A capillary tube whose inside radius is `0.5mm` is dipped in water having surface tension `7.0xx10^(-2) N//m`. To what height is the water raised above the normal water level? Angle of contact of water with glass is `0^(@)`. Density of water is `10^(3)kg//m^(3) and g=9.8 m//s^(2)`.
Text Solution
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Height raised, `h=(2T cos theta)/(r rho g)` Substituting the proper values, we have `h=((2)(7.0xx10^(-2))cos 0^(@))/((0.5xx10^(-3))(10^(3))(9.8))` `=2.86xx10^(-2) m` =2.86 cm
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