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

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six times that on the Earth's surface.

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(B) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is the surface tension,$	heta$ is the angle of contact,$r$ is the radius of the capillary,$ho$ is the density of the liquid,and $g$ is the acceleration due to gravity.From the formula,we see that $h \propto \frac{1}{g}$.Let $g_e$ be the acceleration due to gravity on Earth and $g_m$ be the acceleration due to gravity on the Moon.Given that $g_e = 6g_m$,or $g_m = \frac{g_e}{6}$.Let $h_e$ be the height on Earth and $h_m$ be the height on the Moon.Then,$\frac{h_m}{h_e} = \frac{g_e}{g_m} = \frac{g_e}{g_e / 6} = 6$.Therefore,$h_m = 6h_e$.The rise of the liquid column on the Moon's surface is six times that on the Earth's surface.

$A$ capillary tube is taken from the Earth's surface to the Moon's surface. What happens to the rise of the liquid column on the Moon's surface? (Acceleration due to gravity on the Earth's surface is six times that of the Moon's surface.)

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