A body of mass 200 g is tied to a spring of spring constant 12.5 N m$$^{-1}$$, while the other end of spring is fixed at point $$O$$. If the body moves about $$O$$ in a circular path on a smooth horizontal surface with constant angular speed 5 rad s$$^{-1}$$, then the ratio of extension in the spring to its natural length will be:

Given: mass $$m = 200 \text{ g} = 0.2 \text{ kg}$$, spring constant $$k = 12.5 \text{ N/m}$$, angular speed $$\omega = 5 \text{ rad/s}$$.

Let the natural length of the spring be $$L$$ and the extension be $$x$$. The body moves in a circular path of radius $$r = L + x$$.

The spring force provides the centripetal force:

$$kx = m\omega^2 r = m\omega^2(L + x)$$

$$kx = m\omega^2 L + m\omega^2 x$$

$$kx - m\omega^2 x = m\omega^2 L$$

$$x(k - m\omega^2) = m\omega^2 L$$

$$\frac{x}{L} = \frac{m\omega^2}{k - m\omega^2}$$

Substituting values:

$$m\omega^2 = 0.2 \times 25 = 5 \text{ N/m}$$

$$k - m\omega^2 = 12.5 - 5 = 7.5 \text{ N/m}$$

$$\frac{x}{L} = \frac{5}{7.5} = \frac{2}{3}$$

The ratio of extension to natural length is $$2 : 3$$.

The correct answer is Option 3.