When a meter rod made of silver at $0^{\circ} \mathrm{C}$ is heated to $100{ }^{\circ} \mathrm{C}$, its…
When a meter rod made of silver at $0^{\circ} \mathrm{C}$ is heated to $100{ }^{\circ} \mathrm{C}$, its length increased by $0.19 \mathrm{~cm}$. Then, find the coefficient of volume expansion of the silver.
$0.63 \times 10^{-5 \circ} \mathrm{C}^{-1}$
$1.9 \times 10^{-50} \mathrm{C}^{-1}$
$5.7 \times 10^{-5 \circ} \mathrm{C}^{-1}$
$16.1 \times 10^{-5 \circ} \mathrm{C}^{-1}$
Solution
Given that, length of silver rod, $L=1 \mathrm{~m}$
Initial temperature, $T_1=0^{\circ} \mathrm{C}$
Final temperature, $T_2=100^{\circ} \mathrm{C}$
Increased in length, $\Delta L=0.19 \mathrm{~cm}$
Let $\alpha$ and $\gamma$ be the coefficient of linear and volume expansions.
Then, we know that,
$\Delta L=L \alpha \Delta T$
Substituting the values, we get
$\frac{0.19}{100}=1 \times \alpha\left(T_2-T_1\right)$
$\Rightarrow \quad \frac{0.19}{100}=\alpha(100-0)$
$\Rightarrow \quad \alpha=0.19 \times 10^{-4 \circ} \mathrm{C}^{-1}$
We also know that, volume expansion,
$\begin{aligned} \gamma & =3 \alpha=3 \times 0.19 \times 10^{-4} \\ & =5.7 \times 10^{-50} \mathrm{C}^{-1}\end{aligned}$