Air is pushed in a soap bubble to increase its radius from 'R' to ' $2 \mathrm{R}$ '. In this case, the…
Air is pushed in a soap bubble to increase its radius from 'R' to ' $2 \mathrm{R}$ '. In this case, the pressure inside the bubble
- does not change
- decrease
- becomes zero
- increase
Solution
Excess pressure in a soap bubble is given by
$\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}}$
Hence if radius is increased, the pressure will decrease.
Asked in: MHT CET 2021 (20 Sep Shift 2)
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