A vertical disc has three grooves directed along chords AB, AC and AD. Three bodies begin to slide down the respective grooves from A simultaneously. If AB > AC > AD, the respective time intervals to reach the bottoms of the respective grooves `t_1, t_2 and t_3` are 

Options

A. `t_1 gt t_2 gt t_3`
B. `t_1 lt t_2 lt t_3`
C. `t_1 gt t_2 lt t_3`
D. `t_1 = t_2 = t_3`

Answer

To solve the problem, we need to analyze the motion of the bodies sliding down the grooves of the vertical disc. Let's break down the solution step by step:

### Step 1: Understand the Setup
We have a vertical disc with three grooves (AB, AC, and AD) that are directed along chords of the disc. The lengths of the grooves are given as \( AB > AC > AD \).

### Step 2: Identify the Forces Acting on the Bodies
As the bodies slide down the grooves, the only force acting on them is gravity. The effective acceleration along each groove can be expressed as:
\[ a = g \cos \theta \]
where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of the groove with respect to the vertical.

### Step 3: Determine the Length of Each Groove
The length of each groove can be represented as:
- For groove AB: \( L_1 = 2R \cos \theta_1 \)
- For groove AC: \( L_2 = 2R \cos \theta_2 \)
- For groove AD: \( L_3 = 2R \cos \theta_3 \)

Given that \( AB > AC > AD \), we can infer that \( \theta_1 < \theta_2 < \theta_3 \).

### Step 4: Apply the Kinematic Equation
Using the kinematic equation:
\[ S = ut + \frac{1}{2} a t^2 \]
Since the bodies start from rest, \( u = 0 \), and the equation simplifies to:
\[ S = \frac{1}{2} a t^2 \]

### Step 5: Substitute for Each Groove
For each groove, we can substitute the respective lengths and accelerations:
1. For groove AB:
\[ L_1 = \frac{1}{2} (g \cos \theta_1) t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2L_1}{g \cos \theta_1} \]

2. For groove AC:
\[ L_2 = \frac{1}{2} (g \cos \theta_2) t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{2L_2}{g \cos \theta_2} \]

3. For groove AD:
\[ L_3 = \frac{1}{2} (g \cos \theta_3) t_3^2 \]
Rearranging gives:
\[ t_3^2 = \frac{2L_3}{g \cos \theta_3} \]

### Step 6: Analyze the Time Intervals
From the equations derived, we can see that the time taken for each body to reach the bottom depends on the length of the groove and the angle. However, since the acceleration due to gravity \( g \) and the factor of \( \frac{2}{g} \) is constant for all grooves, the time intervals \( t_1, t_2, t_3 \) will depend primarily on the cosine of the angles.

### Step 7: Conclusion
Since the angles \( \theta_1, \theta_2, \theta_3 \) will affect the time taken, and given that \( AB > AC > AD \) implies that \( \theta_1 < \theta_2 < \theta_3 \), we can conclude:
- \( t_1 < t_2 < t_3 \)

Thus, the respective time intervals to reach the bottoms of the respective grooves are:
\[ t_1 < t_2 < t_3 \]