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JEE Main 2020 (3 Sep, Shift 1) · Angle of Contact & Capillarity

Angle of Contact & Capillarity question from JEE Main 2020 (3 Sep, Shift 1)

Q57JEE Main 2020 · 3 SEP · SHIFT 1Medium

Angle of Contact & Capillarity

When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to 0°, the surface tension of the liquid, in milliNewton m–1, is [ρ\rho(liquid) = 900 kgm–3, g = 10 ms–2]

(Give answer in closest integer) _____.

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

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Answer: 101

  1. h=2Trρg⇒T=hrρg2h = \frac{2T}{r\rho g} \Rightarrow T = \frac{hr\rho g}{2}
  2. T=0.15×1.5×10−4×900×102=0.10125 N/mT = \frac{0.15 \times 1.5\times10^{-4} \times 900 \times 10}{2} = 0.10125\text{ N/m}
  3. T≈101 mN/mT \approx 101\text{ mN/m}
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