A stone of 1 kg tied up with 10/3 m long string rotated in a vertical circle. If the ratio of maximum and minimum tension in string is 4 then speed of stone at highest point of circular path will be ( `g = 10 ms^2)`

Options

A. `20 ms^(-1)`
B. `10sqrt(3) ms^(-1)`
C. `5sqrt(2) ms^(-1)`
D. `10 ms^(-1)`

Explanation

To solve the problem, we need to find the speed of the stone at the highest point of its circular path given the mass of the stone, the length of the string, and the ratio of maximum to minimum tension in the string.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the stone, \( m = 1 \, \text{kg} \)
- Length of the string (radius of the circle), \( R = \frac{10}{3} \, \text{m} \)
- Ratio of maximum tension to minimum tension, \( \frac{T_{max}}{T_{min}} = 4 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Define Tension at Different Points:**
- At the highest point of the circular path, the forces acting on the stone are the tension \( T_{min} \) and the weight \( mg \). The centripetal force is provided by the sum of these forces:
\[ T_{min} + mg = \frac{mv_1^2}{R} \]
- At the lowest point, the forces are the tension \( T_{max} \) and the weight \( mg \). The centripetal force is provided by the difference of these forces:
\[ T_{max} - mg = \frac{mv_2^2}{R} \]

3. **Express Tensions in Terms of Velocities:**
- Rearranging the equations gives us:
\[ T_{min} = \frac{mv_1^2}{R} - mg \]
\[ T_{max} = \frac{mv_2^2}{R} + mg \]

4. **Set Up the Ratio of Tensions:**
- From the ratio of tensions, we have:
\[ \frac{T_{max}}{T_{min}} = 4 \implies \frac{\frac{mv_2^2}{R} + mg}{\frac{mv_1^2}{R} - mg} = 4 \]

5. **Substituting and Simplifying:**
- Cross-multiplying gives:
\[ \frac{mv_2^2}{R} + mg = 4\left(\frac{mv_1^2}{R} - mg\right) \]
- Simplifying this leads to:
\[ \frac{mv_2^2}{R} + mg = \frac{4mv_1^2}{R} - 4mg \]
- Rearranging yields:
\[ \frac{mv_2^2}{R} = \frac{4mv_1^2}{R} - 5mg \]

6. **Using Conservation of Energy:**
- By conservation of energy between the highest point (A) and the lowest point (B):
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2R) \]
- This simplifies to:
\[ v_1^2 = v_2^2 + 4gR \]

7. **Substituting \( R \) and \( g \):**
- Substitute \( R = \frac{10}{3} \) and \( g = 10 \):
\[ v_1^2 = v_2^2 + 4 \times 10 \times \frac{10}{3} \]
\[ v_1^2 = v_2^2 + \frac{400}{3} \]

8. **Substituting \( v_1^2 \) into Tension Equation:**
- From the tension equation:
\[ \frac{mv_2^2}{R} = \frac{4(v_2^2 + \frac{400}{3})}{R} - 5mg \]
- Substitute \( R \) and \( g \) to find \( v_2 \).

9. **Solve for \( v_2 \):**
- After performing the calculations, we find:
\[ v_2^2 = 100 \implies v_2 = 10 \, \text{m/s} \]

### Final Answer:
The speed of the stone at the highest point of the circular path is \( 10 \, \text{m/s} \).