Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15kg weight is attached to the mid point which is no longer remains horizontal. The minimum tension required to completely straighten the rope is:

Options

A. 15kg
B. 15/2 kg
C. 5 kg
D. Infinitely large

Explanation

To solve the problem of finding the minimum tension required to completely straighten the rope when a 15 kg weight is attached to its midpoint, we can follow these steps:

### Step 1: Identify the Weight
The weight attached to the midpoint of the rope is given as 15 kg. We can calculate the weight (force due to gravity) using the formula:
\[ \text{Weight} = m \cdot g \]
where \( m = 15 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

Calculating the weight:
\[ \text{Weight} = 15 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 147 \, \text{N} \]
For simplicity, we can round this to 150 N.

### Step 2: Set Up the Forces
When the weight is attached to the midpoint, the rope will form an angle \( \theta \) with the horizontal. The tension \( T \) in the rope will have 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 object. Therefore, we can write:
\[ 2T \sin \theta = 150 \, \text{N} \]

### Step 3: Solve for Tension
From the equation above, we can express the tension \( T \):
\[ T \sin \theta = \frac{150}{2} = 75 \, \text{N} \]
Thus,
\[ T = \frac{75}{\sin \theta} \]

### Step 4: Find Minimum Tension
To find the minimum tension, we need to maximize \( \sin \theta \). The maximum value of \( \sin \theta \) is 1, which occurs when \( \theta = 90^\circ \). Therefore, substituting this into our equation gives:
\[ T_{\text{min}} = \frac{75}{1} = 75 \, \text{N} \]

### Conclusion
The minimum tension required to completely straighten the rope is:
\[ \boxed{75 \, \text{N}} \]