A stone of mass 1 kg tied to a string 4 m long and is rotated at constant speed of 40 m/s a vertical circle. The ratio of the tension at the top and the bottom, is `(g=10" m"//"s"^(2))`

Options

A. `11 : 12`
B. `39 : 41`
C. `41: 39`
D. `12 : 11`

Answer

To find the ratio of the tension at the top (T_p) and the bottom (T_b) of a vertical circle in which a stone of mass 1 kg is rotated at a constant speed of 40 m/s, we can follow these steps:

### Step 1: Identify the forces acting on the stone at the top and bottom of the circle.

At the top of the circle:
- The tension (T_p) acts downward.
- The weight (mg) also acts downward.

At the bottom of the circle:
- The tension (T_b) acts upward.
- The weight (mg) acts downward.

### Step 2: Write the equations for the forces at the top and bottom of the circle.

**At the top:**
The net force towards the center (centripetal force) is given by:
\[ T_p + mg = \frac{mv^2}{L} \]
Where:
- m = mass of the stone = 1 kg
- v = speed = 40 m/s
- L = length of the string = 4 m
- g = acceleration due to gravity = 10 m/s²

Rearranging gives:
\[ T_p = \frac{mv^2}{L} - mg \]

**At the bottom:**
The net force towards the center is given by:
\[ T_b - mg = \frac{mv^2}{L} \]
Rearranging gives:
\[ T_b = \frac{mv^2}{L} + mg \]

### Step 3: Substitute the values into the equations.

**Calculate T_p:**
1. Calculate \( \frac{mv^2}{L} \):
\[ \frac{mv^2}{L} = \frac{1 \cdot (40)^2}{4} = \frac{1600}{4} = 400 \, \text{N} \]
2. Calculate \( mg \):
\[ mg = 1 \cdot 10 = 10 \, \text{N} \]
3. Substitute into the equation for T_p:
\[ T_p = 400 - 10 = 390 \, \text{N} \]

**Calculate T_b:**
1. Using the same \( \frac{mv^2}{L} \) and \( mg \):
2. Substitute into the equation for T_b:
\[ T_b = 400 + 10 = 410 \, \text{N} \]

### Step 4: Calculate the ratio of T_p to T_b.

The ratio is given by:
\[ \frac{T_p}{T_b} = \frac{390}{410} \]
This can be simplified:
\[ \frac{T_p}{T_b} = \frac{39}{41} \]

### Final Answer:
The ratio of the tension at the top to the tension at the bottom is \( \frac{39}{41} \).

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