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(C) The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$.Since the m

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$3 : 1$

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(C) The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$.Since the material of the capillaries,the liquid,and the contact angle are the same,all parameters except $h$ and $r$ are constant.Therefore,$h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.Given $h_1 = 22 \ cm$ and $h_2 = 66 \ cm$.We need to find the ratio of their radii,$\frac{r_1}{r_2}$.From the relation $h_1 r_1 = h_2 r_2$,we get $\frac{r_1}{r_2} = \frac{h_2}{h_1}$.Substituting the values,$\frac{r_1}{r_2} = \frac{66}{22} = \frac{3}{1}$.Thus,the ratio of their radii is $3 : 1$.

Two capillaries of the same material but of different radii are dipped in a liquid. In one of the capillaries,the liquid rises to a height of $22 \ cm$ and in the other to $66 \ cm$. The ratio of their radii is

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