The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane : (i) a ring of radius R, (ii) a solid cylinder of radius `R/2` and (iii) a solid sphere of radius `R/4` If, in each case, the speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is

Options

A. `4:3:2`
B. `14:15:20`
C. `20:15:14`
D. `2:3:4`

Explanation

To solve the problem, we need to analyze the motion of each object as it rolls up the incline. We will use the principle of conservation of energy to find the maximum heights each object can reach based on their initial kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Energy Conservation**:
- At the bottom of the incline, each object has kinetic energy (both translational and rotational) and zero potential energy (since it's at the datum level).
- At the maximum height, the kinetic energy will be converted entirely into potential energy.

2. **Kinetic Energy at the Bottom**:
- The total kinetic energy \( KE \) at the bottom for each object can be expressed as:
\[ KE = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- Using the relation \( v = r \omega \), we can express \( \omega \) in terms of \( v \) and \( r \).

3. **Moment of Inertia for Each Object**:
- For the ring (radius \( R \)):
\[ I = mR^2 \quad \Rightarrow \quad KE = \frac{1}{2} mv^2 + \frac{1}{2} mR^2 \left(\frac{v}{R}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]
- For the solid cylinder (radius \( \frac{R}{2} \)):
\[ I = \frac{1}{2} m\left(\frac{R}{2}\right)^2 = \frac{1}{8} mR^2 \quad \Rightarrow \quad KE = \frac{1}{2} mv^2 + \frac{1}{2} \cdot \frac{1}{8} mR^2 \left(\frac{v}{\frac{R}{2}}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]
- For the solid sphere (radius \( \frac{R}{4} \)):
\[ I = \frac{2}{5} m\left(\frac{R}{4}\right)^2 = \frac{2}{80} mR^2 = \frac{1}{40} mR^2 \quad \Rightarrow \quad KE = \frac{1}{2} mv^2 + \frac{1}{2} \cdot \frac{1}{40} mR^2 \left(\frac{v}{\frac{R}{4}}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{10} mv^2 = \frac{3}{5} mv^2 \]

4. **Potential Energy at Maximum Height**:
- At maximum height \( h \), the kinetic energy is converted to potential energy \( PE = mgh \).
- Setting \( KE \) equal to \( PE \):
- For the ring:
\[ mv^2 = mgh_1 \quad \Rightarrow \quad h_1 = \frac{v^2}{g} \]
- For the solid cylinder:
\[ \frac{3}{4} mv^2 = mgh_2 \quad \Rightarrow \quad h_2 = \frac{3v^2}{4g} \]
- For the solid sphere:
\[ \frac{3}{5} mv^2 = mgh_3 \quad \Rightarrow \quad h_3 = \frac{3v^2}{5g} \]

5. **Finding the Ratios of Heights**:
- The heights are:
- \( h_1 = \frac{v^2}{g} \)
- \( h_2 = \frac{3v^2}{4g} \)
- \( h_3 = \frac{3v^2}{5g} \)
- To find the ratio \( h_1 : h_2 : h_3 \):
\[ h_1 : h_2 : h_3 = \frac{v^2}{g} : \frac{3v^2}{4g} : \frac{3v^2}{5g} \]
- Simplifying gives:
\[ 1 : \frac{3}{4} : \frac{3}{5} \quad \Rightarrow \quad 20 : 15 : 12 \]

### Final Ratio:
Thus, the ratio of the maximum heights they climb is:
\[ 20 : 15 : 12 \]