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A glass capillary tube is partially dipped in water vertically and water rises to a height h in the tube. The tube is now pushed down so that length of tube above the free surface of water is h/2. Then new angle of contact is

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$30^{o}$

$45^{o}$

$60^{o}$

$0^{o}$

answer is C.

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Detailed Solution

Initial angle of contact is $0^{o}$ and final angle of contact is θ
$\begin{array}{l} ∴ h = \frac{2 Tcos 0^{o}}{pgr} and \frac{h}{2} = \frac{2 Tcosθ}{pgr} \\ ∴ \frac{h}{2} = hcosθ \Rightarrow θ = 60^{0} \end{array}$

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