A solid cylinder is placed on the end of an inclined plane. It is found that the plane can be tipped at an angle `theta` before the cylinder starts to slide. When the cylinder turns on its sides and is allowed to roll, it is found that the steepest angle at which the cylinder performs pure rolling is `phi`. The ratio `tanphi//tantheta` is

Options

A. `3`
B. `1/3`
C. `1`
D. `1/2`

Explanation

To solve the problem, we need to analyze the forces and torques acting on the solid cylinder when it is placed on an inclined plane. We will derive the relationship between the angles `theta` and `phi` and find the ratio \( \frac{\tan \phi}{\tan \theta} \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
When the inclined plane is at an angle \( \theta \), the gravitational force acting on the cylinder can be resolved into two components: one parallel to the incline \( mg \sin \theta \) and one perpendicular to the incline \( mg \cos \theta \).

2. **Condition for Sliding**:
The cylinder will start to slide when the component of gravitational force parallel to the incline exceeds the static friction force. The maximum static friction force is given by \( f_s = \mu_s N \), where \( N = mg \cos \theta \) is the normal force and \( \mu_s \) is the coefficient of static friction. Thus, we have:
\[ mg \sin \theta = \mu_s mg \cos \theta \]
Simplifying, we get:
\[ \tan \theta = \mu_s \]

3. **Condition for Pure Rolling**:
When the cylinder rolls without slipping, the condition for pure rolling is that the frictional force must provide the necessary torque for rolling. The torque \( \tau \) about the center of mass is given by:
\[ \tau = f_r \cdot r \]
where \( f_r \) is the frictional force and \( r \) is the radius of the cylinder. The moment of inertia \( I \) of a solid cylinder about its center is \( \frac{1}{2} m r^2 \).

4. **Using Newton's Second Law for Rotation**:
The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by:
\[ a = r \alpha \]
Therefore, we can write:
\[ \tau = I \alpha \implies f_r \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \implies f_r = \frac{1}{2} m a \]

5. **Setting Up the Equation for Pure Rolling**:
The net force acting on the cylinder when it is rolling is:
\[ mg \sin \phi - f_r = ma \]
Substituting \( f_r = \frac{1}{2} m a \):
\[ mg \sin \phi - \frac{1}{2} m a = ma \implies mg \sin \phi = \frac{3}{2} m a \]
Thus, we have:
\[ a = \frac{2}{3} g \sin \phi \]

6. **Relating the Angles**:
From the earlier conditions, we have:
\[ \tan \phi = \frac{a}{g} = \frac{2}{3} \sin \phi \]
And from the sliding condition:
\[ \tan \theta = \mu_s \]

7. **Finding the Ratio**:
Now, we relate the two angles:
\[ \frac{\tan \phi}{\tan \theta} = \frac{\frac{2}{3} \sin \phi}{\mu_s} \]
Given that \( \tan \theta = \mu_s \), we can express the ratio as:
\[ \frac{\tan \phi}{\tan \theta} = \frac{2}{3} \cdot \frac{\sin \phi}{\mu_s} \]

8. **Final Ratio**:
After substituting and simplifying, we find:
\[ \frac{\tan \phi}{\tan \theta} = \frac{2}{7} \]

### Conclusion:
Thus, the ratio \( \frac{\tan \phi}{\tan \theta} \) is \( \frac{2}{7} \).