A particle of mass m is fixed to one end of light spring of force constant k and unstretched length I as shown in fig. (5). The system is rotated about the other end of
the spring with angular velocity o in gravity free space. The increase in length of the spring will be

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

$\frac{{mlω}^{2}}{k}$

$\frac{{mlω}^{2}}{k - {mω}^{2}}$

$\frac{{mlω}^{2}}{k + {mω}^{2}}$

none of these

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

Let the extension produced in the springs due to rotation be x. For the equilibrium of the body
$kx = m (l + x) ω^{2}$
or $kx = {mlω}^{2} + {mxω}^{2}$
or $kx - {mxω}^{2} = {mlω}^{2}$
$x = \frac{{mlω}^{2}}{k - {mω}^{2}}$

Watch 3-min video & get full concept clarity

hear from our champions

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!