A body of mass `m` slides down an smooth incline and reaches the bottom with a velocity, Now smooth incline surface is made rough and the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been:

Options

A. `sqrt(2v)`
B. `v`
C. `(sqrt(2/5))v`
D. `v//sqrt(2)`

Explanation

To solve the problem of a ring rolling down a rough incline, we can follow these steps:

### Step 1: Understand the Initial Conditions
When a body of mass `m` slides down a smooth incline, it reaches the bottom with a certain velocity. The velocity can be derived from energy conservation principles. The potential energy at the top is converted into kinetic energy at the bottom.

### Step 2: Calculate the Velocity on a Smooth Incline
For a smooth incline, the potential energy (PE) at height `h` is given by:
\[ PE = mgh \]
The kinetic energy (KE) at the bottom is:
\[ KE = \frac{1}{2} mv^2 \]
By conservation of energy:
\[ mgh = \frac{1}{2} mv^2 \]
This simplifies to:
\[ v^2 = 2gh \]
Thus, the velocity \( v \) at the bottom of the incline is:
\[ v = \sqrt{2gh} \]

### Step 3: Transition to a Rough Incline with a Ring
When the incline is made rough and the mass is in the form of a ring, the ring rolls down the incline. The forces acting on the ring include gravitational force, normal force, and friction.

### Step 4: Analyze Forces on the Ring
The gravitational force acting down the incline is:
\[ F_{\text{gravity}} = mg \sin \theta \]
The frictional force provides the torque necessary for rolling. The equation of motion for the ring can be expressed as:
\[ mg \sin \theta - F_{\text{friction}} = ma \]
Where \( a \) is the linear acceleration of the center of mass of the ring.

### Step 5: Relate Linear and Angular Acceleration
For a ring, the moment of inertia \( I \) is given by:
\[ I = mR^2 \]
The frictional force \( F_{\text{friction}} \) also contributes to the angular acceleration \( \alpha \):
\[ F_{\text{friction}} \cdot R = I \alpha \]
Since \( \alpha = \frac{a}{R} \), we can substitute:
\[ F_{\text{friction}} \cdot R = mR^2 \cdot \frac{a}{R} \]
Thus:
\[ F_{\text{friction}} = \frac{ma}{R} \]

### Step 6: Substitute Back into the Equation of Motion
Now substituting \( F_{\text{friction}} \) into the equation of motion:
\[ mg \sin \theta - \frac{ma}{R} = ma \]
This simplifies to:
\[ mg \sin \theta = ma + \frac{ma}{R} \]
Factoring out \( m \):
\[ g \sin \theta = a \left(1 + \frac{1}{R}\right) \]

### Step 7: Solve for Acceleration
Now, we can express the acceleration \( a \):
\[ a = \frac{g \sin \theta}{1 + \frac{1}{R}} \]

### Step 8: Calculate the Final Velocity
Using the kinematic equation \( v^2 = u^2 + 2as \) (where initial velocity \( u = 0 \)):
\[ v^2 = 0 + 2as \]
Substituting for \( a \):
\[ v^2 = 2s \cdot \frac{g \sin \theta}{1 + \frac{1}{R}} \]
Thus, the final velocity \( v \) at the bottom of the incline is:
\[ v = \sqrt{\frac{2gs \sin \theta}{1 + \frac{1}{R}}} \]

### Step 9: Compare with the Smooth Incline
From the smooth incline case, we had:
\[ v_{\text{smooth}} = \sqrt{2gs \sin \theta} \]
Thus, the ratio of the velocities is:
\[ \frac{v_{\text{ring}}}{v_{\text{smooth}}} = \frac{\sqrt{\frac{2gs \sin \theta}{1 + \frac{1}{R}}}}{\sqrt{2gs \sin \theta}} = \frac{1}{\sqrt{1 + \frac{1}{R}}} \]

### Conclusion
The final velocity of the ring at the bottom of the rough incline is given by:
\[ v_{\text{ring}} = \frac{v_{\text{smooth}}}{\sqrt{1 + \frac{1}{R}}} \]