A particel slides from rest from the topmost point of a vertical circle of radirs `r` along a smooth chord making an angle `theta` with the vertical. The time of descent is .

Options

A. Least for `theta =0`
B. Maximum for `theta =0`
C. Least for `theta =45`
D. Independent of `theta`

Explanation

To solve the problem of a particle sliding from rest from the topmost point of a vertical circle of radius \( r \) along a smooth chord making an angle \( \theta \) with the vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The particle starts from rest at the topmost point of a vertical circle of radius \( r \).
- It slides down along a chord that makes an angle \( \theta \) with the vertical.

2. **Identifying Key Points**:
- Let \( O \) be the center of the circle, \( P \) be the topmost point, and \( Q \) be the point where the particle reaches after sliding down the chord.
- The vertical distance from \( P \) to \( Q \) can be expressed in terms of \( r \) and \( \theta \).

3. **Finding the Length of the Chord**:
- The length \( L \) of the chord \( PQ \) can be calculated using trigonometry. In the triangle formed by points \( P \), \( O \), and \( Q \), we can express \( L \) as:
\[ L = 2r \cos(\theta) \]

4. **Determining the Acceleration**:
- The component of gravitational acceleration acting along the chord is given by:
\[ a = g \cos(\theta) \]
- Here, \( g \) is the acceleration due to gravity.

5. **Using the Equations of Motion**:
- Since the particle starts from rest, its initial velocity \( u = 0 \).
- We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Substituting \( u = 0 \) and \( s = L \):
\[ L = \frac{1}{2} a t^2 \]
- Substituting \( a = g \cos(\theta) \) and \( L = 2r \cos(\theta) \):
\[ 2r \cos(\theta) = \frac{1}{2} (g \cos(\theta)) t^2 \]

6. **Solving for Time \( t \)**:
- Rearranging the equation gives:
\[ 2r \cos(\theta) = \frac{1}{2} g \cos(\theta) t^2 \]
- Dividing both sides by \( \cos(\theta) \) (assuming \( \theta \neq 90^\circ \)):
\[ 2r = \frac{1}{2} g t^2 \]
- Multiplying both sides by 2:
\[ 4r = g t^2 \]
- Finally, solving for \( t \):
\[ t^2 = \frac{4r}{g} \quad \Rightarrow \quad t = 2 \sqrt{\frac{r}{g}} \]

### Final Result:
The time of descent \( t \) is given by:
\[ t = 2 \sqrt{\frac{r}{g}} \]
This shows that the time of descent is independent of the angle \( \theta \).