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Lower end of a vertical capillary tube is immersed in water and the mass of water that rises in the tube is m. Now the tube is tilted so that it makes an angle of $30^{o}$ with the horizontal. Then mass of water that rises in the tube is

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$2 \sqrt{3} m$

$\sqrt{3} m$

$\frac{2 m}{\sqrt{3}}$

answer is B.

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Detailed Solution

If h be the capillary rise in the vertical tube, then

$m = {πr}^{2} . h . ρ = {πr}^{2} . \frac{2 T}{ρ rg} . ρ = \frac{2 πrT}{g}$

When the tube is tilted, let x be the length and m' be the mass of water column in the tube

$∴ 2 πrT = m' gsin 30^{o} \Rightarrow m' = \frac{4 πrT}{g} = 2 m$

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