A ring takes time `t_(1)` in slipping down an inclined plane of length L, whereas it takes time `t_(2)` in rolling down the same plane. The ratio of `t_(1) and t_(2)` is -

Options

A. `sqrt(2):1`
B. `1:sqrt(2)`
C. `1:2 `
D. `2:1 `

Answer

To solve the problem of finding the ratio of the time taken by a ring to slip down an inclined plane versus the time taken to roll down the same plane, we can follow these steps:

### Step 1: Understand the scenario
We have a ring that can either slip or roll down an inclined plane of length \( L \). We need to find the ratio of the time taken to slip (\( t_1 \)) to the time taken to roll (\( t_2 \)).

### Step 2: Analyze the slipping case
When the ring is slipping down the incline, the only force causing it to accelerate is the component of its weight acting down the incline. The acceleration \( a_1 \) can be expressed as:
\[ a_1 = g \sin \theta \]
where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of inclination.

Using the equation of motion, we have:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ L = \frac{1}{2} a_1 t_1^2 \]
Substituting \( a_1 \):
\[ L = \frac{1}{2} (g \sin \theta) t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2L}{g \sin \theta} \]
Taking the square root:
\[ t_1 = \sqrt{\frac{2L}{g \sin \theta}} \]

### Step 3: Analyze the rolling case
When the ring rolls down the incline, both translational and rotational motions are involved. The effective acceleration \( a_2 \) for rolling can be expressed as:
\[ a_2 = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \]
For a ring, the moment of inertia \( I \) is \( mR^2 \). Thus:
\[ a_2 = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]

Using the same equation of motion:
\[ L = \frac{1}{2} a_2 t_2^2 \]
Substituting \( a_2 \):
\[ L = \frac{1}{2} \left(\frac{g \sin \theta}{2}\right) t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{4L}{g \sin \theta} \]
Taking the square root:
\[ t_2 = \sqrt{\frac{4L}{g \sin \theta}} = 2\sqrt{\frac{L}{g \sin \theta}} \]

### Step 4: Find the ratio \( \frac{t_1}{t_2} \)
Now we can find the ratio of the times:
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{2L}{g \sin \theta}}}{2\sqrt{\frac{L}{g \sin \theta}}} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the ratio of the time taken to slip down the incline to the time taken to roll down the incline is:
\[ \frac{t_1}{t_2} = \frac{1}{\sqrt{2}} \]