Verified Solution • 2025–26 Session moderate

QJEEM22Phb2d7: JEE Main 2022 Physics - Class 11 JEEM22Phb2d7 Solution: A water drop of radius is broken into 729 equal droplets. If…

A water drop of radius 1 cm1 \mathrm{~cm} is broken into 729 equal droplets. If surface tension of water is 75 dyne/ cm\mathrm{cm} , then the gain in surface energy upto first decimal place will be :

(Given π=3.14\pi=3.14 )

Multiple Choice Options

  • A 8.5×10−4 J8.5 \times 10^{-4} \mathrm{~J}
  • B 8.2×10−4 J8.2 \times 10^{-4} \mathrm{~J}
  • C 7.5×10−4 J7.5 \times 10^{-4} \mathrm{~J}
  • D 5.3×10−4 J5.3 \times 10^{-4} \mathrm{~J}

Step-by-Step Verified Solution

729×43πr3=43πR3729 \times {4 \over 3}\pi {r^3} = {4 \over 3}\pi {R^3}

⇒R=9r\Rightarrow R = 9r ........ (1)

ΔU=S×ΔA\Delta U = S \times \Delta A ..... (2)

⇒ΔU=S×{−4πR2+729×4πr2}\Rightarrow \Delta U = S \times \{ - 4\pi {R^2} + 729 \times 4\pi {r^2}\}

=S×4π{729r2−81r2}= S \times 4\pi \{ 729{r^2} - 81{r^2}\}

=7.5×10−4 J= 7.5 \times {10^{ - 4}}\,J

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring