Question

In a capillary tube having area of cross-section $A$ , the water rises to a height $h$ . If the cross-sectional area is reduced to $\frac{A}{9}$ , the rise of water in the capillary tube is

• A. $4h$
• B. $3h$
• C. $2h$
• D. $h$

Answer 1

Answer: (B)

$A_{1} = \pi r_{1}^{2}$
$A_{2} = \pi r_{2}^{2}$
$\frac{\pi r_{1}^{2}}{9} = \pi r_{2}^{2}$
$\frac{r_{1}^{2}}{r_{2}^{2}} = 9 \Rightarrow \frac{r_{1}}{r_{2}} = 3$
$r h =$ constant
$r_{1} h_{1} = r_{2} h_{2}$
$\frac{r_{1}}{r_{2}} = \frac{h_{2}}{h_{1}}$
$3 = \frac{h_{2}}{h_{1}} = \frac{h_{2}}{h_{1}}$
$h_{2} = 3 h_{1} = 3 h$