MCQ
The surface tension of a soap solution is $2 \times {10^{ - 2}}N/m.$ To blow a bubble of radius $1 \,cm,$ the work done is
- A$4\pi \times {10^{ - 6}}J$
- B$8\pi \times {10^{ - 6}}J$
- C$12\pi \times {10^{ - 6}}J$
- ✓$16\pi \times {10^{ - 6}}J$
✓
Answer
Correct option: D.
$16\pi \times {10^{ - 6}}J$
d
(d)$W = 8\pi {R^2}T = 8 \times \pi \times {({10^{ - 2}})^2} \times 2 \times {10^{ - 2}}$$ = 16\pi \times {10^{ - 6}}J$
(d)$W = 8\pi {R^2}T = 8 \times \pi \times {({10^{ - 2}})^2} \times 2 \times {10^{ - 2}}$$ = 16\pi \times {10^{ - 6}}J$
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