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

If a cylinder of diameter $1.0 \, cm$ at $30^{\circ}C$ is to be slid into a hole of diameter $0.9997 \, cm$ in a steel plate at the same temperature,then the minimum required rise in the temperature of the plate is ........ $^{\circ}C$ (Coefficient of linear expansion of steel $\alpha = 12 \times 10^{-6} \, ^{\circ}C^{-1}$).

The length of a steel rod is $5 \ cm$ longer than a copper rod at all temperatures. What are the lengths of the steel and copper rods? (The coefficients of linear expansion for steel and copper are $1.1 \times 10^{-5} /{ }^{\circ} C$ and $1.7 \times 10^{-5} /{ }^{\circ} C$,respectively.)

The temperature of a metallic sphere of radius $R$ is increased by a small amount $\Delta T$. If the linear coefficient of thermal expansion of the metal is $\alpha$, the approximate increase in the volume of the sphere is :

Two rods,one of aluminum and another of steel,have initial lengths $l_1$ and $l_2$ respectively. They are joined together to form a single rod of length $(l_1 + l_2)$. If the coefficients of linear expansion for aluminum and steel are $\alpha_a$ and $\alpha_s$ respectively,and the increase in length for both rods is the same when the temperature of both is increased by $t \ ^\circ C$,then $\frac{l_1}{l_1 + l_2} = \dots$

Write the relation between the coefficient of linear expansion $(\alpha_l)$,coefficient of area expansion $(\alpha_A)$,and coefficient of volume expansion $(\alpha_V)$.