A metal wire of length L and density d floats | vedclass

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Question

(A) For the wire to float, the downward force due to gravity must be balanced by the upward force due to surface tension and buoyancy.$1$. The weight

Vedclass · Accepted Answer

$\sqrt{\frac{2T}{\pi dg}}$

Vedclass · Answer

(A) For the wire to float, the downward force due to gravity must be balanced by the upward force due to surface tension and buoyancy.$1$. The weight of the wire is $W = mg = (	ext{Volume} 	imes 	ext{density}) 	imes g = (\pi r^2 L) d g$.$2$. The upward force due to surface tension acts along the two sides of the wire. The force is $F_s = 2 	imes (T 	imes L) = 2TL$.$3$. For the wire to just float, the weight must be equal to the surface tension force (ignoring buoyancy as the wire is thin and floating on the surface): $\pi r^2 L d g = 2TL$.$4$. Solving for $r$: $r^2 = \frac{2T}{\pi dg}$.$5$. Therefore, the maximum radius is $r = \sqrt{\frac{2T}{\pi dg}}$.

$A$ metal wire of length $L$ and density $d$ floats horizontally on the free surface of water. What is the maximum radius of the wire such that it does not sink in water? [$T$ = surface tension of water, $g$ = gravitational acceleration]

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