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

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$3: 2$

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(B) The height of capillary rise $h$ of a liquid in a capillary tube is given by the formula:$h = \frac{2 T \cos 	heta}{r ho g}$Since $T$,$	heta$,$ho$,and $g$ are constant for the same liquid and tube material,we have:$h \propto \frac{1}{r}$Since the diameter $D = 2r$,we can also write $h \propto \frac{1}{D}$.Given that $h_P = \frac{2}{3} h_Q$,we have $\frac{h_P}{h_Q} = \frac{2}{3}$.Using the inverse proportionality $h \propto \frac{1}{D}$,we get:$\frac{h_P}{h_Q} = \frac{D_Q}{D_P} = \frac{2}{3}$Therefore,the ratio of their diameters is $\frac{D_P}{D_Q} = \frac{3}{2}$ or $3: 2$.

Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2}{3}$ of the height in capillary tube $Q$. The ratio of their diameters is

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