$A$ particle of mass $m$ is fixed to one end of a light spring having force constant $k$ and unstretched length $\ell$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity-free space. Then the stretch in the spring is

  • A
    $\frac{m \ell \omega^{2}}{k+m \omega^{2}}$
  • B
    $\frac{m \ell \omega^{2}}{k-m \omega^{2}}$
  • C
    $\frac{m \ell \omega^{2}}{k-m \omega}$
  • D
    $\frac{m \ell \omega^{2}}{k+m \omega}$

Explore More

Similar Questions

$A$ particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance $r$. The speed of the particle is

One end of a massless spring of spring constant $k$ and natural length $l_{0}$ is fixed,while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through the fixed end,then the elongation of the spring will be

Three particles of equal mass are tied to a string and rotated in a horizontal plane as shown. What is the ratio of tensions in the three parts of the string?

Difficult
View Solution

$A$ motorcycle driver doubles their velocity when taking a turn. The force exerted outwardly will be

$A$ train is moving towards north. At one place it turns towards north-east. Here,we observe that:

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