Verified Solution • 2025–26 Session moderate

QJEEM22Ph16b2: JEE Main 2022 Physics - Class 11 JEEM22Ph16b2 Solution: A water drop of diameter 2 cm is broken into 64 equal droplets. …

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :

Multiple Choice Options

  • A 2.8 ×\times 10 −- 4 J
  • B 1.5 ×\times 10 −- 3 J
  • C 1.9 ×\times 10 −- 4 J
  • D 9.4 ×\times 10 −- 5 J

Step-by-Step Verified Solution

r′=r4r' = {r \over 4}

⇒ΔE=T(ΔS)\Rightarrow \Delta E = T(\Delta S)

=T×4π(nr′2−r2), n=64= T \times 4\pi (nr{'^2} - {r^2}),\,n = 64

=T×4π×(4−1)r2= T \times 4\pi \times (4 - 1){r^2}

⇒ΔE=0.075×4×3.142(3)×10−4 \Rightarrow \Delta E = 0.075 \times 4 \times 3.142(3) \times {10^{ - 4}}\, J

=2.8×10−4= 2.8 \times {10^{ - 4}} J

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