MCQMediumJEE Main 2023 · 1 February, Shift 1Surface Tension & Capillarity

Physics Question from JEE Main 2023 · 1 February, Shift 1

A mercury drop of radius 10−3 m10^{-3} \, \text{m} is broken into 125125 equal size droplets. Surface tension of mercury is 0.45 N m−10.45 \, \text{N m}^{-1}. The gain in surface energy is:

  • A

    2.26×10−5 J2.26 \times 10^{-5} \, \text{J}

  • B

    28×10−5 J28 \times 10^{-5} \, \text{J}

  • C

    17.5×10−5 J17.5 \times 10^{-5} \, \text{J}

  • D

    5×10−5 J5 \times 10^{-5} \, \text{J}

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