Question

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

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

Answer 1

Answer: (A)

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