The mass per unit length of a rod of length $l$ is given by $\lambda = \frac{M_0 x}{l}$,where $M_0$ is a constant and $x$ is the distance from one end of the rod. The position of the center of mass of the rod is:

  • A
    $\frac{4l}{3}$
  • B
    $\frac{l}{3}$
  • C
    $\frac{2l}{3}$
  • D
    $\frac{5l}{3}$

Explore More

Similar Questions

The figure shows two cylindrical rods whose centers of mass are marked as $A$ and $B$. The line $AB$ divides the region into two parts: one containing point $O$ (region $1$) and the other containing point $O'$ (region $2$). Choose the correct option regarding the center of mass of the combined system.

Obtain an expression for the position vector of the centre of mass of a system of $n$ particles in two dimensions.

Difficult
View Solution

In general form,what are the coordinates of the centre of mass of a rigid body?

Two identical uniform rectangular blocks (with longest side $L$) and a solid sphere of radius $R$ are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure. If the mass of each block is $M$ and of the sphere is $M/2$,then the maximum distance $x$ that can be achieved is

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$,where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ is:

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