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. ` sqrt ((2 sigma )/( pi rho l g ) ) `
B. `sqrt ((2 sigma l)/( pi rho g )) `
C. `sqrt ((sigma )/(pi rho g )) `
D. `sqrt ((2 sigma )/(pi rho g )) `

Explanation

To determine 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. **Understand the Forces Acting on the Needle**:
- The needle experiences a downward gravitational force due to its weight, \( mg \).
- The upward force due to surface tension acts along the length of the needle. For a needle of length \( L \), the surface tension force can be expressed as \( 2\sigma L \) (considering both sides of the needle).

2. **Establish the Balance of Forces**:
- For the needle to float, the upward force due to surface tension must balance the downward gravitational force:
\[ 2\sigma L \cos \theta = mg \]
- Here, \( \theta \) is the angle between the needle and the surface of the liquid.

3. **Express the Mass of the Needle**:
- The mass \( m \) of the needle can be expressed in terms of its density \( \rho \) and volume. For a cylindrical needle, the volume \( V \) is given by:
\[ V = \pi r^2 L \]
- Therefore, the mass is:
\[ m = \rho V = \rho \pi r^2 L \]

4. **Substitute the Mass into the Force Balance Equation**:
- Substituting \( m \) into the force balance equation gives:
\[ 2\sigma L \cos \theta = \rho \pi r^2 L g \]

5. **Cancel Common Terms**:
- Cancel \( L \) from both sides (assuming \( L \neq 0 \)):
\[ 2\sigma \cos \theta = \rho \pi r^2 g \]

6. **Rearrange to Solve for \( r^2 \)**:
- Rearranging the equation for \( r^2 \):
\[ r^2 = \frac{2\sigma \cos \theta}{\rho \pi g} \]

7. **Determine the Maximum Radius**:
- To find the maximum radius \( r \), we need to maximize \( r^2 \). The maximum value of \( \cos \theta \) is 1 (when the needle is horizontal):
\[ r^2_{\text{max}} = \frac{2\sigma \cdot 1}{\rho \pi g} = \frac{2\sigma}{\rho \pi g} \]
- Therefore, the maximum radius \( r \) is:
\[ r_{\text{max}} = \sqrt{\frac{2\sigma}{\rho \pi g}} \]

### Final Answer:
The needle will float on a liquid of surface tension \( \sigma \) if its radius \( r \) is less than or equal to:
\[ r \leq \sqrt{\frac{2\sigma}{\rho \pi g}} \]