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Question

A thin metal disc of radius 'r', floats on water surface and bends the surface downwards along the perimeter making an angle θ with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is 'T', then the weight of metal disc is :

A
2πrT+W
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B
2πrTcosθW
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C
2πrTcosθ+W
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D
W2πrTcosθ
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Solution

The correct option is C 2πrTcosθ+W
The FBD as shown has three forces acting on the disc:
  • weight mg of the metal disc downward
  • force due to surface tension(T) upward at an angle θ with the vertical
  • buoyancy force upward
Equating all the forces,
mg=2πrTcosθ+W
Note that the horizontal component of the surface tension force Tsinθ will cancel by symmetry
134181_71719_ans.JPG

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