JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Rotational Mechanics

Take the axis EX through E, perpendicular to EG and lying in the plane of the triangle. The particle at E is on the axis and adds nothing. The particle at G is a distance $a$ from the axis, adding $m{a}^2$. The particle at F, dropped onto EG, has its foot at the midpoint of EG, a distance $\frac{a}{2}$ from E, so its perpendicular distance from EX is $\frac{a}{2}$ and it adds $m{\left(\frac{a}{2}\right)}^2=\frac{m{a}^2}{4}$. The total is $I=m{a}^2+\frac{m{a}^2}{4}=\frac{5m{a}^2}{4}=\frac{25}{20}m{a}^2$, so $N=25$.