Two uniform solid spheres having unequal radii are released from rest from the same height on a rough incline. the spheres roll without slipping

Options

A. the heavier sphere reaches the bottom first
B. the bigger sphere reaches the bottom first
C. the two spheres reach the bottom together
D. the information given is not sufficient to tell which sphere will reach the bottom first

Explanation

To solve the problem of two uniform solid spheres rolling down a rough incline without slipping, we will use the work-energy theorem and the relationship between linear and angular motion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two solid spheres of different radii released from the same height on a rough incline.
- They roll without slipping, meaning we can relate their linear velocity (v) and angular velocity (ω) using the equation \( v = r \omega \).

2. **Applying the Work-Energy Theorem**:
- According to the work-energy theorem, the potential energy at the height (mgh) is converted into kinetic energy (both translational and rotational) as the spheres roll down the incline.
- The total kinetic energy (K.E) when the spheres reach the bottom is given by:
\[ K.E = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]

3. **Substituting the Moment of Inertia**:
- Substitute \( I \) into the kinetic energy equation:
\[ K.E = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 \]
- Using the relationship \( \omega = \frac{v}{r} \), we can rewrite \( \omega^2 \) as \( \frac{v^2}{r^2} \):
\[ K.E = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
- This simplifies to:
\[ K.E = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]

4. **Setting Up the Energy Equation**:
- Now, equate the potential energy at the height to the total kinetic energy:
\[ mgh = \frac{7}{10} mv^2 \]
- Cancel \( m \) from both sides:
\[ gh = \frac{7}{10} v^2 \]

5. **Solving for Linear Velocity (v)**:
- Rearranging gives:
\[ v^2 = \frac{10}{7} gh \]
- Taking the square root:
\[ v = \sqrt{\frac{10}{7} gh} \]

6. **Acceleration of the Spheres**:
- The acceleration \( a \) of the spheres can be derived from the forces acting on them:
\[ a = \frac{g \sin \theta}{1 + k} \]
- Here, \( k \) is the rotational inertia factor, which is the same for both spheres since they are solid spheres.

7. **Conclusion**:
- Since both spheres have the same acceleration and start from the same height, they will reach the bottom of the incline at the same time.

### Final Answer:
Both spheres will reach the bottom together.