A needle of length l m and mass m kg is placed horizontal on wter surface habing surface tension T Find T in terms of m, l (g acceleration due to gravity)

Options

A. `T=(mg)/(2l)`
B. `T=(mg)/(l)`
C. `T=(3mg)/(2l)`
D. `T=(m)/(2l)`

Explanation

To solve the problem, we need to analyze the forces acting on the needle placed horizontally on the surface of the water due to surface tension.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Needle:**
- The needle has a weight \( mg \) acting downwards due to gravity, where \( m \) is the mass of the needle and \( g \) is the acceleration due to gravity.
- The surface tension \( T \) acts along the surface of the water at both ends of the needle. Since the needle is horizontal, the surface tension will have a vertical component that helps balance the weight of the needle.

2. **Determine the Components of Surface Tension:**
- The surface tension acts at an angle \( \theta \) with respect to the horizontal. The vertical component of the surface tension at each end of the needle can be expressed as \( T \sin(\theta) \).
- Since there are two ends of the needle, the total upward force due to surface tension is \( 2T \sin(\theta) \).

3. **Set Up the Force Balance Equation:**
- For the needle to be in equilibrium, the total upward force due to surface tension must equal the downward gravitational force:
\[ 2T \sin(\theta) = mg \]

4. **Approximate for Small Angles:**
- If the angle \( \theta \) is very small (which is often the case for a needle floating on water), we can use the approximation \( \sin(\theta) \approx \theta \) and \( \cos(\theta) \approx 1 \).
- Therefore, we can rewrite the equation as:
\[ 2T \theta = mg \]

5. **Relate the Angle to the Length of the Needle:**
- The angle \( \theta \) can be approximated as the ratio of the length of the needle \( l \) to the distance from the center of the needle to the surface (which is negligible for our calculations). Thus, we can consider \( \theta \) as \( \frac{l}{L} \) where \( L \) is some characteristic length (but for simplicity, we can assume it to be negligible).
- However, since we are looking for \( T \) in terms of \( m \), \( l \), and \( g \), we can directly relate:
\[ \theta \approx 1 \quad (\text{as } \theta \text{ approaches } 0) \]

6. **Solve for Surface Tension \( T \):**
- Rearranging the equation gives:
\[ T = \frac{mg}{2\theta} \]
- Substituting \( \theta \approx 1 \):
\[ T = \frac{mg}{2l} \]

### Final Result:
Thus, the surface tension \( T \) in terms of \( m \), \( l \), and \( g \) is:
\[ T = \frac{mg}{2l} \]