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(B) The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{\Delta V}{V \Delta T}$.
Given, $\frac{\Delta V}{V} = 0.12 \% = \frac{0

Vedclass · Accepted Answer

$2 	imes 10^{-5}$

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(B) The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{\Delta V}{V \Delta T}$.
Given, $\frac{\Delta V}{V} = 0.12 \% = \frac{0.12}{100}$ and $\Delta T = 20 ^oC$.
Substituting the values, we get $\gamma = \frac{0.12}{100 	imes 20} = \frac{0.12}{2000} = 6 	imes 10^{-5} 	ext{ } ^\circ	ext{C}^{-1}$.
We know that the relationship between the coefficient of linear expansion $\alpha$ and the coefficient of volume expansion $\gamma$ is $\gamma = 3\alpha$.
Therefore, $\alpha = \frac{\gamma}{3} = \frac{6 	imes 10^{-5}}{3} = 2 	imes 10^{-5} 	ext{ } ^\circ	ext{C}^{-1}$.

If the volume of a block of metal changes by $0.12 \%$ when it is heated through $20 ^oC$, the coefficient of linear expansion (in $^oC^{-1}$) of the metal is

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