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MHT CET 2024 (10 May, Shift 2) · Angle of Contact & Capillarity

Angle of Contact & Capillarity question from MHT CET 2024 (10 May, Shift 2)

Q37MHT CET 2024 · 10 MAY · SHIFT 2Medium

Angle of Contact & Capillarity

A glass capillary of radius 0.35 mm is inclined at 60∘60^\circ with the vertical in water. The height of the water column in the capillary is (surface tension of water =7×10−2 N/m= 7 \times 10^{-2}\text{ N/m}, acceleration due to gravity, g=10 m/s2,cos⁡0∘=1,cos⁡60∘=0.5g = 10\text{ m/s}^2, \cos 0^\circ = 1, \cos 60^\circ = 0.5)

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Answer: (b)

h (vertical)l = h/cosθθwater surface
  1. Vertical rise is fixed: h=2Trρg=2×7×10−20.35×10−3×103×10=0.04 m=4 cmh = \frac{2T}{r\rho g} = \frac{2 \times 7\times10^{-2}}{0.35\times10^{-3} \times 10^3 \times 10} = 0.04\text{ m} = 4\text{ cm}.
  2. In a tube inclined at θ\theta to the vertical, the length of the column is l=hcos⁡θl = \frac{h}{\cos\theta}.
  3. l=40.5=8 cml = \frac{4}{0.5} = 8\text{ cm}
  4. So the answer is (b) 8 cm8\text{ cm}.
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