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  • The density of a metal at a normal pressure is . Its density when it is subjected to an excess

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

Option: 1

\mathrm{\frac{1}{1-p / B}}


Option: 2

\mathrm{1+\frac{p}{B}}


Option: 3

\mathrm{\frac{1}{1+p / B}}


Option: 4

\mathrm{1+B / p}


Answers (1)

best_answer

Let V be the volume of the metal at normal pressure. From \mathrm{B=-\frac{p}{\Delta V / V}}, the decrease in the volume when the metal is subjected to an excess pressure p is given by

                  \mathrm{ |\Delta V|=\frac{p V}{B} }

New volume of metal is \mathrm{V^{\prime}=V-\frac{p V}{B}=V\left(1-\frac{p}{B}\right)}

Now mass of the metal is \mathrm{m=\rho V.} Therefore, its new density is

\mathrm{ \rho^{\prime}=\frac{m}{V^{\prime}}=\frac{\rho V}{V\left(1-\frac{p}{B}\right)}=\frac{\rho}{1-\frac{p}{B}} }
or     \mathrm{ \frac{\rho^{\prime}}{\rho}=\frac{1}{1-p / B} \text {, which is choice (a). } }
 

Posted by

Deependra Verma

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