A solid cylinder is released from rest from the top of an inclined plane of inclination 30° and length 60 cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is _____ m s$$^{-1}$$.
(Given $$g = 10$$ m s$$^{-2}$$)

We have a solid cylinder rolling without slipping down an inclined plane of inclination $$\theta = 30°$$ and length $$l = 60$$ cm $$= 0.6$$ m, with $$g = 10$$ m/s$$^2$$.

The height of the incline is:

$$h = l \sin\theta = 0.6 \times \sin 30° = 0.6 \times 0.5 = 0.3 \text{ m}$$

For a solid cylinder rolling without slipping, using energy conservation:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Now, for a solid cylinder $$I = \frac{1}{2}mr^2$$ and $$\omega = \frac{v}{r}$$ (rolling condition), so:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

Solving for $$v$$:

$$v = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4 \times 10 \times 0.3}{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2 \text{ m/s}$$

Hence, the answer is $$2$$ m/s.