We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron $\left( {{\beta _{{\text{v brass }}}} = 6 \times {{10}^{ - 5}}/K} \right.$ and $\left. {{\beta _{{\text{viron }}}} = 3.55 \times {{10}^{ - 5}}/K} \right)$ to create a volume of $100\,cc$ . How do you think you can achieve this.

Let, $\mathrm{V}_{i o}$ and $\mathrm{V}_{b o}$ are volume of iron and brass at $0^{\circ} \mathrm{C}$ respectively.

$\mathrm{V}_{i}, \mathrm{~V}_{b}$ are volume of iron and brass respectively of $\Delta \mathrm{T}^{\circ} \mathrm{C} . \gamma_{i}$ and $\gamma_{b}$ are coefficient of volume expansion of them respectively.

According to question,

$\mathrm{V}_{i o}-\mathrm{V}_{b o}=\mathrm{V}_{i}-\mathrm{V}_{b}=100 \mathrm{cc}$

$\therefore \mathrm{V}_{i}=\mathrm{V}_{i o}\left(1+\gamma_{i} \Delta \mathrm{T}\right)$ and $\mathrm{V}_{b}=\mathrm{V}_{b o}\left(1+\gamma_{b} \Delta \mathrm{T}\right)$

$\therefore \quad$ From equation $(1)$,

$\mathrm{V}_{i}-\mathrm{V}_{b}=\mathrm{V}_{i \mathrm{o}}\left(1+\gamma_{i} \Delta \mathrm{T}\right)-\mathrm{V}_{i b}\left(1+\gamma_{b} \Delta \mathrm{T}\right)$

$\mathrm{V}_{i}-\mathrm{V}_{b}=\mathrm{V}_{i \mathrm{o}}-\mathrm{V}_{i b}+\mathrm{V}_{i \mathrm{o}} \gamma_{i} \Delta \mathrm{T}-\mathrm{V}_{i b} \gamma_{b} \Delta \mathrm{T}$

But $\mathrm{V}_{i}-\mathrm{V}_{b}=\mathrm{V}_{i \mathrm{o}}-\mathrm{V}_{i b}$

$\therefore \mathrm{V}_{i \mathrm{o}} \gamma_{i} \Delta \mathrm{T}=\mathrm{V}_{i b} \gamma_{b} \Delta \mathrm{T}$ $\therefore \frac{\mathrm{V}_{i o}}{\mathrm{~V}_{b o}}=\frac{\gamma_{b}}{\gamma_{i}}$ for midpoint $=\frac{6 \times 10^{-5}}{3.55 \times 10^{-5}}=\frac{6}{3.55}$ $\therefore \mathrm{V}_{i 0}=\frac{6}{3.55} \mathrm{~V}_{b 0}$ $\therefore$ From equation $\mathrm{V}_{i \mathrm{o}}-\mathrm{V}_{i b}=100$, $\frac{6}{3.55} \mathrm{~V}_{b o}-\mathrm{V}_{b o}=100$