`A` and `B` are two solid spheres of equal masses. A rolls down an inclined plane without slipping from a height `H`. `B` falls vertically from the same height. Then on reaching the ground.

Options

A. both cannot do work
B. A can do more work than `B`
C. `B` can do more work than `A`
D. both `A` and `B` will have different linear speeds

Answer

To solve the problem, we need to analyze the motion of both spheres, A and B, as they descend from height H.

### Step-by-Step Solution:

1. **Understanding the Motion of Sphere A**:
- Sphere A rolls down an inclined plane without slipping.
- As it rolls, it converts its potential energy (due to height H) into two forms of kinetic energy: translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its rotation).

2. **Understanding the Motion of Sphere B**:
- Sphere B falls vertically from the same height H.
- It converts its potential energy entirely into translational kinetic energy as it falls straight down.

3. **Potential Energy at Height H**:
- Both spheres start with the same potential energy when at height H, given by:
\[ PE = mgh \]
where \( m \) is the mass of the sphere, \( g \) is the acceleration due to gravity, and \( h \) is the height.

4. **Kinetic Energy of Sphere A**:
- When Sphere A reaches the bottom, its total kinetic energy is the sum of translational and rotational kinetic energy:
\[ KE_A = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- For a solid sphere, the moment of inertia \( I \) is given by \( I = \frac{2}{5} m r^2 \) and the relationship between linear velocity \( v \) and angular velocity \( \omega \) is \( v = r \omega \).
- Substituting \( I \) and \( \omega \) into the kinetic energy equation gives:
\[ KE_A = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

5. **Kinetic Energy of Sphere B**:
- Sphere B, falling vertically, has only translational kinetic energy:
\[ KE_B = \frac{1}{2} mv_B^2 \]
- Since both spheres start with the same potential energy, we equate the potential energy to the kinetic energy at the bottom:
\[ mgh = KE_A \quad \text{and} \quad mgh = KE_B \]

6. **Comparing Kinetic Energies**:
- Since both spheres have the same potential energy at height H, we can set:
\[ mgh = \frac{7}{10} mv_A^2 \quad \text{and} \quad mgh = \frac{1}{2} mv_B^2 \]
- This leads to:
\[ v_A^2 = \frac{10gh}{7} \quad \text{and} \quad v_B^2 = \frac{2gh}{1} \]
- Thus, we can see that \( v_A \) and \( v_B \) will be different.

7. **Conclusion**:
- Sphere A and Sphere B have different linear speeds when they reach the ground, but they both have the same total kinetic energy derived from their potential energy.

### Final Answer:
Both A and B have different linear speeds.