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 will follow these steps:

### Step 1: Understand the Forces at Play
At the highest point of the circular path, the forces acting on the stone are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_{min} \) in the string also acting downwards.

At the lowest point, the forces are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_{max} \) in the string acting upwards.

### Step 2: Write the Equations for Centripetal Force
At the highest point, the centripetal force is provided by the sum of the tension and the weight of the stone:
\[ T_{min} + mg = \frac{mv^2}{L} \]
At the lowest point, the centripetal force is provided by the difference between the tension and the weight:
\[ T_{max} - mg = \frac{mu^2}{L} \]

### Step 3: Relate Maximum and Minimum Tension
We are given that the ratio of maximum to minimum tension is 4:
\[ \frac{T_{max}}{T_{min}} = 4 \implies T_{max} = 4T_{min} \]

### Step 4: Substitute Tension Values
Substituting \( T_{max} \) in terms of \( T_{min} \) into the centripetal force equation at the lowest point:
\[ 4T_{min} - mg = \frac{mu^2}{L} \]

### Step 5: Solve for Tension at the Highest Point
From the equation for the highest point, we can express \( T_{min} \):
\[ T_{min} = \frac{mv^2}{L} - mg \]

### Step 6: Substitute \( T_{min} \) into the Equation for \( T_{max} \)
Now, substitute this expression for \( T_{min} \) into the equation for \( T_{max} \):
\[ 4\left(\frac{mv^2}{L} - mg\right) - mg = \frac{mu^2}{L} \]
Expanding this gives:
\[ \frac{4mv^2}{L} - 4mg - mg = \frac{mu^2}{L} \]
\[ \frac{4mv^2}{L} - 5mg = \frac{mu^2}{L} \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 4mv^2 - 5mgL = mu^2 \]

### Step 8: Use Work-Energy Theorem
Using the work-energy theorem between the highest and lowest points:
\[ W = -mg \cdot 2L = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]
This leads to:
\[ -mg \cdot 2L = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]

### Step 9: Solve for \( v \)
Now we have two equations:
1. \( 4mv^2 - 5mgL = mu^2 \)
2. \( -2mgL = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \)

By substituting the values of \( m = 1 \, kg \), \( g = 10 \, m/s^2 \), and \( L = \frac{10}{3} \, m \) into these equations, we can solve for \( v \).

### Step 10: Final Calculation
Substituting the values:
1. \( 4v^2 - 5(10)(\frac{10}{3}) = u^2 \)
2. \( -2(10)(\frac{10}{3}) = \frac{1}{2}v^2 - \frac{1}{2}u^2 \)

After solving these equations, we find:
\[ v = \sqrt{30} \approx 10 \, m/s \]

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

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