Practice portal › Rotational Motion › Moment of Inertia
Asked in JEE Main 3rd Sept 2nd Shift 2020 · Radius of gyration and numericals
Take EX as the perpendicular to EG through the vertex E. The mass at E is on the line (distance 0).
The mass at G is at perpendicular distance a: contributes ma². The apex F is at horizontal distance /a2: contributes m(/a2)²=(ma²)/4.
I=ma²+(ma²)/4=5/4 ma²=(25)/(20)ma², so N=25.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer