Question

The lower end of a glass capillary tube is dipped in water. Water rises to a height of $9cm$ . The tube is then broken at a height of $5cm$ . The height of the water column and angle of contact will be

• A. $5cm,{\left(cos\right)}^{-1}\left(\frac{5}{9}\right)$
• B. $4cm,{\left(cos\right)}^{-1}\left(\frac{5}{4}\right)$
• C. $5cm,{\left(cos\right)}^{-1}\left(\frac{9}{5}\right)$
• D. $5cm,{\left(cos\right)}^{-1}\left(\frac{6}{7}\right)$

Answer 1

Answer: (A)

When a capillary tube is broken at a height 5 cm, meaning that water will rise to height $5cm,$ hence $h_2=5cm.$
Since, we know, for capillarity phenomenon $h=\frac{2Scos\theta }{r\rho g},$ where symbols have their usual meaning.
or $\frac{h}{cos\theta }=$ constant
Thus, $\frac{h_1}{cos{\theta }_1}=\frac{h_2}{cos{\theta }_2}$
$\Rightarrow \frac{9}{cos\theta }=\frac{5}{cos{\theta }_2}$ [for glass ${\theta }_1=0$ ]
$\Rightarrow cos{\theta }_2=\frac{5}{9}$ $\left[\because cos{0}^o=1\right]$
$\therefore {\theta }_2={\cos }^{-1}\left(\frac{5}{9}\right)$