Q: When a long glass capillary tube of radius 0.015 cm is dipped in a liquid , the liquid rises to a height of 15 cm within it . If the contact angle between the liquid and glass to close to 0° , the surface tension of the liquid in millinewton m^{-1} is (ρ_{liquid} = 900 kg/m^{-3} , g = 10 m/s^{2}) (give Answer in closest integer) ….

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Sol: Here r = 0.015 cm , ρ_{liquid} = 900 kg/m^{-3} , h = 15 cm g = 10 m/s^2 , θ = 0°

Height to which liquid will rise , $\displaystyle h = \frac{2 S cos\theta}{\rho g r }$

$\displaystyle S = \frac{h \rho g r}{2 cos\theta}$

$\displaystyle S = \frac{0.15 \times 900 \times 10 \times 0.015 \times 10^{-2}}{2 cos0^o}$

S = 0.10125 N/m = 101.25 mN/m