A body is moving in a vertical circle of radius V by a string. If the ratio of maximum to minimum speed is `sqrt(3):1` , the ratio of maximum to minimum tensions in the string is:

Options

A. `3:1`
B. `5:1`
C. `7:1`
D. `9:1`

Answer

To solve the problem of finding the ratio of maximum to minimum tensions in the string when a body is moving in a vertical circle, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( v_1 \) = speed at the lowest point
- \( v_2 \) = speed at the highest point
- \( r \) = radius of the vertical circle
- \( g \) = acceleration due to gravity

Given that the ratio of maximum to minimum speed is:
\[ \frac{v_1}{v_2} = \sqrt{3}:1 \]

### Step 2: Use Conservation of Energy
Using the conservation of mechanical energy between the lowest and highest points:
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mgh \]
Where \( h \) is the height difference between the lowest and highest points, which is \( 2r \):
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2r) \]
This simplifies to:
\[ v_1^2 = v_2^2 + 4gr \]

### Step 3: Substitute the Speed Ratio
From the speed ratio, we have:
\[ v_1 = \sqrt{3}v_2 \]
Substituting this into the energy equation:
\[ (\sqrt{3}v_2)^2 = v_2^2 + 4gr \]
This leads to:
\[ 3v_2^2 = v_2^2 + 4gr \]
Rearranging gives:
\[ 2v_2^2 = 4gr \quad \Rightarrow \quad v_2^2 = 2gr \]
Now substituting back for \( v_1 \):
\[ v_1^2 = 3v_2^2 = 3(2gr) = 6gr \]

### Step 4: Calculate Tensions
Now, we can calculate the tensions at the lowest and highest points.

**Tension at the lowest point \( T_1 \)**:
\[ T_1 = \frac{mv_1^2}{r} + mg = \frac{m(6gr)}{r} + mg = 6mg + mg = 7mg \]

**Tension at the highest point \( T_2 \)**:
\[ T_2 = \frac{mv_2^2}{r} - mg = \frac{m(2gr)}{r} - mg = 2mg - mg = mg \]

### Step 5: Find the Ratio of Tensions
Now, we find the ratio of maximum to minimum tension:
\[ \frac{T_1}{T_2} = \frac{7mg}{mg} = 7 \]
Thus, the ratio of maximum to minimum tensions in the string is:
\[ \frac{T_1}{T_2} = 7:1 \]

### Final Answer
The ratio of maximum to minimum tensions in the string is \( 7:1 \).

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