Physics · 2020
JEE Main · 9 January 2020, Shift 2 · Q6
A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed…
A spring mass system (mass m, spring constant k and natural length $\displaystyle l$ ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity $\displaystyle \omega,\left(\mathrm{k} \gg \mathrm{m} \omega^2\right)$ the relative change in the length of the spring is best given by the option :
Official answer
(1)
$\displaystyle \frac{m \omega^2}{k}$
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JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.