$A$ clock which keeps correct time at $20\,^{\circ}C$ has a pendulum rod made of brass. How many seconds will it gain or lose per day when the temperature falls to $0\,^{\circ}C$? $(\alpha = 18 \times 10^{-6}/^{\circ}C)$

$A$ pendulum clock keeps correct time at $0\,^{\circ}C$. The thermal coefficient of linear expansion of the material of the pendulum is $\alpha$. If the temperature rises to $t\,^{\circ}C$,then the clock loses per day by (in seconds):

Two different wires having lengths $L_{1}$ and $L_{2}$ and respective temperature coefficients of linear expansion $\alpha_{1}$ and $\alpha_{2}$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is

$A$ uniform metal rod of length $L$ and mass $M$ is rotating about an axis passing through one of the ends and perpendicular to the rod with angular speed $\omega$. If the temperature increases by $t^{\circ} C$, then the change in its angular velocity is proportional to which of the following? (Coefficient of linear expansion of rod $= \alpha$)

Three rods of equal length $l$ are joined to form an equilateral triangle $PQR.$ $O$ is the midpoint of $PQ.$ The distance $OR$ remains the same for a small change in temperature. The coefficient of linear expansion for $PR$ and $RQ$ is the same,i.e.,$\alpha_2,$ but that for $PQ$ is $\alpha_1.$ Then:

In the figure, which strip, brass or steel, has a higher coefficient of linear expansion?