A rope of mass 0.1 kg is connected at the same height of two opposite walls. It is allowed to hang under its own weight. At the contact point between the rope and the wall, the rope makes an angle ` theta = 10^(@)` with respect to horizontal. The tension in the rope at its midpoint between the wall is

Options

A. 2.78 N
B. 2.56 N
C. 2.82 N
D. 2 .71 N

Explanation

To solve the problem of finding the tension in the rope at its midpoint, we will follow these steps:

### Step 1: Understand the Setup
We have a rope of mass \( m = 0.1 \, \text{kg} \) hanging between two walls, making an angle \( \theta = 10^\circ \) with the horizontal. The rope is in static equilibrium under the influence of its weight.

### Step 2: Identify Forces Acting on the Rope
The weight of the rope acts downwards at its midpoint. The weight \( W \) of the rope can be calculated using the formula:
\[ W = mg \]
where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

### Step 3: Calculate the Weight of the Rope
Substituting the values:
\[ W = 0.1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.98 \, \text{N} \]

### Step 4: Analyze the Tension Components
At the midpoint of the rope, the tension \( T \) has two components:
- A vertical component \( T \sin \theta \)
- A horizontal component \( T \cos \theta \)

Since the rope is in equilibrium, the vertical components of the tension must balance the weight of the rope:
\[ 2T \sin \theta = W \]
Here, \( 2T \sin \theta \) accounts for the contributions from both sides of the rope.

### Step 5: Substitute the Weight into the Equation
Now, substituting \( W \):
\[ 2T \sin \theta = 0.98 \, \text{N} \]

### Step 6: Solve for Tension \( T \)
Rearranging the equation gives:
\[ T = \frac{0.98}{2 \sin \theta} \]

### Step 7: Calculate \( \sin \theta \)
For \( \theta = 10^\circ \), we can use a calculator or trigonometric tables to find:
\[ \sin 10^\circ \approx 0.1736 \]

### Step 8: Substitute \( \sin \theta \) into the Tension Equation
Now, substituting \( \sin 10^\circ \):
\[ T = \frac{0.98}{2 \times 0.1736} \approx \frac{0.98}{0.3472} \approx 2.83 \, \text{N} \]

### Step 9: Conclusion
Thus, the tension in the rope at its midpoint is approximately:
\[ T \approx 2.83 \, \text{N} \]