The coefficient of volume expansion of a material is $5 \times 10^{-4} \mathrm{o}^{-1}$. The fractional…
The coefficient of volume expansion of a material is $5 \times 10^{-4} \mathrm{o}^{-1}$. The fractional change in its density for a $40^{\circ} \mathrm{C}$ rise in temperature is nearly.
$0.01$
$0.02$
$0.03$
$0.04$
Solution
As density, $\rho \propto \frac{1}{V}$
Where, $V=$ volume
We have, $\frac{\rho_2}{\rho_1}=\frac{V_1}{V_2}=\frac{V_1}{V_1(1+r \Delta \theta)}$
$\begin{aligned} & r=\text { coefficient of cubical expansion } \\ & \Delta \theta=\text { rise of temperature } \\ & \text { Since, } \rho_2=\rho_1(1-r \Delta \theta) \\ & \text { Here, } r=5 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\end{aligned}$
$\Delta \theta=40^{\circ} \mathrm{C}$
So, in given case
$\frac{\rho_2}{\rho_1}=1-5 \times 10^{-4} \times 40$
$=1-2 \times 10^{-2}=0.08$
Hence, fractional change in density $=\frac{\Delta \rho}{\rho_1}$
$\Rightarrow \frac{\Delta \rho}{\rho_1}=\frac{\rho_2-\rho_1}{\rho_1}=\frac{\rho_2}{\rho_1}-1$
$\Rightarrow$ Fractional change is
$\frac{\Delta \rho}{\rho_1}=0.08-1=-0.02$
Here, negative sign shows decrease in density.