Question

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. $\frac{40}{39}$

Answer 1

Answer: (A)

Here, $p=20,000 \,Ncm ^{-2}=2 \times 10^{8} Nm ^{-2}$
As, $k=\frac{p V}{\Delta V}$
or $\Delta V=\frac{p V}{k}$
$=\frac{2 \times 10^{8} \times V}{8 \times 10^{9}}=\frac{V}{40}$
New volume of the metal,
$V'=V-\Delta V=V-\frac{V}{40}=\frac{39 V}{40}$
New mass of the metal
$=V' \times \rho=\frac{39 V}{40} \rho'=V \times 11 $
or $\rho' =\frac{40 \times 11}{39}=\frac{440}{39} \, g\, cm ^{-3}$