A spring of unstretched length l has a mass m with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity $\mathrm{\nu }$ is: 

Let linear mass density $\mathrm{\omega }=\frac{\mathrm{m}}{\mathrm{l}}$

dx element's velocity and x's distance from rigid support are,

$\mathrm{v}\text{'}=\frac{\mathrm{vx}}{\mathrm{l}}$

Now the mass of this element of thickness $\left(\mathrm{dx}\right)=\mathrm{\omega dx}=\mathrm{dm}$

Kinetic energy of particle of thickness dx = $\frac{1}{2}dm{v}^2$

$\mathrm{KE}={\int }_0^{\mathrm{L}}\frac{1}{2}\mathrm{\omega dx}{\left(\frac{\mathrm{vx}}{\mathrm{L}}\right)}^2$

$\mathrm{KE}=\frac{1}{2}\frac{\mathrm{m}}{\mathrm{L}}\frac{{\mathrm{v}}^2}{{\mathrm{L}}^2}{\int }_0^{\mathrm{L}}{\mathrm{x}}^2\mathrm{dx}$

$\mathrm{KE}=\frac{{\mathrm{mv}}^2}{2{\mathrm{L}}^3}{{\left[\frac{{\mathrm{x}}^3}{3}\right]}^{\mathrm{L}}}_0$

$\mathrm{KE}=\frac{1}{6}{\mathrm{mv}}^2$

Hence the correct answer is $\frac{1}{6}m{v}^2.$