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

  \(r \cos \theta\)
• Correct answer
  D

  \(\frac{r}{\cos \theta}\)

Solution

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, \(\frac{r}{R}=\cos \theta \Rightarrow R=\frac{r}{\cos \theta}\)

Hence, the correct answer is option: 'D'.