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MHT CET 2021 (24 Sep, Shift 1) · Force due to Surface Tension

Force due to Surface Tension question from MHT CET 2021 (24 Sep, Shift 1)

Q16MHT CET 2021 · 24 SEP · SHIFT 1Difficult

Force due to Surface Tension

A thin metal disc of radius 'r' floats on water surface and bends the surface downwards along the perimeter making an angle 'θ\theta' with the vertical edge of the disc. If the weight of water displaced by the disc is 'WW', the weight of the metal disc is [T = surface tension of water]

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Answer: (a)

  1. Upward forces on the disc: buoyancy WW and vertical component of surface tension along the rim.
  2. Surface tension makes angle θ\theta with the vertical: vertical part =2πrTcos⁡θ= 2\pi r T\cos\theta.
  3. Weight of disc =2πrTcos⁡θ+W= 2\pi r T\cos\theta + W. So the answer is (a).
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