JEE Class main Answered

A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular speed ω, in gravity free space. The increase in length of the spring will be

The elastic force of spring gives the centrifugal force required  

Thus, 

kx = mω²r

x is the increase in the length of the spring, ω is the angular speed and r is the total length of the spring. 

Thus, 

kx = mω²(l+x) ....(as r = l+x)

kx = mω²l + mω²x 

kx - mω²x  = mω²l

x(k - mω²) = mω²l

The increase in length of the spring will be mω²l/(k - mω²)

The correct option is (1)