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A water drop is divided into eight equal droplets then the pressure difference between the inner and outer side of the big drop will be

Options

  1. Asame as for smaller droplet
  2. B1 / 2 of that for smaller droplet
  3. C1 / 4 of that for smaller droplet
  4. DTwice that for smaller droplet

Correct answer

B. 1 / 2 of that for smaller droplet

Step-by-step solution

When break into eight drop radius of the small drop is r = R n 1 / 3 ⇒   r = R 8 1 / 3 = R 2 For large drop Δ P 1 = 2 T R for small drop Δ P 2 = 2 T r = 2 T R / 2 =   2 2 T R

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