Two discs have the same thickness. Their radii are $R_1$ and $R_2$ and their densities are $d_1$ and $d_2$. The moment of inertia of the second disc is greater than the first if:

The mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma(r) = A + Br$. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is

$A$ thin uniform wire of mass $m$ and linear mass density $\rho$ is bent in the form of a circular loop. The moment of inertia of the loop about its diameter is

$A$ solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio of $I_1/I_2$ is given by

Two solid spheres ($A$ and $B$) are made of metals of different densities $\rho_A$ and $\rho_B$ respectively. If their masses are equal, the ratio of their moments of inertia $(I_B / I_A)$ about their respective diameter is

$A$ solid sphere of radius $10 \ cm$ is rotating about an axis which is at a distance $15 \ cm$ from its centre. The radius of gyration about this axis is $\sqrt{n} \ cm$. The value of $n$ is