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JEE Main 2020 (9 Jan, Shift 2) · Force due to Surface Tension

Force due to Surface Tension question from JEE Main 2020 (9 Jan, Shift 2)

Q19JEE Main 2020 · 9 JAN · SHIFT 2Difficult

Force due to Surface Tension

A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ\rho and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

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Answer: (d)

  1. Weight = buoyancy (half volume) + upward surface force 2πrT2\pi r T.
  2. 43πr3dg=23πr3ρg+2πrT⇒23r2g(2d−ρ)=2T\frac{4}{3}\pi r^3 d g = \frac{2}{3}\pi r^3\rho g + 2\pi r T \Rightarrow \frac{2}{3}r^2 g(2d - \rho) = 2T.
  3. r=3T(2d−ρ)gr = \sqrt{\dfrac{3T}{(2d - \rho)g}}. So the answer is (d).
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