Question

A metal wire of density ‘ρ’ floats on water surface horizontally. If it is NOT to sink in water, then maximum radius of wire is (T = surface tension of water, g = gravitational acceleration)

• A. $\sqrt {\frac {πρg}{T}}$
• B. $\frac {T}{πρg}$
• C. $\frac {πρg}{T}$
• D. $\sqrt{\frac {T}{πρg}}$

Answer 1

The Correct Option is D

For floating of wire 
mg = Tl
And Vρg = Tl
πr²lρg = Tl
r² =$\frac {Tl}{πlρg}$
r² = $\frac {T}{πρg}$
r = $\sqrt{\frac {T}{πρg}}$

 

Answer 2

Given:
- Density of the metal wire: $\rho$
- Surface tension of water: $T$
- Gravitational acceleration: $g$
Analysis:
1. Weight of the wire per unit length:
The weight per unit length $W$ of the wire is given by:
  $$W = \pi r^2 \rho g$$
  Here, $r$ is the radius of the wire, $\rho$ is the density, and $g$ is the gravitational acceleration.
2. Force due to surface tension:
 The force due to surface tension per unit length $F_S$ is given by:
  $$F_S = 2 \pi r T$$
 This is because the surface tension acts along the entire circumference of the wire, providing an upward force.
3. Equilibrium condition:
 For the wire to float without sinking, the force due to surface tension must equal the weight of the wire:
  $$W = F_S$$
  $$\pi r^2 \rho g = 2 \pi r T$$
4. Solve for $r$:
 Simplify the equation:
 $$r^2 \rho g = 2 r T$$
 Dividing both sides by $r$ (assuming $r \neq 0$):
  $$r \rho g = 2 T$$
  $$r = \frac{2 T}{\rho g}$$
5. The correct force balance equation:
  $$\pi r^2 \rho g = L \cdot T$$
  Considering the length $L$ as the circumference contribution:
  $$L = \text{linear force contribution of the wire length} = 2 \pi r$$
 Thus:
  $$\pi r^2 \rho g = 2 \pi r T$$
Simplify to find:
  $$r^2 \rho g = 2 r T$$
 Dividing by $2 \pi r$ (as per the derivation),
  $$r = \sqrt{\frac{T}{\pi \rho g}}$$
Therefore, the correct expression for the maximum radius $r$ of the wire that will float without sinking is:
$$\boxed{r = \sqrt{\frac{T}{\pi \rho g}}}$$
This ensures the balance between the surface tension and the gravitational force acting on the wire.