The density of a metal at normal pressure is | vedclass

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Question

(A) The Bulk modulus $B$ is defined as $B = -V \frac{dp}{dV}$.For a small change in pressure $p$,the change in volume $\Delta V$ is given by $\Delta V

Vedclass · Accepted Answer

$\frac{1}{1 - \frac{p}{B}}$

Vedclass · Answer

(A) The Bulk modulus $B$ is defined as $B = -V \frac{dp}{dV}$.For a small change in pressure $p$,the change in volume $\Delta V$ is given by $\Delta V = -\frac{pV}{B}$.The new volume $V^{\prime}$ is $V^{\prime} = V + \Delta V = V - \frac{pV}{B} = V(1 - \frac{p}{B})$.Since mass $m$ remains constant,the new density $\rho^{\prime}$ is $\rho^{\prime} = \frac{m}{V^{\prime}} = \frac{m}{V(1 - \frac{p}{B})}$.Substituting $\rho = \frac{m}{V}$,we get $\rho^{\prime} = \frac{\rho}{1 - \frac{p}{B}}$.Therefore,the ratio $\frac{\rho^{\prime}}{\rho} = \frac{1}{1 - \frac{p}{B}}$.

The density of a metal at normal pressure is $\rho$. Its density when it is subjected to an excess pressure $p$ is $\rho^{\prime}$. If $B$ is the Bulk modulus of the metal,the ratio $\frac{\rho^{\prime}}{\rho}$ is:

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