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MHT CET 2021 (22 Sep, Shift 1) · Force due to Surface Tension

Force due to Surface Tension question from MHT CET 2021 (22 Sep, Shift 1)

Q17MHT CET 2021 · 22 SEP · SHIFT 1Difficult

Force due to Surface Tension

A drop of liquid of density 'ρ\rho' is floating half immersed in a liquid of density 'dd'. If 'TT' is the surface tension, then the diameter of the drop of the liquid is

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Answer: (d)

  1. Weight = buoyancy + surface force: 43πr3ρg=23πr3dg+2πrT\frac{4}{3}\pi r^3\rho g = \frac{2}{3}\pi r^3 d g + 2\pi r T.
  2. r2=3Tg(2ρ−d)r^2 = \dfrac{3T}{g(2\rho - d)}.
  3. Diameter D=2r=12Tg(2ρ−d)D = 2r = \sqrt{\dfrac{12T}{g(2\rho - d)}}. So the answer is (d).
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