Q. A solid block of silver with density $10.5\ \times 10^{3}\ kg\ m^{- 3}$ is subjected to an external pressure of $10^{7}\ Nm^{- 2}.$ If the bulk modulus of silver is $17\ \times 10^{10}\ Nm^{- 2},$ the change in density of silver ( in kg $m^{- 3})$ is
A) 0.61
B) 1.7
C) 6.1
D) 17 $\times 10^{3}$
Explanation
(a) Decrease in volume, $\mathrm{\Delta}V = \ \frac{\mathrm{\Delta}p \times V}{K}$
Final volume $V^{'} = \ V - \mathrm{\Delta}V = V - \frac{V\mathrm{\Delta}p}{K} = V(1 - \mathrm{\Delta}p/K$)
Or $\frac{m}{\rho^{'}} = \frac{m}{\rho}\ \left( 1 - \frac{\mathrm{\Delta}p}{K} \right)$
Or $\rho^{'} = \ \frac{\rho}{(1 - \frac{\mathrm{\Delta}\rho}{K})\ }$
Or $\rho = \ \frac{10.5\ \times 10^{3}}{(1 - 10^{7}/17 \times 10^{10})}$
= $10500.61\ kg\ m^{- 3}\ $
So $\rho^{'} - \rho = 10500.61 - 10500 = 0.61\ kg\ m^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.