Verified Solution • 2025–26 Session moderate

QJEEM23Ph2bbb: JEE Main 2023 Physics - Class 11 JEEM23Ph2bbb Solution: A mercury drop of radius is broken into 125 equal size…

A mercury drop of radius 10−3 m10^{-3}~\mathrm{m} is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm−10.45~\mathrm{Nm}^{-1} . The gain in surface energy is :

Multiple Choice Options

  • A 28×10−5 J28\times10^{-5}~\mathrm{J}
  • B 17.5×10−5 J17.5\times10^{-5}~\mathrm{J}
  • C 5×10−5 J5\times10^{-5}~\mathrm{J}
  • D 2.26×10−5 J2.26\times10^{-5}~\mathrm{J}

Step-by-Step Verified Solution

Initial surface energy =0.45×4π(10−3)2=0.45 \times 4 \pi\left(10^{-3}\right)^2

amp;43π(10−3)3=125×4π3Rnew 3∴amp;10−3=5Rnew ∴amp;Rnew =10−35 m\begin{aligned}& \frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3 \\\\\therefore & 10^{-3}=5 R_{\text {new }} \\\\\therefore & R_{\text {new }}=\frac{10^{-3}}{5} \mathrm{~m}\end{aligned}

So, final surface energy =0.45×125×4π(10−35)2=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2

Increase in energy =0.45×4π×(10−3)2[12525−1]=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]

amp;=4×0.45×4π×10−6amp;=2.26×10−5 J\begin{aligned}& =4 \times 0.45 \times 4 \pi \times 10^{-6} \\\\& =2.26 \times 10^{-5} \mathrm{~J}\end{aligned}

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