Question

Liquid rises to a height of $2\text{cm}$ in a capillary tube. The angle of contact between the solid and the liquid is zero. The tube is depressed more now so that the top of the capillary is only $1\text{cm}$ above the liquid. Then, the apparent angle of contact between the solid and the liquid is:

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$


We know that,
$hR=L{R}^{\prime }$
Also,
$R=\frac{r}{\cos \theta }$
where $r=$ radius of capillary
$\theta =$ contact angle

Let new contact angle be ${\theta }^{\prime }$.
$\therefore \frac{hr}{\cos \theta }=\frac{Lr}{\cos {\theta }^{\prime }}$
$\Rightarrow \frac{h}{\cos \theta }=\frac{L}{\cos {\theta }^{\prime }}$
$\Rightarrow \frac{2}{\cos 0^{\circ }}=\frac{1}{\cos {\theta }^{\prime }}$

$\Rightarrow {\theta }^{\prime }={60}^{\circ }$
$\therefore$ Apparent angle of contact between solid and liquid is ${60}^{\circ }$.