A boy of mass 40 kg is hanging from the horizontal branch of a tree. The tension in his arms is minimum when the angle between the arms is:-

Options

A. `0^(@)`
B. `90^(@)`
C. `120^(@)`
D. `180^(@)`

Explanation

To solve the problem of finding the angle between the boy's arms when the tension in his arms is minimum, we can follow these steps:

### Step 1: Understand the Forces Acting on the Boy
The boy is hanging from a horizontal branch, and the forces acting on him include:
- The gravitational force (weight) acting downwards, which is equal to \( mg \), where \( m = 40 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \). Therefore, \( mg = 40 \times 10 = 400 \, \text{N} \).
- The tension in his arms, which we will denote as \( T \).

### Step 2: Define the Angle Between the Arms
Let the angle between the boy's arms be \( 2\theta \). This means that each arm makes an angle \( \theta \) with the vertical.

### Step 3: Resolve the Tension into Components
The tension \( T \) in each arm can be resolved into two components:
- A vertical component: \( T \cos \theta \)
- A horizontal component: \( T \sin \theta \)

### Step 4: Set Up the Vertical Force Balance
Since the boy is in equilibrium (not moving), the total upward force must equal the downward force due to gravity. Therefore, we can write:
\[ 2T \cos \theta = mg \]
Substituting \( mg = 400 \, \text{N} \):
\[ 2T \cos \theta = 400 \]

### Step 5: Solve for Tension
From the above equation, we can solve for the tension \( T \):
\[ T = \frac{400}{2 \cos \theta} = \frac{200}{\cos \theta} \]

### Step 6: Minimize the Tension
To find the angle that minimizes the tension, we need to maximize \( \cos \theta \). The maximum value of \( \cos \theta \) is 1, which occurs when \( \theta = 0^\circ \).

### Step 7: Conclusion
Thus, the angle between the boy's arms is:
\[ 2\theta = 2 \times 0^\circ = 0^\circ \]
This means the arms are straight down, and the tension in the arms is at its minimum.

### Final Answer
The angle between the arms is \( 0^\circ \).

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