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JEE Main 2019 (12 Jan, Shift 2) · Excess Pressure: Drops & Bubbles

Excess Pressure: Drops & Bubbles question from JEE Main 2019 (12 Jan, Shift 2)

Q56JEE Main 2019 · 12 JAN · SHIFT 2Difficult

Excess Pressure: Drops & Bubbles

A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume, with time, at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by:

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Answer: (b)

  1. V=ktV = kt and V=43πr3⇒r∝t1/3V = \frac{4}{3}\pi r^3 \Rightarrow r \propto t^{1/3}.
  2. P=P0+4Tr=P0+Ct1/3P = P_0 + \frac{4T}{r} = P_0 + \frac{C}{t^{1/3}}
  3. So PP vs 1t1/3\frac{1}{t^{1/3}} is a straight line with intercept P0P_0.
  4. So the answer is (b).
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