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To what depth, relative to the surface does mercury dip down in a capillary tube of radius $1.0 mm$ surface tension $T=0.5 N/m$ density of mercury $=13.6 \times 10^{3}\;kg/m^3$ $\;g=9.8 m/s^2$

\[\begin {array} {1 1} (a)\;+3.75\;mm \\ (b)\;-3.75\;mm \\ (c)\;2.22\;mm \\ (d)\;-2.22\;mm \end {array}\]

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1 Answer

$h= \large\frac{2T \cos \theta}{r \rho g}$
$h= \large\frac{2 \times 0.5 \times \cos 120^{\circ}}{10^{-3} \times 13.6 \times 10^{3} \times 9.8}$
$\quad= -3.75 \times 10^{-3}$
$\quad= -3.75 \;mm$
Hence b is the correct answer.

 

answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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