The coefficient of volume expansion of a mate | vedclass

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(B) The density $ho$ is inversely proportional to volume $V$,i.e.,$ho \propto \frac{1}{V}$.For a small change in temperature $\Delta 	heta$,the v

Vedclass · Accepted Answer

$0.02$

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(B) The density $ho$ is inversely proportional to volume $V$,i.e.,$ho \propto \frac{1}{V}$.For a small change in temperature $\Delta 	heta$,the volume changes as $V_2 = V_1(1 + \gamma \Delta 	heta)$,where $\gamma$ is the coefficient of volume expansion.Thus,the new density $ho_2$ is given by $ho_2 = \frac{m}{V_2} = \frac{m}{V_1(1 + \gamma \Delta 	heta)} = ho_1(1 + \gamma \Delta 	heta)^{-1}$.Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$,we get $ho_2 \approx ho_1(1 - \gamma \Delta 	heta)$.The fractional change in density is $\frac{\Delta ho}{ho_1} = \frac{ho_2 - ho_1}{ho_1} = -\gamma \Delta 	heta$.Given $\gamma = 5 	imes 10^{-4} {^{\circ}C}^{-1}$ and $\Delta 	heta = 40^{\circ}C$.Substituting the values: $\frac{\Delta ho}{ho_1} = -(5 	imes 10^{-4} {^{\circ}C}^{-1}) 	imes (40^{\circ}C) = -200 	imes 10^{-4} = -0.02$.The magnitude of the fractional change in density is $0.02$.

The coefficient of volume expansion of a material is $5 \times 10^{-4} {^{\circ}C}^{-1}$. The fractional change in its density for a $40^{\circ}C$ rise in temperature is nearly.

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