Draw the graph of the coefficient of volume expansion $(\alpha_V)$ versus temperature $(T)$ for a solid.

$A$ uniform rod of length $L$ pivoted at one end $P$ is freely rotated in a horizontal plane with an angular velocity $\omega$ about a vertical axis passing through $P$. If the temperature of the system is increased by $\Delta T$, the angular velocity becomes $\frac{\omega}{2}$. If the coefficient of linear expansion of the rod is $\alpha$ (where $\alpha \ll 1$), then $\Delta T$ will be:

$A$ clock which keeps correct time at $20^{\circ} C$, is subjected to $40^{\circ} C$. If the coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /{ }^{\circ} C$, how much time will it gain or lose per day?

Two uniform brass rods $A$ and $B$ of length $l$ and $2l$ and their radii $2r$ and $r$ respectively are heated to the same temperature. The ratio of the increase in the volume of rod $A$ to that of rod $B$ is

Two rods of different metals have coefficients of linear expansion $\alpha_1$ and $\alpha_2$ respectively. Their respective lengths are $L_1$ and $L_2$. At all temperatures,$(L_2 - L_1)$ remains the same. The correct relation is:

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains,say $10 \ cm$. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that the difference between their lengths remains constant. If $\alpha_{\text{iron}} = 1.2 \times 10^{-5} \ K^{-1}$ and $\alpha_{\text{brass}} = 1.8 \times 10^{-5} \ K^{-1}$,what should be the length of each strip?