Question

A glass capillary tube of internal radius r=0.25 mm is immersed in water. The top end of the tube projects by 2 cm above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water=0.07 N $m^{-1}$ and its contact angle is 0.

• A. $si{n}^{-1}\left(0.5\right)$
• B. $co{s}^{-1}\left(0.5\right)$)
• C. $co{s}^{-1}5/14$
• D. $si{n}^{-1}5/14$

Answer 1

The correct option is C $co{s}^{-1}5/14$

$h=\frac{2T\cos \theta }{r1g}\Rightarrow 2\times {10}^{-2}=\frac{2\times 7\times {10}^{-2}\times \cos \theta }{0.25\times {10}^{-3}\times {10}^3\times 10}$

$\Rightarrow \cos \theta =\frac{25}{70}\Rightarrow \theta ={\cos }^{-1}\left(\frac{5}{14}\right)$