A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^\circ$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $\frac{2-\sqrt{3}}{\sqrt{10}}\,\text{s}$, then the height of the top of the inclined plane, in metres, is ________. Take $g = 10\,\text{ms}^{-2}$

Step 1: Identify the given parameters.

The given values for the inclined plane and the rolling bodies are:

$$\theta = 60^\circ$$ $$\Delta t = \frac{2-\sqrt{3}}{\sqrt{10}}\,\text{s}$$ $$g = 10\,\text{ms}^{-2}$$

Step 2: Determine the acceleration of a body rolling without slipping.

For a body of mass $M$, radius $R$, and moment of inertia $I$ rolling down an incline, the equations of motion are:

$$mg\sin\theta - f = ma$$ $$fR = I\alpha = I\frac{a}{R}$$

Solving for acceleration $a$ yields:

$$a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}}$$

Step 3: Calculate the acceleration for the ring and the disc.

For the ring, $I = MR^2$:

$$a_r = \frac{g\sin\theta}{2}$$

For the disc, $I = \frac{1}{2}MR^2$:

$$a_d = \frac{2}{3}g\sin\theta$$

Step 4: Express the time taken to reach the bottom of the incline.

The length of the path along the incline $L$ is related to the height $h$ by:

$$L = \frac{h}{\sin\theta}$$

Using the equation of motion from rest, $L = \frac{1}{2}at^2$:

$$t = \sqrt{\frac{2L}{a}}$$

Step 5: Find the time expressions for both bodies.

Substitute the accelerations into the time formula:

$$t_r = \sqrt{\frac{4h}{g\sin^2\theta}}$$ $$t_d = \sqrt{\frac{3h}{g\sin^2\theta}}$$

Step 6: Substitute the angle of inclination.

For $\theta = 60^\circ$, we have $\sin^2\theta = \frac{3}{4}$:

$$t_r = \sqrt{\frac{16h}{3g}} = \frac{4}{\sqrt{3}}\sqrt{\frac{h}{g}}$$ $$t_d = \sqrt{\frac{4h}{g}} = 2\sqrt{\frac{h}{g}}$$

Step 7: Express the time difference.

Subtract the two times to find the difference:

$$\Delta t = t_r - t_d = \sqrt{\frac{h}{g}}\left(\frac{4}{\sqrt{3}} - 2\right)$$ $$\Delta t = \sqrt{\frac{h}{g}}\frac{2(2-\sqrt{3})}{\sqrt{3}}$$

Step 8: Solve for the height $h$.

Equate the expression to the given time difference:

$$\sqrt{\frac{h}{g}}\frac{2(2-\sqrt{3})}{\sqrt{3}} = \frac{2-\sqrt{3}}{\sqrt{10}}$$ $$\sqrt{\frac{h}{g}} = \frac{\sqrt{3}}{2\sqrt{10}}$$

Square both sides and substitute $g = 10\,\text{ms}^{-2}$:

$$\frac{h}{g} = \frac{3}{40}$$ $$h = \frac{3 \times 10}{40} = 0.75\,\text{m}$$

Answer: 0.75