A uniform solid cylinder of length $L$ and radius $R$ has moment of inertia about its axis equal to $I_1$. A small co-centric cylinder of length $L / 2$ and radius $R / 3$ carved from this cylinder has moment of inertia about its axis equals to $I_2$. The ratio $I_1 / I_2$ is $\_\_\_\_$ .

Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is $\_\_\_\_$ .

Two masses $m$ and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius $r=0.1 \mathrm{~m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is

$\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. (Assume string does not slip and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

A circular disc has radius $R_1$ and thickness $T_1$. Another circular disc made of the same material has radius $R_2$ and thickness $T_2$. If the moment of inertia of both discs are same and $\frac{R_1}{R_2}=2$ then $\frac{T_1}{T_2}=\frac{1}{\alpha}$. The value of $\alpha$ is $\_\_\_\_$ .