Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $a$. The moment of inertia of this system about an axis passing along one side of the triangle is:

The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is $I$. Now,the rod is bent into a ring and its moment of inertia about its diameter is $I_{1}$. Then $\frac{I}{I_{1}}$ is:

The ratio of masses and radii of two circular rings are $1 : 2$ and $2 : 1$ respectively. What is the ratio of their moments of inertia about their central axes?

$A$ wheel of mass $10 \ kg$ has a moment of inertia of $160 \ kg \ m^2$ about its axis. Its radius of gyration is ...... $m$.

$A$ thin metal rod of mass $M$ and length $L$ is cut into four equal parts by cutting it perpendicular to its length. If the moment of inertia of the rod about an axis passing through its centre and perpendicular to its length is $I$,then the moment of inertia of each part about an axis passing through its own centre and perpendicular to its length is:

The moment of inertia of a meter scale of mass $0.6 \ kg$ about an axis perpendicular to the scale and located at the $20 \ cm$ position on the scale in $kg-m^2$ is: (Breadth of the scale is negligible)