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JEE Mains · Physics · STD 11 - 9.2 surface tension

Surface tension of a soap bubble is \(2.0 \times 10^{-2}\; Nm ^{-1}\). Work done to increase the radius of soap bubble from \(3.5\; cm\) to \(7\; cm\) will be \(.........\times 10^{-4}\,J\) [Take \(\pi=\frac{22}{7}\) ]

  1. A \(0.72\)
  2. B \(5.76\)
  3. C \(18.48\)
  4. D \(9.24\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(18.48\)

Step-by-step Solution

Detailed explanation

Surface area of soap bubble \(=2 \times 4 \pi R^2\) Work done \(=\) change in surface energy \(\times T_S\) \(= T _{ S } \times 8 \pi \times\left( R _2^2- R _1^2\right)\) \(=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}\)…
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