Question

A capillary tube is taken from the earth to the surface of the moon. The rise of the liquid column on the moon (acceleration due to gravity on the earth is 6 times that of the moon)is

• A. six times that on the earths surface
• B. $\frac{1}{6}$ that on the earths surface
• C. equal to that on the earths surface
• D. zero

Answer 1

Answer: (A)

From the relation $h=\frac{2T\cos \theta }{rdg}$ where r = radius of capillary. h = rise or fall of the liquid. g = acceleration due to gravity. d = density of the liquid. $\therefore$ $h\propto \frac{1}{g}$ $\Rightarrow$ $\frac{h_2}{h_1}=\frac{g_1}{g_2}$ According to the question, On earth, $h_1=h,g_1=g$ On moon, $h_2=?g_2=\frac{g}{6}$ $\frac{h_2}{h}=\frac{g}{g/6}$ $\Rightarrow$ $h_2=6h$ Hence, the rise of the liquid column on the moon becomes six time that on the earths surface.