A circular disc reaches from top to bottom of an inclined plane of length $$L$$. When it slips down the plane, it takes time $$t_1$$. When it rolls down the plane, it takes time $$t_2$$. The value of $$\frac{t_2}{t_1}$$ is $$\sqrt{\frac{3}{x}}$$. The value of $$x$$ will be ___.

1. Analyze Case 1: Slipping Down the Inclined Plane

When the disc slips down a frictionless inclined plane of angle $$\theta$$, it does not rotate. It experiences only linear translation due to the component of gravity down the incline.

• Acceleration ($$a_1$$):

  $$a_1 = g \sin\theta$$

• Time taken ($$t_1$$): Using the second equation of motion ($$L = \frac{1}{2}a_1 t_1^2$$):

  $$t_1 = \sqrt{\frac{2L}{a_1}} = \sqrt{\frac{2L}{g \sin\theta}}$$

2. Analyze Case 2: Pure Rolling Down the Inclined Plane

When the disc rolls down a rough inclined plane without slipping, both torque (from friction) and translational forces act on it. The linear acceleration for a rolling object with radius $$R$$ and radius of gyration $$K$$ is given by:

$$a_2 = \frac{g \sin\theta}{1 + \frac{K^2}{R^2}}$$

For a solid circular disc, the moment of inertia about its central axis is $$I = \frac{1}{2}MR^2 = MK^2$$. Therefore, its shape factor is:

$$\frac{K^2}{R^2} = \frac{1}{2}$$

Substituting this back into the acceleration formula:

$$a_2 = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3}g \sin\theta$$

Using the equation of motion for the same distance $$L$$ ($$L = \frac{1}{2}a_2 t_2^2$$):

$$t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L}{\frac{2}{3}g \sin\theta}} = \sqrt{\frac{3L}{g \sin\theta}}$$

3. Determine the Ratio and Find $$x$$

Now, we take the ratio of the rolling time ($$t_2$$) to the slipping time ($$t_1$$):

$$\frac{t_2}{t_1} = \frac{\sqrt{\frac{3L}{g \sin\theta}}}{\sqrt{\frac{2L}{g \sin\theta}}} = \sqrt{\frac{3}{2}}$$

Comparing this result to the given expression in the problem statement:

$$\sqrt{\frac{3}{2}} = \sqrt{\frac{3}{x}}$$

By direct comparison of the denominators, we find:

$$x = 2$$

Final Answer: 2