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JEE Main 2023 (15 Apr, Shift 1) · Excess Pressure: Drops & Bubbles

Excess Pressure: Drops & Bubbles question from JEE Main 2023 (15 Apr, Shift 1)

Q36JEE Main 2023 · 15 APR · SHIFT 1Medium

Excess Pressure: Drops & Bubbles

There is an air bubble of radius 1.0 mm1.0\text{ mm} in a liquid of surface tension 0.075 N m−10.075\text{ N m}^{-1} and density 1000 kg m−31000\text{ kg m}^{-3} at a depth of 10 cm10\text{ cm} below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is _________ Pa(g=10 m s−2)\text{Pa}\left(g = 10\text{ m s}^{-2}\right)

Numerical answer type. Work it out, then open the answer.

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Answer: 1150

  1. Air bubble in liquid: Pin−P0=hρg+2TrP_{in} - P_0 = h\rho g + \frac{2T}{r}.
  2. hρg=0.1×1000×10=1000 Pah\rho g = 0.1 \times 1000 \times 10 = 1000\text{ Pa}
  3. 2Tr=2×0.07510−3=150 Pa\frac{2T}{r} = \frac{2 \times 0.075}{10^{-3}} = 150\text{ Pa}
  4. Excess pressure =1000+150=1150 Pa= 1000 + 150 = 1150\text{ Pa}
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