We would like to make a vessel whose volume does not change with temperature. We can use brass and iron $\left( {{\gamma _{{\text{brass}}}} = 6 \times {{10}^{ - 5}}/K} \right.$ and $\left. {{\gamma _{{\text{iron}}}} = 3.55 \times {{10}^{ - 5}}/K} \right)$ to create a volume of $100 \, cc$. How can you achieve this?

(N/A) Let $V_{i,0}$ and $V_{b,0}$ be the volumes of iron and brass at $0^{\circ} C$ respectively.
Let $V_i$ and $V_b$ be the volumes of iron and brass at temperature $\Delta T^{\circ} C$ respectively.
Let $\gamma_i$ and $\gamma_b$ be the coefficients of volume expansion for iron and brass respectively.
The net volume of the vessel is $V_c = V_i - V_b = 100 \, cc$.
For the volume to remain constant with temperature,the change in volume must be zero:
$\Delta V_c = \Delta V_i - \Delta V_b = 0$
$V_{i,0} \gamma_i \Delta T - V_{b,0} \gamma_b \Delta T = 0$
$V_{i,0} \gamma_i = V_{b,0} \gamma_b$
$\frac{V_{i,0}}{V_{b,0}} = \frac{\gamma_b}{\gamma_i} = \frac{6 \times 10^{-5}}{3.55 \times 10^{-5}} = \frac{6}{3.55} \approx 1.69$
Given $V_{i,0} - V_{b,0} = 100 \, cc$,we substitute $V_{i,0} = \frac{6}{3.55} V_{b,0}$:
$\left( \frac{6}{3.55} - 1 \right) V_{b,0} = 100$
$\left( \frac{6 - 3.55}{3.55} \right) V_{b,0} = 100$
$\frac{2.45}{3.55} V_{b,0} = 100$
$V_{b,0} = \frac{355}{2.45} \approx 144.9 \, cc$
$V_{i,0} = 100 + 144.9 = 244.9 \, cc$.