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Surface Energy

The radii of ...

Question

The radii of two soap bubbles are r₁ and r₂. In isothermal conditions, two meet together in vaccum. Then the radius of the resultant bubble is given by

r =(r₁ + r₂)/2

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r =r₁(r₁r₂+r₂)

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r² =r₁²+r₂²

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r =r₁+r₂

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Solution

The correct option is C
r² =r₁²+r₂²

In isothermal Condition.

$Δ P_{1} V_{1} + Δ P_{2} V_{2} = Δ P V Δ P_{1} \frac{4}{3} π r_{1}^{3} + Δ P_{2} \frac{4}{3} π r_{2}^{3} = Δ P \frac{4}{3} π r^{3} Δ P_{1} r_{1}^{3} + Δ P_{2} r_{2}^{3} = Δ P r^{3} [w h e r e^{'} r^{'} i s t h e r a d i u s o f n e w b u b b l e] \frac{4 T}{r_{1}} (r_{1}^{3}) + \frac{4 T}{r_{2}} (r_{2}^{3}) = \frac{4 T}{r} (r^{3}) \Rightarrow r_{1}^{2} + r_{2}^{2} = r^{2} \Rightarrow r = \sqrt{r_{1}^{2} + r_{2}^{2}}$

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