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 velocity ω , in gravity free space. The increase in length of the spring will be

Let us analyze the system step-by-step.

Let the unstretched length of the spring be $l$ and its spring constant be $k$.
When the system is rotated with an angular velocity $\omega$ in gravity-free space, the spring stretches due to the centrifugal force acting on the particle of mass $m$ in the rotating frame of reference (or because the spring provides the necessary centripetal force in the inertial frame).

Let the increase in the length of the spring be $x$.
Thus, the stretched length of the spring becomes:
$r=l+x$
This stretched length $r$ is the radius of the circular path in which the mass $m$ rotates.

In the steady state, the restoring force exerted by the stretched spring balances the centripetal force required for circular motion:
$F=m{\omega }^{2}r$
Here, the spring force is given by Hooke's Law:
$F=kx$

Equating the two forces:
$kx=m{\omega }^2(l+x)$

Now, we expand and solve for the extension $x$:
$kx=m{\omega }^{2}l+m{\omega }^{2}x$
Rearranging the terms containing $x$ to one side:
$kx-m{\omega }^{2}x=m{\omega }^{2}l$
Factoring out $x$:
$x(k-m{\omega }^2)=m{\omega }^{2}l$
Dividing both sides by $k-m{\omega }^2$ gives:
$x=\frac{m{\omega }^{2}l}{k-m{\omega }^2}$

Note: The correct option corresponding to this expression is typically represented as $\frac{m{\omega }^{2}l}{k-m{\omega }^2}$, which matches the provided answer value in the options when properly formatted (represented in the raw text option as "mω²l/k-mω²" meaning $m{\omega }^{2}l/(k-m{\omega }^2)$).

Therefore, the increase in length of the spring is:
$\frac{m{\omega }^{2}l}{k-m{\omega }^2}$