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JEE Mains · Physics · STD 11 - Surface tension
  • JEE Mains 2023
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  • MCQ

A mercury drop of radius \(10^{-3}\,m\) is broken into \(125\) equal size droplets. Surface tension of mercury is \(0.45\,Nm ^{-1}\). The gain in surface energy is \(......\times 10^{-5}\,J\)

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Answer & Solution

Correct Answer

(A) \(2.26\)

Step-by-step Solution

Detailed explanation

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

\(\frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3\)

\(10^{-3}=5 R_{\text {new }}\)

\(R_{\text {new }}=\frac{10^{-3}}{5}\,m\)

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

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

\(=4 \times 0.45 \times 4 \pi \times 10^{-6}\)

\(=2.26 \times 10^{-5}\,J\)

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