A solid iron sphere A rolls down an inclined plane, while another hollow sphere B with the same mass and same radius also rolls down the inclided plane. If `V_(A)` and `V_(B)` are their velocities a the bottom of the inclined plane. Then

A. `V_(A)gtV_(B)`
B. `V_(A)=V_(B)`
C. `V_(A)ltV_(B)`
D. `V_(A)gt = ltV_(B)`

Options

A. `V_(A)gtV_(B)`
B. `V_(A)=V_(B)`
C. `V_(A)ltV_(B)`
D. `V_(A)gt = ltV_(B)`

Explanation

To solve the problem, we need to analyze the motion of both the solid sphere A and the hollow sphere B as they roll down the inclined plane.

### Step-by-Step Solution:

1. **Identify the Objects**: We have two spheres:
- Sphere A: A solid iron sphere
- Sphere B: A hollow sphere
Both spheres have the same mass (m) and radius (r).

2. **Understand Rolling Motion**: When an object rolls down an incline without slipping, both translational and rotational motion must be considered. The total acceleration of the rolling object depends on its moment of inertia (I).

3. **Moment of Inertia**:
- For a solid sphere, the moment of inertia \( I_A = \frac{2}{5} m r^2 \).
- For a hollow sphere, the moment of inertia \( I_B = \frac{2}{3} m r^2 \).

4. **Acceleration of Rolling Objects**: The acceleration \( a \) of a rolling object down an incline can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of the incline.

5. **Calculate Acceleration for Each Sphere**:
- For sphere A (solid):
\[ a_A = \frac{g \sin \theta}{1 + \frac{\frac{2}{5} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5g \sin \theta}{7} \]
- For sphere B (hollow):
\[ a_B = \frac{g \sin \theta}{1 + \frac{\frac{2}{3} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{2}{3}} = \frac{g \sin \theta}{\frac{5}{3}} = \frac{3g \sin \theta}{5} \]

6. **Compare Accelerations**: Now we compare the accelerations:
- \( a_A = \frac{5g \sin \theta}{7} \)
- \( a_B = \frac{3g \sin \theta}{5} \)

To compare \( a_A \) and \( a_B \), we can find a common denominator:
\[ a_A = \frac{5g \sin \theta}{7} \quad \text{and} \quad a_B = \frac{3g \sin \theta}{5} \]
The common denominator is 35:
\[ a_A = \frac{25g \sin \theta}{35} \quad \text{and} \quad a_B = \frac{21g \sin \theta}{35} \]
Thus, \( a_A > a_B \).

7. **Velocity at the Bottom of the Incline**: Since both spheres start from rest and roll down the same distance, the final velocities \( V_A \) and \( V_B \) can be derived from the kinematic equation:
\[ V^2 = U^2 + 2as \]
Since \( U = 0 \):
\[ V_A = \sqrt{2a_A s} \quad \text{and} \quad V_B = \sqrt{2a_B s} \]
Since \( a_A > a_B \), it follows that \( V_A > V_B \).

8. **Conclusion**: Therefore, we conclude that the velocity of the solid sphere A at the bottom of the incline is greater than that of the hollow sphere B:
\[ V_A > V_B \]

### Final Answer:
The correct option is **A. \( V_A > V_B \)**.