A thin circular ring first slips down a smooth incline then rolls down a rough incline of identical geometry from same height. Ratio of time taken in the two motion is :

Options

A. `(1)/(2)`
B. 1
C. `(1)/(sqrt(2))`
D. `(1)/(4)`

Answer

To solve the problem of comparing the time taken for a thin circular ring to slide down a smooth incline versus rolling down a rough incline, we can break it down into a few steps.

### Step 1: Analyze the Motion on the Smooth Incline
When the ring slips down the smooth incline, it does not roll. The only force acting on it is the component of gravitational force along the incline.

1. **Identify Forces**: The gravitational force acting down the incline is \( mg \sin \theta \), where \( m \) is the mass of the ring, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline.
2. **Calculate Acceleration**: The net force acting on the ring is \( mg \sin \theta \). Using Newton's second law, \( F = ma \), we can write:
\[ mg \sin \theta = ma \implies a = g \sin \theta \]

3. **Determine Time Taken**: Using the kinematic equation for distance \( s = ut + \frac{1}{2} a t^2 \) (where \( u = 0 \)), we have:
\[ s = \frac{1}{2} g \sin \theta t^2 \]
Rearranging gives:
\[ t^2 = \frac{2s}{g \sin \theta} \implies t = \sqrt{\frac{2s}{g \sin \theta}} \]

### Step 2: Analyze the Motion on the Rough Incline
When the ring rolls down the rough incline, it experiences both translational and rotational motion.

1. **Identify Forces**: The gravitational force component along the incline is still \( mg \sin \theta \). However, now we have to consider the moment of inertia of the ring.
2. **Calculate Acceleration**: The ring rolls without slipping, so we need to account for both translational and rotational motion. The net force is still \( mg \sin \theta \), but the friction force provides the torque necessary for rotation.
- The moment of inertia \( I \) of a thin ring about its axis is \( I = m r^2 \).
- The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by \( a = r \alpha \).
- Using Newton's second law for rotation, \( \tau = I \alpha \), we have:
\[ f r = I \alpha \implies f r = m r^2 \frac{a}{r} \implies f = \frac{ma}{2} \]
3. **Net Force**: The net force acting on the ring is:
\[ mg \sin \theta - f = ma \implies mg \sin \theta - \frac{ma}{2} = ma \]
Rearranging gives:
\[ mg \sin \theta = \frac{3ma}{2} \implies a = \frac{2g \sin \theta}{3} \]

4. **Determine Time Taken**: Using the same kinematic equation:
\[ s = \frac{1}{2} a t^2 \implies s = \frac{1}{2} \left(\frac{2g \sin \theta}{3}\right) t^2 \implies s = \frac{g \sin \theta}{3} t^2 \]
Rearranging gives:
\[ t^2 = \frac{3s}{g \sin \theta} \implies t = \sqrt{\frac{3s}{g \sin \theta}} \]

### Step 3: Calculate the Ratio of Times
Now we can find the ratio of the time taken on the smooth incline \( t_1 \) to the time taken on the rough incline \( t_2 \):
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{2s}{g \sin \theta}}}{\sqrt{\frac{3s}{g \sin \theta}}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \]

### Final Answer
The ratio of the time taken in the two motions is:
\[ \frac{t_1}{t_2} = \frac{\sqrt{2}}{\sqrt{3}} \]