Question

A liquid flows through two capillary tubes fitted horizontally side by side to the bottom of a vessel containing liquid. Their lengths are $l$ and $2l$ and radii are $r$ and $2r$ respectively. If '$V$' is the volume of the liquid that flows through the first tube in one minute, the time required for the same volume of liquid to flow through the second tube is:

• A. $8minute$
• B. $\frac{1}{8}$ minute
• C. $\frac{1}{4}$ minute
• D. $4minute$

Answer 1

The correct option is B $\frac{1}{8}$ minute

Consider the situation as two cases:
(i) fluid to come out of vessel $t_{1a}$
(ii) fluid to flow through the tube $t_{2a}$
(i) Since efflux velocity $,u$ in both cases is same$,u_a=u_b$
$\frac{V}{t_{1a}}$ =$A_a{u}_a$

$\frac{V}{t_{1b}}$ =$A_b{u}_b$

$\frac{t_{1a}}{t_{1b}}$ = $(\frac{r_b}{r_a}{)}^2$

(ii) Consider a dx element of tube, since velocity is same,time taken is proportional to length
$\frac{t_{2a}^{\prime }}{t_{2b}^{\prime }}$ = $\left(\frac{l_a}{l_b}\right)$

But since area of cross section is different, dx is different in the tubes, dV is same
$⟹$ $A_{a}d{x}_a=A_{b}d{x}_b$
$\frac{d{x}_a}{d{x}_b}$ = $\frac{A_b}{A_a}$= 4

$\frac{t_{2a}}{t_{2b}}$ = $\frac{t_{2a}^{\prime }}{t_{2b}^{\prime }}$ $\times$ $\frac{d{x}_a}{d{x}_b}$ = $\left(\frac{l_a}{l_b}\right)$$\times 4$

$⟹$$\frac{t_a}{t_b}$ =$\frac{t_{1a}}{t_{1b}}$ $\times$ $\frac{t_{2a}}{t_{2b}}=4\times \frac{4}{2}=8$

$t_b=\frac{t_a}{8}$= $\frac{1}{8}$minute