Q. The bulk modulus of a metal is $8 \times 10^{9}\ Nm^{- 2}\ and$ its density is 11 g$cm^{- 2}.$ The density of this metal under a pressure of 20,000 N $cm^{- 2}$ will be (in g$cm^{- 3})$
A) $\frac{440}{39}$
B) $\frac{431}{39}$
C) $\frac{451}{39}$
D) $\frac{40}{39}$
Explanation
(a) Here, $p = 20,000\ Ncm^{- 2} = 2 \times 10^{8}\ Nm^{- 2}$
$K = \frac{pV}{\mathrm{\Delta}V}$
$\mathrm{\Delta}V = \frac{pV}{k}\ $
= $\frac{2 \times 10^{8} \times V}{8 \times 10^{9}} = \frac{V}{40}$
New volume of the metal,
$V' = V - \mathrm{\Delta}V = V - \frac{V}{40} = \frac{39V}{40}\ $
New mass of the metal
= $V^{'} \times \rho = \frac{39V}{40}\rho' = \ V \times 11$
Or $\rho' = \ \frac{440}{39}$ g${cm}^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.