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(B) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.The magnitude of the fractional change in volume is given by $\left| \frac

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$1.67$

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(B) The bulk modulus $B$ is defined as $B = -\frac{\Delta P}{\Delta V / V}$.The magnitude of the fractional change in volume is given by $\left| \frac{\Delta V}{V} ight| = \frac{\Delta P}{B}$.Substituting the given values,$\Delta P = 4 	imes 10^9 \; Pa$ and $B = 8 	imes 10^{10} \; Pa$:$\left| \frac{\Delta V}{V} ight| = \frac{4 	imes 10^9}{8 	imes 10^{10}} = \frac{1}{20} = 0.05$.For a cube of side length $\ell$,the volume is $V = \ell^3$. Differentiating,we get $\frac{\Delta V}{V} = 3 \frac{\Delta \ell}{\ell}$.Therefore,the fractional change in length is $\frac{\Delta \ell}{\ell} = \frac{1}{3} \left| \frac{\Delta V}{V} ight| = \frac{1}{3} 	imes \frac{1}{20} = \frac{1}{60}$.The percentage change in length is $\frac{\Delta \ell}{\ell} 	imes 100\% = \frac{1}{60} 	imes 100\% = \frac{10}{6}\% \approx 1.67\%$.

$A$ 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,$B = 8 \times 10^{10} \; Pa$)

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