$A$ pendulum clock shows the correct time at $20^{\circ}C$. During summer days,when the temperature rises to $40^{\circ}C$,what will be the time change in the clock in one day (in $.64$)? (Given: $\alpha = 10^{-5} {^{\circ}C}^{-1}$)

Solids expand on heating because ................

$A$ pendulum clock loses $12 \ s$ a day if the temperature is $40^{\circ} C$ and gains $4 \ s$ a day if the temperature is $20^{\circ} C$. The temperature at which the clock will show the correct time is: (in $^{\circ} C$)

$A$ 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 $= \alpha$)

$A$ cuboid $ABCDEFGH$ is anisotropic with $\alpha_x = 1 \times 10^{-5} /^{\circ}C$,$\alpha_y = 2 \times 10^{-5} /^{\circ}C$,$\alpha_z = 3 \times 10^{-5} /^{\circ}C$. The coefficient of superficial expansion of the faces can be:

$A$ bimetallic strip consists of metals $X$ and $Y$. It is mounted rigidly at the base as shown. The metal $X$ has a higher coefficient of linear expansion compared to that for metal $Y$. When the bimetallic strip is placed in a cold bath: