Question

The density of a metal at normal pressure is $\rho$, Its density when it is subjected to an excess pressure $P$ is $\rho$. If bulk modulus of metal is $2P$ then $\frac{\rho }{\rho }$ is:

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $3$

Answer 1

Answer: (B)

$\frac{1}{2}$

As we know that

$K=-V\frac{dP}{dV}$

but for constant mass $\frac{dV}{V}=\frac{d\rho }{\rho }$

so, $K=-\rho \frac{dP}{d\rho }$

given that K=2P and when dP=P then above equation becomes,

$2P=-\rho \frac{P}{d\rho }$

$\Rightarrow d\rho =-\frac{\rho }{2}$

$\Rightarrow {\rho }^{\prime }-\rho =-\frac{\rho }{2}$

(where, ${\rho }^{\prime }$=final density and $\rho$= initial density)

$\Rightarrow \frac{{\rho }^{\prime }}{\rho }=\frac{1}{2}$