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 at the center of the disc. If the disc together with spring mass system, rotates about it’s axis with an angular velocity $ω, (k > > m ω^{2})$ the relative change in the length of the spring is best given by the option

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{m ω^{2}}{k}$

$\sqrt{\frac{2}{3}} (\frac{m ω^{2}}{k})$

$\frac{2 m ω^{2}}{k}$

$\frac{m ω^{2}}{3 k}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Spring force=centrifugal force

$k x = m (l + x) ω^{2}$
$\Rightarrow k x - m x ω^{2} = m l ω^{2} \Rightarrow x = \frac{m l ω^{2}}{k - m ω^{2}} \Rightarrow x = \frac{m l ω^{2}}{k} [k > > m ω^{2}]$
Relative change is $\frac{x}{l} = \frac{m ω^{2}}{k}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET