Question

In a capillary tube having area of cross-section 'A', water rises to a height 'h'. If cross-sectional area is reduced to $\frac{{}^{\prime }{A}^{\prime }}{9}$, 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
$\frac{h_2}{h_1}=3\Rightarrow h_2=3h_1$
$h_2=3h$