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MHT CET202211 Aug 2022Morning ShiftPhysicsMechanical Properties of FluidsActual

The radii of two mercury drops are R₁ and R₂ . Under isothermal conditions, a single drop of radius R is formed from them. The relations between R, R₁ and R₂ is

Options

  1. AR^2=R₁^2+R₂^2
  2. BR=R₁+R₂
  3. CR= R₁+R₂ 2
  4. DR^3=R₁^3+R₂^3

Correct answer

D. R^3=R₁^3+R₂^3

Step-by-step solution

Total volume remains same, aligned & 4 3 R^3= 4 3 R₁^3+ 4 3 R₂^3 & R^3=R₁^3+R₂^3 aligned

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