Consider three solid spheres, sphere (i) has radius r and mass m, sphere (ii) has radius r and mass 3 m, sphere (iii) has radius 3r and mass m, All can be placed at the same point on the same inclined plane, where they will roll without slipping to the bottom, If allowed to roll down the incline, then at the bottom of the incline

Options

A. sphere (i) will have the largest speed
B. sphere (ii) will havethe largest speed
C. sphere (iii) will have the largest kinetic energy
D. all the sphere will have equal speeds.

Explanation

To solve the problem, we will analyze the motion of the three solid spheres as they roll down an inclined plane using the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify the Initial Potential Energy**:
Each sphere starts from the same height \( h \) on the inclined plane. The potential energy (PE) at the top for each sphere can be expressed as:
\[ \text{PE} = mgh \]
For sphere (i), it is \( mgh \); for sphere (ii), it is \( 3mgh \); and for sphere (iii), it is \( mgh \).

2. **Set Up the Energy Conservation Equation**:
As the spheres roll down the incline, their potential energy converts into kinetic energy (KE). The total kinetic energy at the bottom consists of translational kinetic energy and rotational kinetic energy:
\[ \text{KE} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. 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:
\[ \omega = \frac{v}{r} \]

3. **Calculate for Each Sphere**:
- **Sphere (i)**: Mass \( m \), Radius \( r \)
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
Simplifying gives:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
\[ mgh = \frac{7}{10} mv^2 \]
Cancel \( m \):
\[ gh = \frac{7}{10} v^2 \implies v^2 = \frac{10gh}{7} \implies v = \sqrt{\frac{10gh}{7}} \]

- **Sphere (ii)**: Mass \( 3m \), Radius \( r \)
\[ 3mgh = \frac{1}{2} (3m) v^2 + \frac{1}{2} \left(\frac{2}{5} (3m) r^2\right) \left(\frac{v}{r}\right)^2 \]
Simplifying gives:
\[ 3mgh = \frac{3}{2} mv^2 + \frac{3}{5} mv^2 \]
\[ 3mgh = \frac{21}{10} mv^2 \]
Cancel \( 3m \):
\[ gh = \frac{7}{10} v^2 \implies v^2 = \frac{10gh}{7} \implies v = \sqrt{\frac{10gh}{7}} \]

- **Sphere (iii)**: Mass \( m \), Radius \( 3r \)
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m (3r)^2\right) \left(\frac{v}{3r}\right)^2 \]
Simplifying gives:
\[ mgh = \frac{1}{2} mv^2 + \frac{6}{5} mv^2 \]
\[ mgh = \frac{7}{10} mv^2 \]
Cancel \( m \):
\[ gh = \frac{7}{10} v^2 \implies v^2 = \frac{10gh}{7} \implies v = \sqrt{\frac{10gh}{7}} \]

4. **Conclusion**:
From the calculations, we find that all three spheres have the same final speed at the bottom of the incline:
\[ v = \sqrt{\frac{10gh}{7}} \]
Thus, the answer is that all spheres will have equal speed at the bottom of the incline.