Question

A metal wire of length 'L' and density 'd' floats horizontally on the free surface of water. The maximum radius of the wire does not sink in water is
[ \(T\) = surface tension of water , \(g\) = gravitational acceleration]

• A. \(\sqrt{\frac{2dg}{πT}}\)
• B. \(\sqrt{\frac{2T}{πdg}}\)
• C. \(\sqrt{\frac{πdg}{2T}}\)
• D. \(\sqrt{\frac{T}{πdg}}\)

Hint:

The wire floats when the upward surface tension force equals its weight; the force acts along both sides of the wire.

Answer 1

The Correct Option is B

Step 1: Understand the concept
A thin wire floats because surface tension pulls on the water surface along its length. The surface acts on both sides of the wire, so the upward force is \(2TL\) (taking the force as vertical in the limiting case).

Step 2: Weight of the wire
The volume is \(\pi r^2L\) and the density is \(d\), so the weight is \(\pi r^2L\,d\,g\).

Step 3: Balance
\[ \pi r^2 L d g = 2TL \]

Step 4: Solve
\[ r^2 = \frac{2T}{\pi d g} \Rightarrow r = \sqrt{\frac{2T}{\pi d g}} \]
Option (B). The length \(L\) cancels, so the maximum radius does not depend on it. Option (D) would result if only one side of the wire were counted.

Final Answer:
The maximum radius is sqrt(2T/(pi d g)). This is option (B). \[ \boxed{\text{(B) }\sqrt{\frac{2T}{\pi d g}}} \]