The density of a metal at normal pressure P i | vedclass

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Question

(D) The bulk modulus $K$ is defined as $K = -V \frac{dp}{dV}$.For a small change in pressure $p$,we have $K = -V \frac{p}{\Delta V}$,which gives $\Del

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$\frac{1}{1-\frac{P}{K}}$

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(D) The bulk modulus $K$ is defined as $K = -V \frac{dp}{dV}$.For a small change in pressure $p$,we have $K = -V \frac{p}{\Delta V}$,which gives $\Delta V = -\frac{pV}{K}$.The new volume $V^{\prime}$ is $V^{\prime} = V + \Delta V = V - \frac{pV}{K} = V(1 - \frac{p}{K})$.Since density $\varrho = \frac{m}{V}$,the new density is $\varrho^{\prime} = \frac{m}{V^{\prime}} = \frac{m}{V(1 - \frac{p}{K})}$.Substituting $\varrho = \frac{m}{V}$,we get $\varrho^{\prime} = \frac{\varrho}{1 - \frac{p}{K}}$.Therefore,the ratio $\frac{\varrho^{\prime}}{\varrho} = \frac{1}{1 - \frac{p}{K}}$.

The density of a metal at normal pressure $P$ is $\varrho$. When it is subjected to an excess pressure $p$,the density becomes $\varrho^{\prime}$. If $K$ is the bulk modulus of the metal,then the ratio $\frac{\varrho^{\prime}}{\varrho}$ is

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