Two soap bubbles coalesce to form a single bubble. If V is the… | JEE Main 2014 (12 Apr, Shift 1)

Q59JEE Main 2014 · 12 APR · SHIFT 1Difficult

Excess Pressure: Drops & Bubbles

Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S the change in total surface area, T is the surface tension and P atmospheric pressure, which of the following relation is correct?

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Answer: (b)

Isothermal: $P(R^3 - r_1^3 - r_2^3) = -4T(R^2 - r_1^2 - r_2^2)$ (from Boyle's law with $P_{in} = P + \frac{4T}{r}$ ).
Change in volume $V = \frac{4}{3}\pi(R^3 - r_1^3 - r_2^3)$ , change in area $S = 4\pi(R^2 - r_1^2 - r_2^2)$ .
Substituting: $P \cdot \frac{3V}{4\pi} = -4T \cdot \frac{S}{4\pi}$
$3PV + 4ST = 0$
So the answer is (b).

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Excess Pressure: Drops & Bubbles question from JEE Main 2014 (12 Apr, Shift 1)

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