Question

Three liquids having densities ${\rho }_1,{\rho }_2$ and $\rho {}_3$ $({\rho }_1>{\rho }_2>{\rho }_3)$, having the same value of surface tension $T$, rise to the same height $h$ in three identical capillaries. The angle of contact ${\theta }_1,{\theta }_2$ and ${\theta }_3$ obeys

• A. $\frac{\pi }{2}>{\theta }_1>{\theta }_2>{\theta }_3\ge 0$
• B. $0\le {\theta }_1<{\theta }_2<{\theta }_3<\frac{\pi }{2}$
• C. $\frac{\pi }{2}<{\theta }_1<{\theta }_2<{\theta }_3<\pi$
• D. $\pi >{\theta }_1>{\theta }_2>{\theta }_3>\frac{\pi }{2}$

Answer 1

The correct option is B $0\le {\theta }_1<{\theta }_2<{\theta }_3<\frac{\pi }{2}$

We know capillary rise,
$h=\frac{2T\cos \theta }{r\rho g}$

Liquid is rising in capillary tube, so angle of contact is $0\le \theta <\frac{\pi }{2}$.
For the constant value of surface tension $T$ and radius of capillary $r$ of a fluid.
$h\propto \frac{\cos \theta }{\rho }$
Since, $h_1=h_2=h_3$
$\Rightarrow \frac{\cos {\theta }_1}{{\rho }_1}=\frac{\text{cos}{\theta }_2}{{\rho }_2}=\frac{\text{cos}{\theta }_3}{{\rho }_3}$ .....(1)
Given, ${\rho }_1>{\rho }_2>{\rho }_3$
Hence, to maintain the equal ratio in (1),
$\cos {\theta }_1>\cos {\theta }_2>\cos {\theta }_3$
$\Rightarrow {\theta }_1<{\theta }_2<{\theta }_3$
[ as the value of angle increases from $0$ to $\frac{\pi }{2}$, the value of $\cos \theta$ decreases from $1$ to $0$ ]
Hence, $0\le {\theta }_1<{\theta }_2<{\theta }_3<\frac{\pi }{2}$