The moment of inertia of a semicircular ring of radius $R$ and mass $M$ about an axis passing through point $A$ (one end of the ring) and perpendicular to the plane of the paper is:

  • A
    $\frac{2}{3} MR^2$
  • B
    $MR^2$
  • C
    $\frac{5}{\pi} MR^2$
  • D
    $2MR^2$

Explore More

Similar Questions

What is the moment of inertia of a ring of mass $M$ and radius $R$ about the axis $PQ$ as shown in the figure?

From a disc of mass $M$ and radius $R$,a circular hole of diameter $R$ is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre is

The moment of inertia of a rod of length $l$ about an axis passing through its centre of mass and perpendicular to the rod is $I$. The moment of inertia of a hexagonal shape formed by six such rods,about an axis passing through its centre of mass and perpendicular to its plane,will be (in $I$)

Difficult
View Solution

$A$ solid sphere has mass $M$ and radius $R$. Its moment of inertia about a parallel axis passing through a point at a distance $R/3$ from its centre is:

Moment of inertia of a disc about a diameter is $I$. Find the moment of inertia of the disc about an axis perpendicular to its plane and passing through its rim. (in $I$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo