Physics - Fluid Question with Solution | TestHub

The height of liquid in a capillary tube is given by Jurin's law:

$h=\frac{2T\cos \theta }{r\rho g}$

For water:
$h_1=6\text{ cm}$
$T_1=T$
${\theta }_1=0^{\circ }$
${\rho }_1={\rho }_w$

For the second liquid:
$T_2=2T$
${\theta }_2=60^{\circ }$
${\rho }_2=2{\rho }_w$

Taking the ratio of the heights for the same capillary tube ($r_1=r_2=r$):

$\frac{h_2}{h_1}=\frac{T_2}{T_1}\times \frac{\cos {\theta }_2}{\cos {\theta }_1}\times \frac{{\rho }_1}{{\rho }_2}$

Substituting the given values:

$\frac{h_2}{6}=\left(\frac{2T}{T}\right)\times \left(\frac{\cos 60^{\circ }}{\cos 0^{\circ }}\right)\times \left(\frac{{\rho }_w}{2{\rho }_w}\right)$

$\frac{h_2}{6}=2\times \frac{0.5}{1}\times \frac{1}{2}$

$\frac{h_2}{6}=0.5$

$h_2=3\text{ cm}$

Answer: 3 cm