An iron rod of length $10 \ m$ is heated from $0 \ ^{\circ}C$ to $100 \ ^{\circ}C$. If the coefficient of linear expansion of iron is $10 \times 10^{-6} \ ^{\circ}C^{-1}$,then the increase in the length of the rod is ..... $cm$.

$A$ unit scale is to be prepared whose length does not change with temperature and remains $20\,cm$,using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that the difference between their lengths remains constant. If the length of brass is $40\,cm$,what is the length of iron in $cm$?
($\alpha_{\text{iron}} = 1.2 \times 10^{-5} K^{-1}$ and $\alpha_{\text{brass}} = 1.8 \times 10^{-5} K^{-1}$)

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?

Brass and lead rods of length $80 \ cm$ are connected in parallel at $0^{\circ}C$. If they are heated to $100^{\circ}C$,what will be the difference in length between their ends in $mm$? (Given: $\alpha_{brass} = 18 \times 10^{-6} \ ^{\circ}C^{-1}$ and $\alpha_{lead} = 28 \times 10^{-6} \ ^{\circ}C^{-1}$)

Suppose there is a hole in a copper plate. On heating the plate,the diameter of the hole will:

$A$ bimetallic strip is formed out of two identical strips,one of copper and the other of brass. The coefficients of linear expansion of the two metals are $\alpha_C$ and $\alpha_B$. On heating,the temperature of the strip increases by $\Delta T$ and the strip bends to form an arc of radius of curvature $R$. Then $R$ is: