Three solid cylinders A, B and C each of mass m and radius R are allowed to roll down on three inclined plane A', B' and C' respectively (each of inclination `theta`) such that for A' and A coefficient of friction is zero, for B' and B coefficient of friction is greater than zero but less than `(tan theta)/3` and for C' and C coefficient of friction is more than `(tan theta)/3`. On reaching bottom of the inclined plane

Options

A. C has maximum angular speed
B. B has minimum total kinetic energy
C. A takes minimum time reach the bottom
D. B has minimum translational kinetic energy

Explanation

To solve the problem of three solid cylinders A, B, and C rolling down inclined planes A', B', and C' respectively, we need to analyze the motion of each cylinder based on the coefficient of friction and the forces acting on them.

### Step-by-Step Solution:

1. **Identify Forces Acting on Each Cylinder:**
- For each cylinder, the gravitational force acting down the incline is \( mg \sin \theta \).
- The normal force acting perpendicular to the incline is \( N = mg \cos \theta \).
- The frictional force \( f \) will depend on the coefficient of friction and the normal force.

2. **Analyze Cylinder A (No Friction):**
- Since the coefficient of friction is zero, there is no frictional force acting on cylinder A.
- The acceleration of the center of mass \( a_{cm} \) for cylinder A is given by:
\[ a_{cm} = \frac{mg \sin \theta}{m} = g \sin \theta \]
- The cylinder will slide down without rolling, and its angular acceleration \( \alpha \) is zero.

3. **Analyze Cylinder B (Friction Less than \( \frac{\tan \theta}{3} \)):**
- Here, the frictional force is present but not enough to prevent slipping.
- The net force acting on cylinder B is:
\[ F_{net} = mg \sin \theta - f \]
- The maximum frictional force is \( f = \mu N = \mu mg \cos \theta \), where \( 0 < \mu < \frac{\tan \theta}{3} \).
- The acceleration of the center of mass is:
\[ a_{cm} = \frac{mg \sin \theta - f}{m} = g \sin \theta - \mu g \cos \theta \]

4. **Analyze Cylinder C (Friction Greater than \( \frac{\tan \theta}{3} \)):**
- In this case, the frictional force is sufficient to prevent slipping.
- The cylinder will roll without slipping, and the equations of motion will involve both translational and rotational dynamics.
- The acceleration of the center of mass can be expressed as:
\[ a_{cm} = g \sin \theta - \mu g \cos \theta \]
- The angular acceleration \( \alpha \) can be calculated using the torque due to friction:
\[ \tau = f \cdot R = I \alpha \]
- For a solid cylinder, \( I = \frac{1}{2} m R^2 \), leading to:
\[ f R = \frac{1}{2} m R^2 \alpha \implies f = \frac{1}{2} m R \alpha \]

5. **Determine the Maximum Angular Speed:**
- The maximum angular speed will occur for cylinder C since it has the highest friction, allowing it to roll without slipping.
- The angular speed \( \omega \) can be related to the linear speed \( v_{cm} \) as:
\[ v_{cm} = R \omega \]

6. **Determine Total Kinetic Energy:**
- The total kinetic energy for each cylinder at the bottom of the incline can be expressed as:
\[ KE_{total} = KE_{translational} + KE_{rotational} = \frac{1}{2} mv_{cm}^2 + \frac{1}{2} I \omega^2 \]
- For cylinder B, since it is slipping, it will have a lower total kinetic energy compared to cylinder C.

7. **Determine Time to Reach the Bottom:**
- The time taken to reach the bottom can be calculated using the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
- Since \( a_{cm} \) is highest for cylinder A, it will take the least time to reach the bottom.

### Summary of Results:
- **Cylinder C** has the maximum angular speed.
- **Cylinder B** has the minimum total kinetic energy.
- **Cylinder A** takes the minimum time to reach the bottom.