Question

In a capillary tube having area of cross-section 'A', water rises to a height 'h'. If 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)

For a capillary tube, $m=$ constant
where, $r=$ radius of capillary tube
and $h=$ height of rised water in capillary tube.
According to the question.
$r_{1} h_{1} =r_{2} h_{2} $
$\Rightarrow \,\,\, \frac{r_{1}}{r_{2}}=\frac{h_{2}}{h_{1}}\,\,\,\,\,(i)$
In the first condition,
$A_{1}=\pi r_{1}^{2}\,\,\,\,\,\,...(ii)$
In the second condition,
$A_{2}=\pi r_{2}^{2}$
or$\,\,\,\,\,\,\frac{A}{9}=\pi r_{2}^{2} \,\,\,\,\left[\text { as, } A_{2}=\frac{A}{9}\right]\,\,\,\,\,\,\,...(iii)$
On dividing Eq. (ii) by Eq. (iii), we get
$9=\frac{r_{1}^{2}}{r_{2}^{2}} \text { or } \frac{r_{1}}{r_{2}}=\sqrt{9} \Rightarrow \frac{r_{1}}{r_{2}}=3$
From Eq. (i), we get
$\frac{h_{2}}{h_{1}}=3 \Rightarrow h_{2}=3 h_{1} $
$h_{2}=3 h$