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JEE Mains · Physics · STD 11 - 9.2 surface tension
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 \(......\times 10^{-5}\,J\)
- A \(2.26\)
- B \(28\)
- C \(17.5\)
- D \(5\)
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…
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