A thin metal disc of radius r floats on the w | vedclass

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(C) For the disc to be in equilibrium,the total downward force (weight of the disc) must be equal to the total upward force.The total upward force con

Vedclass · Accepted Answer

$2\pi rT \cos 	heta + W$

Vedclass · Answer

(C) For the disc to be in equilibrium,the total downward force (weight of the disc) must be equal to the total upward force.The total upward force consists of:$1$. The upthrust (buoyant force) acting on the disc,which is equal to the weight of the displaced water,$W$.$2$. The vertical component of the surface tension force acting along the perimeter of the disc.The surface tension force $F_s$ acts along the perimeter $2\pi r$ at an angle $	heta$ with the vertical. The vertical component of this force is $F_{s,v} = T 	imes (2\pi r) 	imes \cos 	heta$.Therefore,the weight of the metal disc $Mg = W + 2\pi rT \cos 	heta$.

$A$ thin metal disc of radius $r$ floats on the water surface and bends the surface downwards along the perimeter,making an angle $\theta$ with the vertical edge of the disc. If the disc displaces a weight of water $W$ and the surface tension of water is $T$,then the weight of the metal disc is:

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