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The lower end of a vertical glass capillary tube of radius 2r is dipped in water. Now a solid glass rod of radius r is coaxially inserted into the tube. It surface tension of water is T and density of water is $ρ$ , then capillary rise (h) is

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$3 T / ρrg$

$\frac{2 T}{ρrg}$

$\frac{3 T}{2 ρrg}$

$\frac{4 T}{ρrg}$

answer is B.

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Detailed Solution

Mass of water in the space between the tube and the rod

$m = π [{(2 r)}^{2} - {(r)}^{2}] ρh = 3 {πr}^{2} . ρh$

vertically upward tension force acting on the water column = $2 πrT + 2 π (2 r) T = 6 πrT$

In equilibrium, $3 {πr}^{2} ρhg = 6 πrT \Rightarrow h = \frac{2 T}{rρg}$

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