Find the moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane.

  • A
    $MR^2$
  • B
    $\frac{1}{2}MR^2$
  • C
    $\frac{2}{5}MR^2$
  • D
    $\frac{2}{3}MR^2$

Explore More

Similar Questions

From a circular ring of mass $M$ and radius $R$,an arc corresponding to a $90^{\circ}$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is $K$ times $MR^{2}$. Then the value of $K$ is

The radius of gyration of a body depends upon

Two uniform thin identical rods $AB$ and $CD$ each of mass $M$ and length $L$ are joined so as to form a cross as shown. The moment of inertia of the cross about a bisector line $EF$ is (Line $EF$ is in the plane of the cross and bisects the angle between the rods).

Difficult
View Solution

Three point masses $m$ are placed at the corners of an equilateral triangle of side length $\ell$. What is the moment of inertia of the system about an axis passing through one side of the triangle?

Four spheres each of diameter $2a$ and mass $M$ are placed with their centres on the four corners of a square of side $b$. Then the moment of inertia of the system about an axis along one of the sides of the square is

Difficult
View Solution

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