A stone of mass 1 kg tied to a light inextensible string of lenth `L=(10)/(3)m`, whirling in a circular path in a vertical plane. The ratio of maximum tension to the minimum tension in the string is 4. If g is taken to be `10ms^(-2)`, the speed of the stone at the highest point of the circle is

Options

A. `10ms^(-1)`
B. `5sqrt2ms^(-1)`
C. `10sqrt3ms^(-1)`
D. `20ms^(-1)`

Explanation

To find the speed of the stone at the highest point of its circular motion, we need to analyze the forces acting on the stone at both the highest and lowest points of the circular path. Let's break down the solution step by step.

### Step 1: Identify the Forces at the Highest Point
At the highest point of the circular path, the forces acting on the stone are:
- The tension in the string (T_a) acting downward.
- The weight of the stone (mg) acting downward.

The net force towards the center of the circle (centripetal force) is given by:
\[ T_a + mg = \frac{mv_a^2}{L} \]
where:
- \( m = 1 \, \text{kg} \) (mass of the stone)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( L = \frac{10}{3} \, \text{m} \) (length of the string, which is the radius of the circular path)
- \( v_a \) is the speed at the highest point.

### Step 2: Identify the Forces at the Lowest Point
At the lowest point of the circular path, the forces acting on the stone are:
- The tension in the string (T_b) acting upward.
- The weight of the stone (mg) acting downward.

The net force towards the center of the circle is given by:
\[ T_b - mg = \frac{mv_b^2}{L} \]
where \( v_b \) is the speed at the lowest point.

### Step 3: Set Up the Equations
From the equations derived in Steps 1 and 2, we can express the tensions:
1. From the highest point:
\[ T_a = \frac{mv_a^2}{L} - mg \]
\[ T_a = \frac{1 \cdot v_a^2}{\frac{10}{3}} - 1 \cdot 10 \]
\[ T_a = \frac{3v_a^2}{10} - 10 \]

2. From the lowest point:
\[ T_b = \frac{mv_b^2}{L} + mg \]
\[ T_b = \frac{1 \cdot v_b^2}{\frac{10}{3}} + 1 \cdot 10 \]
\[ T_b = \frac{3v_b^2}{10} + 10 \]

### Step 4: Use the Ratio of Tensions
We are given that the ratio of maximum tension to minimum tension is 4:
\[ \frac{T_b}{T_a} = 4 \]

Substituting the expressions for \( T_a \) and \( T_b \):
\[ \frac{\frac{3v_b^2}{10} + 10}{\frac{3v_a^2}{10} - 10} = 4 \]

### Step 5: Cross Multiply and Simplify
Cross-multiplying gives:
\[ 10(3v_b^2 + 100) = 4(3v_a^2 - 100) \]
Expanding both sides:
\[ 30v_b^2 + 1000 = 12v_a^2 - 400 \]
Rearranging gives:
\[ 12v_a^2 - 30v_b^2 = 1400 \]

### Step 6: Use Conservation of Energy
Using conservation of energy between the highest and lowest points:
\[ \frac{1}{2}mv_b^2 = \frac{1}{2}mv_a^2 + mg(2L) \]
Substituting values:
\[ \frac{1}{2}v_b^2 = \frac{1}{2}v_a^2 + 10 \cdot \left(2 \cdot \frac{10}{3}\right) \]
\[ \frac{1}{2}v_b^2 = \frac{1}{2}v_a^2 + \frac{200}{3} \]

### Step 7: Solve the System of Equations
Now we have two equations:
1. \( 12v_a^2 - 30v_b^2 = 1400 \)
2. \( \frac{1}{2}v_b^2 = \frac{1}{2}v_a^2 + \frac{200}{3} \)

From the second equation, express \( v_b^2 \):
\[ v_b^2 = v_a^2 + \frac{400}{3} \]

Substituting this into the first equation:
\[ 12v_a^2 - 30\left(v_a^2 + \frac{400}{3}\right) = 1400 \]
\[ 12v_a^2 - 30v_a^2 - 4000 = 1400 \]
\[ -18v_a^2 = 5400 \]
\[ v_a^2 = 300 \]
\[ v_a = \sqrt{300} = 10\sqrt{3} \, \text{m/s} \]

### Final Answer
The speed of the stone at the highest point of the circle is:
\[ v_a = 10\sqrt{3} \, \text{m/s} \]