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 moodulus of metal, $B = 8 \times 10^{10}Pa$ Option: 1 5 Option: 2 0.6 Option: 3 20 Option: 4 1.67

Answers (1)

$\begin{array}{l} \mathrm{B}=\frac{-\frac{\Delta \mathrm{P}}{\Delta \mathrm{V}}}{\mathrm{V}} \\\\ \begin{array}{l} \left|\frac{\Delta \mathrm{V}}{\mathrm{V}}\right|=\frac{\Delta \mathrm{P}}{\mathrm{B}} \\\\ =\frac{4 \times 10^{9}}{8 \times 10^{10}}=\frac{1}{20} \\\\ \frac{\Delta \ell}{\ell}=\frac{1}{3} \times \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{1}{60} \\\\ \text { Percentage change } \\\\ =\frac{\Delta \ell}{\ell} \times 100 \% \\\\ =\frac{100}{60} \%=1.67 \% \end{array} \end{array}$

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