The density of a metal at normal pressure $\mathrm{P}$ is $\varrho .$ When it is subjected to an excess…

The density of a metal at normal pressure $\mathrm{P}$ is $\varrho .$ When it is subjected to an excess pressure, the density becomes $\varrho^{\prime}$. If $\mathrm{B}$ is the bulk modulus of the metal, then the ratio $\frac{\varrho^{\prime}}{\varrho}$ is
  1. $1+\frac{B}{P}$
  2. $1+\frac{\mathrm{P}}{B}$
  3. $\frac{1}{1-\frac{B}{P}}$
  4. $\frac{1}{1-\frac{P}{B}}$

Solution

$\mathrm{B}=-\mathrm{V} \frac{\mathrm{dP}}{\mathrm{dV}}$ Thus, $\Delta \mathrm{V}=-\frac{\mathrm{pV}}{\mathrm{B}}$ or $\mathrm{V}^{\prime}-\mathrm{V}=-\frac{\mathrm{pV}}{\mathrm{B}}$ $\mathrm{Or}, \mathrm{V}^{\prime}=\mathrm{V}\left(1-\frac{\mathrm{p}}{\mathrm{B}}\right)$ Now, $\rho^{\prime}=\frac{m}{V^{\prime}}=\frac{m}{V\left(1-\frac{p}{B}\right)}=\frac{\rho}{\left(1-\frac{p}{B}\right)}$

Asked in: MHT CET 2020 (16 Oct Shift 1)

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