JEE Main 2024 — Rotational Motion Question with Solution

The formula to calculate the moment of inertia of a system of particles about a point can be written as

$I=\sum _{i=1}^n{m}_i{r}_i^2 ...\left(1\right)$

where, $m_i$ is the mass of the $i^{\mathrm{th}}$ particle situated at a distance $r_i$ from the point.

With reference to the above diagram, it can be written that the net moment of inertia of the system of particles is

$\begin{array}{rcl}I& =& m{a}^2+m{a}^2+m(\sqrt{2}a{)}^2\\ & =& 4m{a}^2\\ & =& 4\times 1 \mathrm{kg}\times {\left(2 \mathrm{m}\right)}^2\\ & =& 16 \mathrm{kg} {\mathrm{m}}^2\end{array}$