JEE Main 2023 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 8 April 2023, Shift 1, Physics Q53

Question 53 of 90 in this shift, Physics question 23 of 30, Section B.

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is

\frac{1}{x}\mathrm{MR^2}

, where R is the radius and

M

is the mass of the semicircular ring. The value of

x

will be ________.

Official answer

NTA final key.

Chapter: System of Particles and Rotational Motion
Topic: Moment of Inertia
Idea tested: MOI of Sectors and Arcs

Same topic in other shifts

All Moment of Inertia questions

2 Apr 2026, Shift 2 · Q50Moment of inertia about an axis $AB$ for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass of 10 kg and radius…MediumNumerical valueHas a figure
4 Apr 2026, Shift 1 · Q30A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The smaller part having mass $M/8$ is converted into a sphere…MediumSingle correct
6 Apr 2026, Shift 1 · Q31A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular…MediumSingle correct
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 108 of the 174 JEE Main shifts held from 2017 to 2026.