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Q. Given that the surface tension of water is 75 dyne/cm, its density 1 g/cc and angle of contact zero, the height to which water rises in a capillary tube of 1 mm diameter is: (take $ g=10\text{ }m/{{s}^{2}} $ )

KEAMKEAM 2000

Solution:

Height to which water rises in the capillary tube is $ h=\frac{2T\cos \theta }{r\rho g} $ Given, $ T=75\text{ }dyne/cm,\text{ }\theta =0{}^\circ ,\text{ }r=\frac{1}{2}\times 10\text{ }cm. $ $ g=10\text{ }m/{{s}^{2}},\text{ }\rho =1\text{ }g/cc. $ $ \therefore $ $ h=\frac{2\times 75\times \cos 0{}^\circ \times 2}{10\times 10}=3cm $