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 $\frac{\Delta T}{T} \ll 1$)

What will happen if a rod is tied with fixed supports rigidly at both ends and its temperature is increased?

$A$ student records the initial length $l$,change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows:

$S$.No.	$l (m)$	$\Delta T (^{\circ}C)$	$\Delta l (m)$
$(1)$	$2$	$10$	$4 \times 10^{-4}$
$(2)$	$1$	$10$	$4 \times 10^{-4}$
$(3)$	$2$	$20$	$2 \times 10^{-4}$
$(4)$	$3$	$10$	$6 \times 10^{-4}$

If the first observation is correct,what can you say about observations $(2)$,$(3)$ and $(4)$?

Two rods,one of aluminium and the other of steel,having initial lengths $L_1$ and $L_2$ are connected together to form a single rod of length $(L_1+L_2)$. The coefficients of linear expansion of aluminium and steel are $\alpha_1$ and $\alpha_2$ respectively. If the length of each rod increases by the same amount,when their temperatures are raised by $t^{\circ}C$,then the ratio $\frac{L_1}{L_1+L_2}$ will be

Coefficient of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$. Lengths of brass and steel rods are $l_1$ and $l_2$ respectively. If $(l_2 - l_1)$ is maintained same at all temperatures,which one of the following relations holds good?

Three uniform thin aluminium rods each of length $2 \,m$ form an equilateral triangle $PQR$ as shown in the figure. The midpoint of the rod $PQ$ is at the origin of the coordinate system. If the temperature of the system of rods increases by $50^{\circ} C$, the increase in $y$-coordinate of the centre of mass of the system of the rods is ............ $mm$. (Coefficient of volume expansion of aluminium $= 12 \sqrt{3} \times 10^{-6} \,K^{-1}$)