Question

If B is the bulk modulus of a metal and a pressure p is applied uniformly on all sides of the metal, with density D, then the fractional increases in density is given by

• A. $\frac{B}{p}$
• B. $\frac{P}{B}$
• C. $\frac{pD}{B}$
• D. $\frac{BD}{p}$

Answer 1

The correct option is B $\frac{P}{B}$

Since, Volume of metal will decrease

Hence, change in volume will be negative.

i.e.,

$strain=\frac{-\Delta V}{V}$

We know that

Bulk modules $=\frac{stress}{strain}$

$\Rightarrow B=\frac{P}{-\left(\frac{\Delta V}{V}\right)}$

$\Rightarrow \frac{\Delta V}{V}=\frac{P}{B}$.....(1)

Now, we know that density (D):

$D=\frac{m}{V}$

$\Rightarrow \Delta D=-\frac{m}{V^2}\Delta V$

$\Rightarrow \Delta D=-\left(\frac{\Delta V}{V}\right)$

$\Rightarrow \frac{\Delta V}{V}=-\frac{\Delta D}{D}$....(2)

equating (1)&(2) we get fraction increase in density at.

$\frac{\Delta D}{D}=\frac{P}{B}$