathered} \frac{∆\rho }{\rho }=-\frac{∆V}{V} \\ \frac{∆V}{V}=-\frac{∆\rho }{\rho }·····\left(3\right) \end{gathered}$$

Equate equation $\left(3\right)$ and equation $\left(2\right)$,

$$\begin{gathered} -\frac{∆\rho }{\rho }=-\frac{P}{K} \\ ∆\rho =\frac{\rho P}{K} \end{gathered}$$

Thus, the magnitude of increase in density of the material will be $\frac{\rho P}{K}$.

Hence, option D is the correct answer.