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JEE Main20231 Feb 2023Morning ShiftPhysicsMechanical Properties of FluidsActual

A mercury drop of radius 10 – 3 m is broken into 125 equal size droplets. Surface tension of mercury is 0 . 45 N m – 1 . The gain in surface energy is:

Options

  1. A2 . 26   ×   10 – 5   J
  2. B28   ×   10 – 5   J  
  3. C17 . 5   ×   10 – 5   J  
  4. D5   ×   10 – 5   J

Correct answer

A. 2 . 26   ×   10 – 5   J

Step-by-step solution

As the volume of the mercury will remain same, therefore 4 3 π R 3 = 125 4 3 π r 3 ⇒ R = 5 r Now, increase in surface energy ∆ U = 125 × S × 4 π r 2 - S × 4 π R 2 = 0 . 45 × 4 π × 10 - 3 2 125 25 - 1 = 4 × 0 . 45 × 4 π × 10 - 6 = 2 . 26 × 10 - 5 J

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