Question

The density of a metal at normal pressure \( P \) is \( \rho \). When it is subjected to an excess pressure, the density becomes \( \rho' \). If \( K \) is the bulk modulus of the metal, then the ratio \( \frac{\rho'}{\rho} \) is

• A. \( 1 + \frac{K}{P} \)
• B. \( 1 + \frac{P}{K} \)
• C. \( \frac{1}{1 - \frac{K}{P}} \)
• D. \( \frac{1}{1 - \frac{P}{K}} \)

Hint:

The bulk modulus \( K \) describes how much the volume of a substance changes with pressure. The change in density due to pressure can be calculated using the relation between bulk modulus and pressure.

Answer 1

The Correct Option is D

Step 1: Density and pressure relation.
The density of a material increases when it is subjected to an external pressure. The change in density \( \Delta \rho \) is related to the bulk modulus \( K \) and the excess pressure \( P \) by the equation: \[ \frac{\Delta \rho}{\rho} = \frac{P}{K} \] Thus, the ratio \( \frac{\rho'}{\rho} \) is given by: \[ \frac{\rho'}{\rho} = \frac{1}{1 - \frac{P}{K}} \]
Step 2: Conclusion.
Thus, the correct answer is (D) \( \frac{1}{1 - \frac{P}{K}} \).