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Q. A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of water in the capillary tube is $5\,g$. Another capillary tube of radius $2\,r$ is immersed in water. The mass of water that will rise in this tube is

AIIMSAIIMS 2010Mechanical Properties of Fluids

Solution:

$h=\frac{2S\, cos\, \theta}{r\rho g}$
Mass of water in the first tube,
$m=\pi r^{2}h\rho=\pi r^{2}\times\left(\frac{2S \,cos\, \theta}{r\rho g}\right)\times\rho$
$=\frac{2\pi rS\, cos \,\theta}{g}$
$\therefore \, m \propto r.$ Hence, $\frac{m'}{m}=\frac{2r}{r}=2$
or $\, m'=2m=2\times 5 \,g=10\,g.$