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
Correct Answer: A
Solution :
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.
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