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(D) The height of the rise of liquid due to surface tension in a capillary tube is given by the formula: $h = \frac{2 S \cos 	heta}{r ho g}$.Given

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$0.2 \ mm$

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(D) The height of the rise of liquid due to surface tension in a capillary tube is given by the formula: $h = \frac{2 S \cos 	heta}{r ho g}$.Given values are: height of liquid rise $h = 7.5 \ cm = 7.5 	imes 10^{-2} \ m$,surface tension $S = 7.5 	imes 10^{-2} \ N \ m^{-1}$,angle of contact $	heta = 0^{\circ}$,density of water $ho = 1000 \ kg \ m^{-3}$,and acceleration due to gravity $g = 10 \ m \ s^{-2}$.Rearranging the formula to solve for the radius $r$: $r = \frac{2 S \cos 	heta}{h ho g}$.Substituting the values: $r = \frac{2 	imes (7.5 	imes 10^{-2}) 	imes \cos 0^{\circ}}{(7.5 	imes 10^{-2}) 	imes 1000 	imes 10}$.Since $\cos 0^{\circ} = 1$,we get: $r = \frac{2 	imes 7.5 	imes 10^{-2}}{7.5 	imes 10^{-2} 	imes 10^4} = \frac{2}{10^4} = 2 	imes 10^{-4} \ m$.Converting to millimeters: $r = 2 	imes 10^{-4} 	imes 10^3 \ mm = 0.2 \ mm$.

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