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MHT CET 2025 (19 Apr, Shift 1) · Angle of Contact & Capillarity

Angle of Contact & Capillarity question from MHT CET 2025 (19 Apr, Shift 1)

Q27MHT CET 2025 · 19 APR · SHIFT 1Medium

Angle of Contact & Capillarity

One end of a capillary tube is dipped in water, the rise of water column is ' hh '. The upward force of 98 dyne due to surface tension is balanced by the force due to the weight of the water column. The inner circumference of the capillary is (( surface tension of water =7×10−2 N m−1)= 7 \times 10^{-2}\text{ N m}^{-1})

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Answer: (a)

  1. Upward force F=T×LF = T \times L (θ=0\theta = 0).
  2. L=98×10−57×10−2=1.4×10−2 mL = \frac{98\times10^{-5}}{7\times10^{-2}} = 1.4\times10^{-2}\text{ m}
  3. So the answer is (a) 1.4 cm1.4\text{ cm}.
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