MHT CET · Physics · Mechanical Properties of Solids
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 \(\mathrm{K}\) is the bulk modulus of the metal, then the ratio \(\frac{\varrho^{\prime}}{\varrho}\) is
- A \(1+\frac{K}{P}\)
- B \(1+\frac{\mathrm{P}}{K}\)
- C \(\frac{1}{1-\frac{K}{P}}\)
- D \(\frac{1}{1-\frac{P}{K}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{1-\frac{P}{K}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}=-\mathrm{V} \frac{\mathrm{dP}}{\mathrm{dV}}\)
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, \(\rho^{\prime}=\frac{m}{V^{\prime}}=\frac{m}{V\left(1-\frac{p}{B}\right)}=\frac{\rho}{\left(1-\frac{p}{B}\right)}\)
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, \(\rho^{\prime}=\frac{m}{V^{\prime}}=\frac{m}{V\left(1-\frac{p}{B}\right)}=\frac{\rho}{\left(1-\frac{p}{B}\right)}\)
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