JEE Main 2020 (8 Jan, Shift 1) Physics Q2: Circular Motion

Question

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length $l$. 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 :

Accepted Answer

Option 2: $\frac{ml\omega^2}{k-m\omega^2}$

Answer

$\frac{ml\omega^2}{k+m\omega^2}$

Answer

$\frac{ml\omega^2}{k-m\omega^2}$

Answer

$\frac{ml\omega^2}{k+m\omega}$

Answer

$\frac{ml\omega^2}{k-\omega m}$

JEE Main 8 January 2020, Shift 1, Physics Q2

Same idea in other shifts