MCQ
Work done in increasing the size of a soap bubble from a radius of $3\ cm$ to $5\ cm$ is nearly (Surface tension of soap solution $=$ $0.03$ $Nm^{-1}$)
- A$0.2 $ $\pi mJ$
- B$2 $ $\pi mJ$
- ✓$0.4$ $ \pi mJ$
- D$4$ $ \pi mJ$
✓
Answer
Correct option: C.
$0.4$ $ \pi mJ$
c
$W = T \times change\,in\,surface\,area$
$W = T \times change\,in\,surface\,area$
$W = 2T4\pi \left[ {\left( {{5^2}} \right) - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$
$ = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}J = 0.4\pi \times {10^{ - 3}}J$
$ = 0.4\pi mJ$
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