If water rises in a capillary tube up to 3 , | vedclass

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(A) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.For water,the contact angle $	heta =

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$9.6 	imes 10^{-4} \, m$

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(A) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.For water,the contact angle $	heta = 0^\circ$,so $\cos 0^\circ = 1$.Given: $h = 3 \, cm = 3 	imes 10^{-2} \, m$,$T = 7.2 	imes 10^{-2} \, N/m$,$ho = 10^3 \, kg/m^3$,and $g = 10 \, m/s^2$.Rearranging the formula for radius $r$: $r = \frac{2T}{h ho g}$.Substituting the values: $r = \frac{2 	imes 7.2 	imes 10^{-2}}{3 	imes 10^{-2} 	imes 10^3 	imes 10} = \frac{14.4 	imes 10^{-2}}{3 	imes 10^2} = 4.8 	imes 10^{-4} \, m$.The diameter $d$ is $2r$: $d = 2 	imes 4.8 	imes 10^{-4} \, m = 9.6 	imes 10^{-4} \, m$.

If water rises in a capillary tube up to $3 \, cm$,what is the diameter of the capillary tube? (Surface tension of water $T = 7.2 \times 10^{-2} \, N/m$,density $\rho = 10^3 \, kg/m^3$,$g = 10 \, m/s^2$)

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