The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is $2400 \,g \,cm^2$. The length of the $400 \,g$ rod is nearly: (in $\,cm$)

Five masses,each of $2\, kg$,are placed on a horizontal circular disc that can rotate about a vertical axis passing through its center. All masses are equidistant from the axis at a distance of $10\, cm$. Calculate the moment of inertia of the whole system in $gm-cm^2$. (Assume the disc has negligible mass.)

The moment of inertia of a thin circular lamina of mass $1\,kg$ and diameter $0.2\,m$ rotating about one of its diameters is

The figure shows the variation of the moment of inertia of a uniform rod about an axis passing through its centre and inclined at an angle $\theta$ to the length of the rod. The moment of inertia of the rod about an axis passing through one of its ends and making an angle $\theta = \frac{\pi}{3}$ with the rod will be ........ $kg-m^2$.

What is the moment of inertia of a cylinder of diameter $D$ and length $L$ about an axis passing through its center of gravity and perpendicular to its length?

The radius of gyration of a circular disc of radius $R$ and mass $m$ rotating about its diameter as an axis is: