Question

The radius of the bore of a capillary tube is ' $r$ ' and the angle of contact of the liquid is ' $\theta$ '. When the tube is dipped in the liquid, the radius of curvature of the meniscus of liquid rising in the tube is

• A. $r\sin \theta$
• B. $\frac{r}{\sin \theta }$
• C. $r\cos \theta$
• D. $\frac{r}{\cos \theta }$

Answer 1

Answer: (D)

The given situation is shown in the figure,

where, $r=$ radius of capillary tube,
$R=$ radius of meniscus
and $\theta =$ angle of contact.
From figure, $\frac{r}{R}=\cos \theta$
$\Rightarrow R=\frac{r}{\cos \theta }$