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A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodided (surface tension = 0.05 Nm-1, density = 667 kg m-3 ) Which rises to height h in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary ) make and angle of 60o with one another. Then h is close to (g=10 ms-2).
Option: 1 0.172 m
Option: 2 0.049 m
Option: 3 0.087 m
Option: 4 0.137 m

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So Angle of contact=30^0

So using

\begin{array}{l} h=\frac{2 T \cos \theta}{r \rho g} =\frac{2 \times 0.05 \times\left(\frac{\sqrt{3}}{2}\right)}{0.15 \times 10^{-3} \times 667 \times 10}=0.087 \mathrm{~m} \end{array}

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avinash.dongre

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