A dry clean steel needle of diameter d and density `rho` when carefully placed on the surface of water remains floating . If T is the surface tension of water , then maximum value for the diameter d of the needle for enabling it to float will be

Options

A. `d=sqrt((8 rho pi)/(T g ))`
B. `d=sqrt((4 rho pi)/(T g ))`
C. `d=sqrt((8 T)/(rho pi g ))`
D. data incomplete

Explanation

To solve the problem, we need to find the maximum diameter \( d \) of a steel needle that can float on the surface of water, given its density \( \rho \) and the surface tension \( T \) of water.

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Needle**:
- The needle floats due to the balance between its weight and the upward force due to surface tension.
- The weight of the needle \( W \) is given by \( W = m \cdot g \), where \( m \) is the mass of the needle and \( g \) is the acceleration due to gravity.

2. **Calculating the Volume and Mass of the Needle**:
- The volume \( V \) of the needle can be expressed as:
\[ V = A \cdot L = \left(\frac{\pi d^2}{4}\right) \cdot L \]
- The mass \( m \) of the needle can be calculated using its density \( \rho \):
\[ m = \rho \cdot V = \rho \cdot \left(\frac{\pi d^2}{4}\right) \cdot L \]

3. **Expressing the Weight of the Needle**:
- The weight \( W \) of the needle is then:
\[ W = m \cdot g = \left(\rho \cdot \frac{\pi d^2}{4} \cdot L\right) \cdot g = \frac{\rho \pi d^2 L g}{4} \]

4. **Calculating the Surface Tension Force**:
- The surface tension force \( F_T \) acting on the needle is given by:
\[ F_T = T \cdot \text{(length in contact with surface)} \]
- Since the needle has two ends in contact with the surface, the effective length in contact is \( 2L \):
\[ F_T = T \cdot 2L \]

5. **Setting Up the Equilibrium Condition**:
- For the needle to float, the upward surface tension force must equal the downward weight of the needle:
\[ F_T = W \]
- Substituting the expressions for \( F_T \) and \( W \):
\[ T \cdot 2L = \frac{\rho \pi d^2 L g}{4} \]

6. **Cancelling \( L \) from Both Sides**:
- Since \( L \) appears on both sides of the equation, we can cancel it out (assuming \( L \neq 0 \)):
\[ 2T = \frac{\rho \pi d^2 g}{4} \]

7. **Rearranging to Solve for \( d^2 \)**:
- Rearranging the equation to isolate \( d^2 \):
\[ d^2 = \frac{8T}{\rho \pi g} \]

8. **Taking the Square Root to Find \( d \)**:
- Finally, taking the square root gives us the maximum diameter \( d \):
\[ d = \sqrt{\frac{8T}{\rho \pi g}} \]

### Final Answer:
The maximum value for the diameter \( d \) of the needle for enabling it to float is:
\[ d = \sqrt{\frac{8T}{\rho \pi g}} \]