MHT CET · Physics · Mechanical Properties of Fluids
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
- A does not change
- B decrease
- C becomes zero
- D increase
Answer & Solution
Correct Answer
(B) decrease
Step-by-step Solution
Detailed explanation
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.
\(
\mathrm{P}=\frac{4 \mathrm{~T}}{\mathrm{R}}
\)
Hence if radius is increased, the pressure will decrease.
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