JEE Main202226 Jul 2022Morning ShiftPhysicsMechanical Properties of FluidsActual
A water drop of radius 1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne cm - 1 , then the gain in surface energy upto first decimal place will be [Given π = 3 . 14 ]
Options
- A8 . 5 × 10 - 4   J
- B8 . 2 × 10 - 4   J
- C7 . 5 × 10 - 4   J
- D5 . 3 × 10 - 4   J
Correct answer
C. 7 . 5 × 10 - 4   J
Step-by-step solution
The volume of the bigger drop will be, V = 4 3 π R 3 . If the radius of smaller drops is r , using the conservation of volume, n 4 3 π r 3 = 4 3 π R 3 ⇒ 729 × 4 3 π r 3 = 4 3 π R 3 ⇒ r = R 9 The initial surface energy will be, E i = 4 π R 2 T And final surface energy will be, E f = n 4 π r 2 T = 729 × 4 π R 9 2 T = 36 π R 2 T Therefore, the change in the surface energy will be, ⇒ Δ E = E f - E i = 32 π R 2 T = 32 × 3 . 14 ×