Q 11-06-004NEETNEET 2026Top questionMedium
A thin horizontal disc is rotating about a vertical axis passing through its fixed centre $O$. Its angular momentum is $L_A$ and $L_B$ computed about points $A$ and $B$, respectively, with $OB = 2 \times OA$. The value of $\dfrac{L_A}{L_B}$ is :
Answer: (C) $1$
The angular momentum about any point P is
$$\vec{L}_P = \vec{L}_{cm} + \vec{r}_{P\to cm} \times M\vec{v}_{cm}$$
The disc rotates about its fixed centre, so its centre of mass is at rest: $\vec{v}_{cm} = 0$. The second term vanishes, and the angular momentum is the same ($I\omega$) about every point.
$$\frac{L_A}{L_B} = 1$$
Solution by Sreeraj P, M.Sc Physics