A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. The smallest kinetic energy at the bottom of the incline will be achieved by

Options

A. the solid sphere
B. the hollow sphere
C. the disc
D. all will achieve same kinetic energy.

Answer

To solve the problem, we need to analyze the kinetic energy at the bottom of the incline for a solid sphere, a hollow sphere, and a disc, all having the same mass and radius. The key points to consider are their moments of inertia and how they affect the distribution of kinetic energy between translational and rotational forms.

### Step-by-Step Solution:

1. **Identify the Objects and Their Moments of Inertia:**
- For a solid sphere: \( I_s = \frac{2}{5} m R^2 \)
- For a hollow sphere: \( I_h = \frac{2}{3} m R^2 \)
- For a disc: \( I_d = \frac{1}{2} m R^2 \)

2. **Understand the Energy Distribution:**
- When the objects roll down the incline, they convert potential energy into kinetic energy. The total kinetic energy \( K \) at the bottom consists of translational kinetic energy \( K_t \) and rotational kinetic energy \( K_r \):
\[ K = K_t + K_r = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- Since \( \omega = \frac{v}{R} \), we can rewrite the rotational kinetic energy as:
\[ K_r = \frac{1}{2} I \left(\frac{v}{R}\right)^2 \]

3. **Substitute the Moments of Inertia:**
- For each object, substitute the respective moment of inertia:
- Solid sphere:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]
- Hollow sphere:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{3} m R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2 \]
- Disc:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

4. **Compare the Total Kinetic Energies:**
- Now we can compare the total kinetic energies:
- Solid sphere: \( K = \frac{7}{10} mv^2 \)
- Hollow sphere: \( K = \frac{5}{6} mv^2 \)
- Disc: \( K = \frac{3}{4} mv^2 \)

5. **Determine the Smallest Kinetic Energy:**
- To find the smallest kinetic energy, we need to evaluate which fraction is the smallest:
- \( \frac{7}{10} \approx 0.7 \)
- \( \frac{5}{6} \approx 0.833 \)
- \( \frac{3}{4} = 0.75 \)
- The smallest value is \( \frac{7}{10} mv^2 \) for the solid sphere.

### Conclusion:
The smallest kinetic energy at the bottom of the incline will be achieved by the **solid sphere**.