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

Vedclass · Accepted Answer

$0.9$

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(D) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.Given that the diameter $(2r)$,angle of contact $(	heta)$,and acceleration due to gravity $(g)$ are constant,the height is proportional to the ratio of surface tension to density: $h \propto \frac{T}{ho}$.Therefore,the ratio of the heights is given by: $\frac{h_1}{h_2} = \frac{T_1}{ho_1} 	imes \frac{ho_2}{T_2} = \left(\frac{T_1}{T_2}ight) 	imes \left(\frac{ho_2}{ho_1}ight)$.Given $\frac{T_1}{T_2} = \frac{6}{5}$ and $\frac{ho_1}{ho_2} = \frac{4}{3}$,we have $\frac{ho_2}{ho_1} = \frac{3}{4}$.Substituting these values: $\frac{h_1}{h_2} = \frac{6}{5} 	imes \frac{3}{4} = \frac{18}{20} = 0.9$.

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