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MHT CET 2026 (13 May, Shift 1) · Angle of Contact & Capillarity

Angle of Contact & Capillarity question from MHT CET 2026 (13 May, Shift 1)

Q11MHT CET 2026 · 13 MAY · SHIFT 1Medium

Angle of Contact & Capillarity

When one end of a capillary tube is dipped in water, water rises to a height h. The upward force of 105105 dyne due to surface tension is balanced by the force due to weight of water column. The inner circumference of the capillary is (surface tension of water = 7×10−27 \times 10^{-2} N/m)

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Answer: (a)

  1. Upward force F=T×LF = T \times L, where L=2πrL = 2\pi r is the inner circumference (θ=0\theta = 0).
  2. F=105 dyne=1.05×10−3 NF = 105\text{ dyne} = 1.05\times10^{-3}\text{ N}
  3. L=1.05×10−37×10−2=1.5×10−2 mL = \frac{1.05\times10^{-3}}{7\times10^{-2}} = 1.5\times10^{-2}\text{ m}
  4. So the answer is (a) 1.5 cm1.5\text{ cm}.
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