Five masses,each of $2\, kg$,are placed on a horizontal circular disc,which can be rotated about a vertical axis passing through its centre. All the masses are equidistant from the axis at a distance of $10\, cm$. The moment of inertia of the whole system (in $g\cdot cm^2$) is: (Assume the disc is of negligible mass)

$A$ uniform square plate $S$ (side $c$) and a uniform rectangular plate $R$ (sides $b, a$) have identical areas and masses. Show that:
$(i) \frac{I_{xR}}{I_{xS}} < 1$
$(ii) \frac{I_{yR}}{I_{yS}} > 1$
$(iii) \frac{I_{zR}}{I_{zS}} > 1$

Obtain the expression for the moment of inertia and define it. What are the factors on which the moment of inertia depends? Write its unit and dimensional formula.

$A$ solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :

$A$ ring of mass $m$ and radius $r$ is melted and recast into a sphere. The moment of inertia of the sphere will be:

Three particles are connected by massless rods lying along the $y-$ axis. If the system rotates about the $x-$ axis with an angular speed of $2 \, rad/s$,the moment of inertia $(M.I.)$ of the system is ......... $kg \cdot m^2$.