$A$ cylinder has a piston at a temperature of $30^{\circ} C$. There is an all-round clearance of $0.08 \ mm$ between the piston and the cylinder wall. If the internal diameter of the cylinder is $15 \ cm$,what is the temperature at which the piston will fit into the cylinder exactly (in $^{\circ} C$)? $(\alpha_p = 1.6 \times 10^{-5} /^{\circ} C$ and $\alpha_c = 1.2 \times 10^{-5} /^{\circ} C)$

$A$ 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 $\alpha_{\text{brass}} = 1.8 \times 10^{-5} K^{-1}$)

The coefficients of linear expansion of brass and steel are ${\alpha _1}$ and ${\alpha _2}$ respectively. If we take a brass rod of length ${l_1}$ and a steel rod of length ${l_2}$ at $0^{\circ}C$,their difference in length $({l_2} - {l_1})$ will remain the same at any temperature if:

$A$ 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 $R$. Then $R$ is proportional to

The moment of inertia $I$ of a uniform rod about a perpendicular bisector increases to $I+\Delta I$, if the temperature is increased slightly by $\Delta T$. If the coefficient of linear expansion is $\alpha$, then $\frac{\Delta I}{I}$ is (Assume $\frac{\Delta T}{T} \ll 1$)

$A$ rod of length $10 \, m$ at $0 \, ^oC$ has a linear expansion coefficient $\alpha = (2x^2 + 1) \times 10^{-6} \, ^oC^{-1}$,where $x$ is the distance from one end of the rod. The length of the rod at $10 \, ^oC$ is: (in $, m$)