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Lower end of a vertical capillary tube of radius ‘r’ is dipped in water and water of mass m rises in that tube. When the lower end of another vertical capillary tube of radius r/2 is dipped mass of water that rise in the tube will be

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$\frac{m}{4}$

$\frac{m}{2}$

$\frac{m}{8}$

answer is B.

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Detailed Solution

Capillary rise $h = \frac{2 T \cos 0^{0}}{ρrg}$

$∴$ Mass of water $m = {πr}^{2} . h . ρ . = \frac{2 πrT}{g}$

$\begin{array}{l} ∴ m \propto r \\ ∴ \frac{m^{'}}{m} = \frac{r / 2}{r} = \frac{1}{2} \Rightarrow m^{'} = m / 2 \end{array}$

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