A thin metal disc of radius r floats on water | vedclass

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Question

(A) The weight of the disc is balanced by the upward force due to surface tension and the buoyant force (upthrust) of water.The surface tension $T$ ac

Vedclass · Accepted Answer

$2 \pi r \cos 	heta + W$

Vedclass · Answer

(A) The weight of the disc is balanced by the upward force due to surface tension and the buoyant force (upthrust) of water.The surface tension $T$ acts along the perimeter $2 \pi r$ at an angle $	heta$ with the vertical.The vertical component of the surface tension force is $F_s = T \cdot (2 \pi r) \cdot \cos 	heta$.The buoyant force (upthrust) is equal to the weight of the displaced water,which is given as $W$.For the disc to float in equilibrium,the total downward force (weight of the disc $W_{disc}$) must equal the total upward force.Therefore,$W_{disc} = F_s + W$.$W_{disc} = 2 \pi r T \cos 	heta + W$.

$A$ thin metal disc of radius $r$ floats on water surface and bends the surface downwards along the perimeter,making an angle $\theta$ with the vertical edge of the disc. If the weight of water displaced by the disc is $W$,the weight of the metal disc is [$T =$ surface tension of water].

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