A needle of length l and density ` rho ` will float on a liquid of surface tension ` sigma ` if its radius r is less than or equal to:

Options

A. (a) ` sqrt ((2 sigma )/( pi rho l g ) ) `
B. (b)`sqrt ((2 sigma l)/( pi rho g )) `
C. (c)`sqrt ((sigma )/(pi rho g )) `
D. (d)`sqrt ((2 sigma )/(pi rho g )) `

Explanation

To solve the problem of determining the maximum radius \( r \) of a needle that can float on a liquid with surface tension \( \sigma \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces Acting on the Needle**:
- When the needle is floating on the liquid surface, it experiences two main forces: the gravitational force \( mg \) acting downwards and the upward force due to surface tension.
- The surface tension force acts along the length of the needle and can be resolved into vertical components.

2. **Setting Up the Equilibrium Condition**:
- For the needle to float, the upward force due to surface tension must balance the downward gravitational force. Therefore, we can write:
\[ \sum F_y = 0 \implies 2F \cos(\theta) - mg = 0 \]
- Here, \( F \) is the total force due to surface tension acting on both sides of the needle.

3. **Expressing the Surface Tension Force**:
- The surface tension force \( F \) can be expressed as:
\[ F = \sigma L \]
- where \( L \) is the length of the needle.

4. **Substituting the Forces into the Equilibrium Equation**:
- Substituting \( F \) into the equilibrium equation gives:
\[ 2(\sigma L) \cos(\theta) = mg \]

5. **Relating Mass to Density**:
- The mass \( m \) of the needle can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ m = \rho V = \rho (\pi r^2 L) \]
- Substituting this into the equilibrium equation yields:
\[ 2\sigma L \cos(\theta) = \rho (\pi r^2 L) g \]

6. **Simplifying the Equation**:
- Cancel \( L \) from both sides (assuming \( L \neq 0 \)):
\[ 2\sigma \cos(\theta) = \rho \pi r^2 g \]

7. **Finding Maximum Radius Condition**:
- To find the maximum radius \( r \), we consider the case where \( \cos(\theta) \) is maximized (i.e., \( \cos(\theta) = 1 \)):
\[ 2\sigma = \rho \pi r^2 g \]

8. **Solving for \( r \)**:
- Rearranging the equation gives:
\[ r^2 = \frac{2\sigma}{\rho \pi g} \]
- Taking the square root to find \( r \):
\[ r = \sqrt{\frac{2\sigma}{\rho \pi g}} \]

9. **Final Expression for Maximum Radius**:
- Thus, the maximum radius \( r \) of the needle that can float on the liquid is:
\[ r \leq \sqrt{\frac{2\sigma}{\rho \pi g}} \]