MCQ
The surface tension of soap solution is $3.5 \times 10^{-2}\,Nm ^{-1}$. The amount of work done required to increase the radius of soap bubble from $10\,cm$ to $20\,cm$ is $.....\times 10^{-4}\,J$
- ✓$264$
- B$263$
- C$262$
- D$265$
✓
Answer
Correct option: A.
$264$
a
$W = T .(\Delta A )$
$W = T .(\Delta A )$
$W = T \left(8 \pi\left( r _2^2- r _1^2\right)\right)$
$W =264 \times 10^{-4}\,J$
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