In a U-shaped tube,the radius of one limb is | vedclass

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(C) The height of capillary rise in a tube is given by $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	heta$ is the angle of con

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(C) The height of capillary rise in a tube is given by $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	heta$ is the angle of contact,$r$ is the radius of the capillary tube,and $ho$ is the density of the liquid.Given: $T = 0.03 \ Nm^{-1}$,$ho = 1500 \ kgm^{-3}$,$	heta = 0^\circ$ (so $\cos 	heta = 1$),$g = 10 \ ms^{-2}$,$r_1 = 2 \ mm = 2 	imes 10^{-3} \ m$,and $r_2 = 4 \ mm = 4 	imes 10^{-3} \ m$.The heights in the two limbs are $h_1 = \frac{2T}{r_1 ho g}$ and $h_2 = \frac{2T}{r_2 ho g}$.The difference in liquid levels is $\Delta h = h_1 - h_2 = \frac{2T}{ho g} \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$.Substituting the values: $\Delta h = \frac{2 	imes 0.03}{1500 	imes 10} \left( \frac{1}{2 	imes 10^{-3}} - \frac{1}{4 	imes 10^{-3}} ight)$.$\Delta h = \frac{0.06}{15000} \left( 500 - 250 ight) = \frac{0.06}{15000} 	imes 250 = \frac{0.06}{60} = 0.001 \ m$.Therefore,$\Delta h = 1 \ mm$.

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