Q.In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is $\sqrt{2.5}$. Its radius of gyration about a tangent in its plane (in the same unit) must be (A) $\sqrt{5}$
(B) 2.5
(C) $2\sqrt{2.5}$
(D) $\sqrt{12.5}$

For any rigid body about any axis, the moment of inertia can always be written in the compact form

where $M$ is the body's total mass and $K$, the radius of gyration, is defined as the perpendicular distance from the axis of rotation to an equivalent single point mass — a fictitious point mass equal to the entire mass $M$ of the body — that would, if placed at that one distance, reproduce exactly the same moment of inertia as the actual, spread-out body. As it is a distance, the radius of gyration has SI unit m and dimension $[L]$.

Physical meaning — root-mean-square distance. If a rotating body is modelled as $n$ discrete point masses, all taken (for simplicity) to be of equal individual mass $m$, at perpendicular distances $r_1,r_2,\ldots,r_n$ from the axis, then $I=m(r_1^2+r_2^2+\cdots+r_n^2)=nm\left(\dfrac{r_1^2+r_2^2+\cdots+r_n^2}{n}\right)$; comparing with $I=MK^2$ (where $M=nm$) gives

This is precisely the root-mean-square (rms) distance of the body's mass from the axis of rotation — the radius of gyration is, in a very concrete sense, a single "average" distance that correctly captures how spread out the mass actually is relative to the axis, regardless of the body's specific shape or how irregularly the mass may be distributed. …