Solid Sphere, Ring and Disc having same mass m and radius r the smallest kinetic energy at the bottom of the incline will be achieved by
1)Solid Sphere
2)Ring
3)Disc
4)All will achieve the same kinetic energy

Options

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

Answer

To determine which object (solid sphere, ring, or disc) has the smallest kinetic energy at the bottom of an incline, we need to analyze their moments of inertia and how they relate to kinetic energy.

### Step-by-Step Solution:

1. **Identify the Moments of Inertia:**
- For a solid sphere: \( I_{\text{sphere}} = \frac{2}{5} m r^2 \)
- For a ring: \( I_{\text{ring}} = m r^2 \)
- For a disc: \( I_{\text{disc}} = \frac{1}{2} m r^2 \)

2. **Understand the Kinetic Energy Components:**
The total kinetic energy (KE) of a rolling object is the sum of its translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( \omega \) is the angular velocity.

3. **Relate Angular Velocity to Linear Velocity:**
For rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

4. **Substitute \( \omega \) in the Kinetic Energy Equation:**
Substitute \( \omega \) in the rotational kinetic energy term:
\[ KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} I \frac{v^2}{r^2} \]
Therefore, the total kinetic energy becomes:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \frac{v^2}{r^2} \]

5. **Factor Out \( v^2 \):**
\[ KE = \frac{1}{2} v^2 \left( m + \frac{I}{r^2} \right) \]

6. **Calculate \( \frac{I}{r^2} \) for Each Object:**
- For the solid sphere:
\[ \frac{I_{\text{sphere}}}{r^2} = \frac{\frac{2}{5} m r^2}{r^2} = \frac{2}{5} m \]
- For the ring:
\[ \frac{I_{\text{ring}}}{r^2} = \frac{m r^2}{r^2} = m \]
- For the disc:
\[ \frac{I_{\text{disc}}}{r^2} = \frac{\frac{1}{2} m r^2}{r^2} = \frac{1}{2} m \]

7. **Sum \( m + \frac{I}{r^2} \) for Each Object:**
- Solid sphere: \( m + \frac{2}{5} m = \frac{7}{5} m \)
- Ring: \( m + m = 2m \)
- Disc: \( m + \frac{1}{2} m = \frac{3}{2} m \)

8. **Determine the Smallest Total Kinetic Energy:**
The object with the smallest value of \( m + \frac{I}{r^2} \) will have the smallest kinetic energy:
- Solid sphere: \( \frac{7}{5} m \)
- Disc: \( \frac{3}{2} m = 1.5 m \)
- Ring: \( 2m \)

Since \( \frac{7}{5} m < \frac{3}{2} m < 2m \), the solid sphere has the smallest kinetic energy.

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