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 , (iii) a solid sphere of radius R/4. If, in each case, the speed of the incline is same, theratio of the maximum heights they climb is :

Options

A. `4:3:2`
B. `10:15:7`
C. `14:15:20`
D. `2:3:4`

Answer

To solve the problem, we will use the principle of conservation of energy. The total mechanical energy of each object will be conserved as they roll up the incline. The initial kinetic energy will be converted into potential energy at the maximum height.

### Step-by-Step Solution:

1. **Identify the types of energy involved**:
- Each object has translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its rotation).
- The potential energy at the maximum height is given by \( PE = mgh \), where \( h \) is the height climbed.

2. **Write the expression for total initial kinetic energy**:
- The total kinetic energy \( KE \) for each object can be expressed as:
\[ KE = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
- Here, \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

3. **Relate angular velocity to linear velocity**:
- For rolling without slipping, \( \omega = \frac{v}{r} \), where \( r \) is the radius of the object.

4. **Calculate the moment of inertia for each object**:
- For a ring: \( I = mR^2 \)
- For a solid cylinder: \( I = \frac{1}{2} m \left(\frac{R}{2}\right)^2 = \frac{1}{8} mR^2 \)
- For a solid sphere: \( I = \frac{2}{5} m \left(\frac{R}{4}\right)^2 = \frac{2}{80} mR^2 = \frac{1}{40} mR^2 \)

5. **Substitute the moment of inertia into the kinetic energy expression**:
- For the ring:
\[ KE_{ring} = \frac{1}{2} mv^2 + \frac{1}{2} (mR^2) \left(\frac{v}{R}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]
- For the solid cylinder:
\[ KE_{cylinder} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{8} mR^2\right) \left(\frac{v}{\frac{R}{2}}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{8} mR^2\right) \left(\frac{4v^2}{R^2}\right) = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]
- For the solid sphere:
\[ KE_{sphere} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{40} mR^2\right) \left(\frac{v}{\frac{R}{4}}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{40} mR^2\right) \left(\frac{16v^2}{R^2}\right) = \frac{1}{2} mv^2 + \frac{2}{5} mv^2 = \frac{9}{10} mv^2 \]

6. **Set the kinetic energy equal to potential energy at maximum height**:
- For the ring:
\[ mv^2 = mgh \implies h_{ring} = \frac{v^2}{g} \]
- For the solid cylinder:
\[ \frac{3}{4} mv^2 = mgh \implies h_{cylinder} = \frac{3v^2}{4g} \]
- For the solid sphere:
\[ \frac{9}{10} mv^2 = mgh \implies h_{sphere} = \frac{9v^2}{10g} \]

7. **Calculate the ratio of maximum heights**:
- The ratio of maximum heights climbed by the ring, cylinder, and sphere is:
\[ \text{Ratio} = h_{ring} : h_{cylinder} : h_{sphere} = \frac{v^2}{g} : \frac{3v^2}{4g} : \frac{9v^2}{10g} \]
- Simplifying this gives:
\[ 1 : \frac{3}{4} : \frac{9}{10} = 40 : 30 : 36 \]
- Thus, the final ratio is:
\[ 20 : 15 : 18 \]

### Final Ratio:
The ratio of the maximum heights they climb is \( 20 : 15 : 18 \).