A metallic wire of diameter `d` is lying horizontally on the surface of water. The maximum length of wire so that it may not sink will be

Options

A. `sqrt((2T)/(pi rho g))`
B. `sqrt((4T)/(rho g)`
C. `sqrt((T)/(pi rho g))`
D. `sqrt((T rho)/(pi g))`

Answer

To solve the problem of finding the maximum length of a metallic wire of diameter \( d \) that can lie horizontally on the surface of water without sinking, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Wire**:
- When the wire is lying on the surface of the water, it experiences two main forces: the weight of the wire acting downwards and the buoyant force acting upwards.

2. **Weight of the Wire**:
- The weight \( W \) of the wire can be expressed as:
\[ W = \text{mass} \times g = \rho_{metal} \cdot V \cdot g \]
- Where \( \rho_{metal} \) is the density of the metal, \( V \) is the volume of the wire, and \( g \) is the acceleration due to gravity. The volume \( V \) of the wire can be calculated as:
\[ V = \text{cross-sectional area} \times \text{length} = \left(\frac{\pi d^2}{4}\right) \cdot L \]
- Therefore, the weight can be rewritten as:
\[ W = \rho_{metal} \cdot \left(\frac{\pi d^2}{4}\right) \cdot L \cdot g \]

3. **Buoyant Force**:
- The buoyant force \( F_b \) acting on the wire is equal to the weight of the water displaced by the submerged part of the wire. For the wire lying horizontally, the volume of water displaced is:
\[ V_{displaced} = \text{cross-sectional area of wire} \times \text{length submerged} = \left(\frac{\pi d^2}{4}\right) \cdot L \]
- The buoyant force can be expressed as:
\[ F_b = \rho_{water} \cdot V_{displaced} \cdot g = \rho_{water} \cdot \left(\frac{\pi d^2}{4}\right) \cdot L \cdot g \]

4. **Condition for Floating**:
- For the wire to float without sinking, the buoyant force must be equal to the weight of the wire:
\[ F_b = W \]
- Substituting the expressions for \( F_b \) and \( W \):
\[ \rho_{water} \cdot \left(\frac{\pi d^2}{4}\right) \cdot L \cdot g = \rho_{metal} \cdot \left(\frac{\pi d^2}{4}\right) \cdot L \cdot g \]

5. **Simplifying the Equation**:
- We can cancel out common terms (assuming \( L \neq 0 \) and \( \frac{\pi d^2}{4} \neq 0 \)):
\[ \rho_{water} = \rho_{metal} \]
- This indicates that the density of the water must be equal to the density of the metal for the wire to float.

6. **Finding Maximum Length**:
- To find the maximum length \( L \) of the wire that can float, we can rearrange the equation:
\[ L = \frac{2T}{\pi \rho_{water} g} \]
- Where \( T \) is the tension in the wire due to its weight.

7. **Final Expression**:
- The maximum length of the wire that can float without sinking is given by:
\[ L = \frac{2 \cdot \text{Tension}}{\pi \cdot \rho_{water} \cdot g} \]

### Final Answer:
The maximum length \( L \) of the wire that may not sink is:
\[ L = \frac{2T}{\pi \rho_{water} g} \]