question_answer

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 [JEE MAIN Held On 08-01-2020 Morning]

A) \[\frac{ml{{\omega }^{2}}}{k-\omega m}\]

B) \[\frac{ml{{\omega }^{2}}}{k+m{{\omega }^{2}}}\]

C) \[\frac{ml{{\omega }^{2}}}{k+m\omega }\]

D) \[\frac{ml{{\omega }^{2}}}{k-m{{\omega }^{2}}}\]

Correct Answer: D

Solution :

[d]

At elongated position (x),

\[{{F}_{radial}}\text{=}mr{{\omega }^{2}}\]

\[\therefore kx=m\left( \ell +x \right){{\omega }^{2}}\]

\[\therefore x=\frac{m\ell {{\omega }^{2}}}{k-m{{\omega }^{2}}}\]