Fluid MechanicsHard
Question
Surface tension of two liquids (having same densities), $T_{1}$ and $T_{2}$, are measured using capillary rise method utilizing two tubes with inner radii of $r_{1}$ and $r_{2}$ where $r_{1} > r_{2}$. The measured liquid heights in these tubes are $h_{1}$ and $h_{2}$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights $h_{1}$ and $h_{2}$ and surface tensions $T_{1}$ and $T_{2}$ satisfy the relation :
Options
A.$h_{1} < h_{2}$ and $T_{1} = T_{2}$
B.$h_{1} = h_{2}$ and $T_{1} = T_{2}$
C.$h_{1} > h_{2}$ and $T_{1} = T_{2}$
D.$h_{1} > h_{2}$ and $T_{1} < T_{2}$
Solution
$h = \frac{2T}{\rho gr}$
$$h \propto \frac{1}{r} $$If $r_{1} > r_{2} \Rightarrow {\text{ }h}_{2} > h_{1}$
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