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JEE Main 2021 (17 Mar, Shift 1) · Excess Pressure: Drops & Bubbles

Excess Pressure: Drops & Bubbles question from JEE Main 2021 (17 Mar, Shift 1)

Q49JEE Main 2021 · 17 MAR · SHIFT 1Medium

Excess Pressure: Drops & Bubbles

When two soap bubbles of radii a and b (b > a) coalesce, the radius of curvature of common surface is:

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Answer: (d)

  1. The smaller bubble (aa) has higher pressure, so the common film bulges into the bigger one.
  2. Pressure difference across the film: 4Tr=4Ta−4Tb\frac{4T}{r} = \frac{4T}{a} - \frac{4T}{b}
  3. 1r=b−aab⇒r=abb−a\frac{1}{r} = \frac{b-a}{ab} \Rightarrow r = \frac{ab}{b-a}
  4. So the answer is (d) abb−a\frac{ab}{b-a}.
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