COMEDK · Physics · 11. Mechanical Properties of Fluids
In to a vessel containing pure water a clean glass tube of radius \(3.6 \times 10^{-4} \mathrm{~m}\) is held vertically with 12 cm of the tube above the water level. Now the capillary tube is moved down in to the water so that only 2 cm of its length is above the water surface. Angle of contact \(\theta\) at this position is (given surface tension of water \(=7.2 \times 10^{-2} \mathrm{Nm}^{-1}\) and \(g=10 \mathrm{~ms}^{-2}\))
- A \(\theta=30^{\circ}\)
- B \(\theta=60^{\circ}\)
- C \(\theta=45^{\circ}\)
- D \(\theta=15^0\)
Answer & Solution
Correct Answer
(B) \(\theta=60^{\circ}\)
Step-by-step Solution
Detailed explanation
The capillary rise \(h\) in a tube of radius \(r\) is given by the formula \(h = \dfrac{2T \cos \theta}{r \rho g}\), where \(T\) is the surface tension, \(\theta\) is the angle of contact, \(\rho\) is the density of water, and \(g\) is the acceleration due to gravity. First,…
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