Two different vertical positions (a) & (b) of a capillary tube are shown in figure with the lower end inside water. For position (a), contact angle is 450 & water rises to height h above the surface of water while for position (b) height of the tube outside water is kept insufficient & equal to h2 then the new contact angle becomes –

h=2Tcos⁡45∘rρg......(i)h2=2Tcos⁡θrρg.......(ii)(i) & (ii) ⇒cos⁡θ=12⇒θ=60∘

Two different vertical positions (a) & (b) of a capillary tube are shown in figure with the lower end inside water. For position (a), contact angle is 450 & water rises to height h above the surface of water while for position (b) height of the tube outside water is kept insufficient & equal to h2 then the new contact angle becomes –

h=2Tcos⁡45∘rρg......(i)h2=2Tcos⁡θrρg.......(ii)(i) & (ii) ⇒cos⁡θ=12⇒θ=60∘

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Two different vertical positions (a) & (b) of a capillary tube are shown in figure with the lower end inside water. For position (a), contact angle is 45⁰ & water rises to height h above the surface of water while for position (b) height of the tube outside water is kept insufficient & equal to $\frac{h}{\sqrt{2}}$ then the new contact angle becomes –

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45°

30°

60°

$\cos^{- 1} (\frac{1}{2 \sqrt{2}})$

answer is C.

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Detailed Solution

$\begin{array}{l} h = \frac{2 Tcos 45^{\circ}}{rρg} . . . . . . (i) \\ \frac{h}{\sqrt{2}} = \frac{2 Tcos θ}{rρg} . . . . . . . (ii) \\ (i) & (ii) \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = 60^{\circ} \end{array}$

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