Two particles `1` and `2` are allowed to descend on the two frictionless chord `OA` and `OB` of a vertical circle, at the same instant from point `O`. The ratio of the velocities of the particles `1` and `2` respectively, when they reach on the circumference will be (OB is the diameter).

Options

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

Answer

To solve the problem, we need to analyze the motion of two particles descending along two different chords of a vertical circle. Let's denote the velocities of particles 1 and 2 as \( V_A \) and \( V_B \) respectively when they reach the circumference of the circle.

### Step 1: Analyze Particle 2 on Chord OB
For particle 2 descending along chord OB (which is the diameter of the circle), the only force acting on it is gravity. The distance it descends is equal to the diameter of the circle, which we can denote as \( d \).

Using the third equation of motion:
\[ V_B^2 - U^2 = 2a s \]
where:
- \( V_B \) is the final velocity of particle 2,
- \( U = 0 \) (initial velocity),
- \( a = g \) (acceleration due to gravity),
- \( s = d \) (distance fallen).

Substituting these values, we have:
\[ V_B^2 = 0 + 2gd \implies V_B = \sqrt{2gd} \]

### Step 2: Analyze Particle 1 on Chord OA
For particle 1 descending along chord OA, the situation is slightly different. The particle experiences a component of gravitational force along the chord. If we denote the angle between the vertical and the chord OA as \( \alpha \), the effective acceleration along the chord is \( g \cos(\alpha) \).

The distance fallen by particle 1 along the chord OA is also \( d \), but we need to consider the vertical component of this distance. The vertical distance covered by particle 1 can be expressed as \( d \cos(\alpha) \).

Using the third equation of motion again:
\[ V_A^2 - U^2 = 2a s \]
where:
- \( V_A \) is the final velocity of particle 1,
- \( U = 0 \) (initial velocity),
- \( a = g \cos(\alpha) \) (effective acceleration),
- \( s = d \cos(\alpha) \) (vertical distance).

Substituting these values, we have:
\[ V_A^2 = 0 + 2(g \cos(\alpha))(d \cos(\alpha)) \implies V_A^2 = 2gd \cos^2(\alpha) \]

### Step 3: Calculate the Ratio of Velocities
Now, we can find the ratio of the velocities \( V_A \) and \( V_B \):
\[ \frac{V_A^2}{V_B^2} = \frac{2gd \cos^2(\alpha)}{2gd} = \cos^2(\alpha) \]

Taking the square root to find the ratio of the velocities:
\[ \frac{V_A}{V_B} = \cos(\alpha) \]

### Final Answer
Thus, the ratio of the velocities of the particles when they reach the circumference is:
\[ \frac{V_A}{V_B} = \cos(\alpha) \]