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 \(\pi=\frac{22}{7}\) ]
- A \(0.72\)
- B \(5.76\)
- C \(18.48\)
- D \(9.24\)
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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