JEE Main 2025 · PhysicsMultiple choiceSingle correctMediumApplication

JEE Main 8 April 2025, Shift 2, Physics Q28

Question 28 of 75 in this shift, Physics question 3 of 25, Section A.

A rod of linear mass density '

\lambda

' and length 'L' is bent to form a ring of radius 'R'. Moment of inertia of ring about any of its diameter is.

(1) $\frac{\lambda L^3}{12}$
(2) $\frac{\lambda L^3}{8\pi^2}$ Official answer
(3) $\frac{\lambda L^3}{4\pi^2}$
(4) $\frac{\lambda L^3}{16\pi^2}$

Official answer

Option 2

NTA final key.

Chapter: System of Particles and Rotational Motion
Topic: Moment of Inertia
Idea tested: MOI Change by Bending a Rod

Same topic in other shifts

All Moment of Inertia questions

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2 Apr 2025, Shift 1 · Q32Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is ' $\alpha$ '. Now…MediumSingle correct
2 Apr 2025, Shift 2 · Q28The moment of inertia of a circular ring of mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is :EasySingle correct
7 Apr 2025, Shift 1 · Q46A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.…HardNumerical valueHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.