A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity `sqrt(3 gr` at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

Options

A. `1 : 6`
B. `1 : 4`
C. `1 : 3`
D. `1 : 2`

Answer

To solve the problem, we need to find the ratio of the tensions in the string at the highest and lowest points of a vertical circular motion of a particle.

### Step-by-Step Solution:

1. **Identify Forces at the Highest Point:**
At the highest point of the circular motion, the forces acting on the particle are:
- The gravitational force (weight) acting downward: \( mg \)
- The tension in the string \( T_H \) acting downward
- The centripetal force required for circular motion: \( \frac{mv_H^2}{r} \)

The equation for the forces at the highest point can be written as:
\[ T_H + mg = \frac{mv_H^2}{r} \]

2. **Substituting the Given Velocity:**
We are given that the velocity at the highest point \( v_H = \sqrt{3gr} \). Substituting this into the equation gives:
\[ T_H + mg = \frac{m(\sqrt{3gr})^2}{r} \]
\[ T_H + mg = \frac{m(3gr)}{r} \]
\[ T_H + mg = 3mg \]
Rearranging this gives:
\[ T_H = 3mg - mg = 2mg \]

3. **Identify Forces at the Lowest Point:**
At the lowest point of the circular motion, the forces acting on the particle are:
- The gravitational force (weight) acting downward: \( mg \)
- The tension in the string \( T_L \) acting upward
- The centripetal force required for circular motion: \( \frac{mv_L^2}{r} \)

The equation for the forces at the lowest point can be written as:
\[ T_L - mg = \frac{mv_L^2}{r} \]

4. **Finding Velocity at the Lowest Point Using Energy Conservation:**
Using conservation of mechanical energy between the highest and lowest points:
\[ \frac{1}{2}mv_H^2 + mg(2r) = \frac{1}{2}mv_L^2 + mg(0) \]
Substituting \( v_H = \sqrt{3gr} \):
\[ \frac{1}{2}m(3gr) + 2mgr = \frac{1}{2}mv_L^2 \]
\[ \frac{3mgr}{2} + 2mgr = \frac{1}{2}mv_L^2 \]
\[ \frac{3mgr + 4mgr}{2} = \frac{1}{2}mv_L^2 \]
\[ \frac{7mgr}{2} = \frac{1}{2}mv_L^2 \]
Canceling \( \frac{1}{2}m \) gives:
\[ v_L^2 = 7gr \]

5. **Substituting Back to Find Tension at the Lowest Point:**
Now substituting \( v_L^2 = 7gr \) into the tension equation at the lowest point:
\[ T_L - mg = \frac{m(7gr)}{r} \]
\[ T_L - mg = 7mg \]
Rearranging gives:
\[ T_L = 7mg + mg = 8mg \]

6. **Finding the Ratio of Tensions:**
Now we can find the ratio of the tensions at the highest and lowest points:
\[ \text{Ratio} = \frac{T_H}{T_L} = \frac{2mg}{8mg} = \frac{2}{8} = \frac{1}{4} \]

### Final Answer:
The ratio of the respective tensions in the string at the highest and lowest points is \( 1:4 \).