Medium
From a circular disc of radius $R$ and mass $9M$,a small disc of mass $M$ and radius $R/3$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
- A$M R^2$
- B$\frac{40}{9} M R^2$
- C$4 M R^2$
- D$\frac{4}{9} M R^2$
Explore More
Similar Questions
Two discs have densities $d_1 = 7.2 \ g/cm^3$ and $d_2 = 8.9 \ g/cm^3$. Both discs have the same mass and the same thickness. Find the ratio of their moments of inertia.
Medium
View Solution$A$ disc $X$ of radius $R$ and thickness $t$ is made of an iron plate. Another disc $Y$ of radius $4R$ and thickness $t/4$ is also made of the same iron plate. What is the relationship between their moments of inertia $I_x$ and $I_y$?
Difficult
View Solution$Assertion$ : Radius of gyration of a body is a constant quantity.
$Reason$ : The radius of gyration of a body about an axis of rotation may be defined as the root mean square distance of the particles from the axis of rotation.
$Reason$ : The radius of gyration of a body about an axis of rotation may be defined as the root mean square distance of the particles from the axis of rotation.
If the length of a thin uniform rod is $L$ and the radius of gyration of the rod about an axis perpendicular to its length and passing through one end is $K$, then $K: L=$
The moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to the plane of the ring is $\frac{1}{x} MR^2$,where $R$ is the radius and $M$ is the mass of the semicircular ring. The value of $x$ will be $...........$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo