MCQ
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$5.76$
- ✓$18.48$
- D$9.24$
✓
Answer
Correct option: C.
$18.48$
c
Surface area of soap bubble $=2 \times 4 \pi R^2$
Surface area of soap bubble $=2 \times 4 \pi R^2$
Work done $=$ change in surface energy $\times T_S$
$= T _{ S } \times 8 \pi \times\left( R _2^2- R _1^2\right)$
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}$
$=18.48 \times 10^{-4}\,J$
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