Question

A cube of metal is subjected to a hydrostatic pressure of $4 \,GPa$. The percentage change in the length of the side of the cube is close to (Given bulk modulus of metal, $\left. B =8 \times 10^{10} Pa \right)$

• A. 0.6
• B. 1.67
• C. 5
• D. 20

Answer 1

The Correct Option is B

\(B =-\frac{\Delta P }{\frac{\Delta V }{ V }}\)

 \(\left|\frac{\Delta V }{ V }\right|=\frac{\Delta P }{ B }\) \(=\frac{4 \times 10^{9}}{8 \times 10^{10}}=\frac{1}{20}\)

 \(\frac{\Delta \ell}{\ell}=\frac{1}{3} \times \frac{\Delta V }{ V }=\frac{1}{60}\) 

Percentage change \(=\frac{\Delta \ell}{\ell} \times 100 \%\) \(=\frac{1}{60}\times100\%=1.67 \%\)

So, The Correct Option is (B) : 1.67