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A capillary of shapes as shown is dipped in a liquid. Contact angle between liquid minisues is to be neglected . T is surface tension of liquid, r is radius of meniscus g is acceleration due to gravity $\rho$ is density of the liquid then height h in equillibrium is

$\begin{array}{1 1} a) greater\; then \;\large\frac{2T}{r \rho g} \\ b) equal \;to \large\frac{2T}{r \rho g} \\ c) less \;than \;\large\frac{2T}{r \rho g} \\ d) \text{any value depending actual value of T} \end{array}$

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As weight of liquid in capillary is balanced by surface tension then $T \times 2 \pi r= \pi r^2 h_1 \rho g$
$h_1= \large\frac{2T}{r\rho g}$
But weight of liquid in tapered table as shown is more than uniform tube of radius the r then in order to balance the weight $h < h_1$
$\therefore h < \large\frac{2T}{r \rho g}$
Hence c is the correct answer.


answered Oct 3, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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