Question

A liquid flows through two capillary tubes fitted horizontally side by side to the bottom of a vessel containing liquid. Their lengths are I and 21 and radii are r and 2r respectively. If 'V' is the volume of the liquid that fows through the first tube in one minute, the time required for the same volume of liquid to flow through the second tube is

• A. 8 minute
• B. 1/8 minute
• C. 1/4 minute
• D. 4 minute

Answer 1

Answer: (B)

1/8 minute

Answer 2

Consider the situation as two cases: (i) fluid to come out of vessel $t _{1 a }$ (ii) fluid to flow through the tube $t _{2 a }$ (i) Since efflux velocity, $u$ in both cases is same, $u _{ a }= u _{ b }$ $\frac{ V }{ t _{ la }}= A _{ a } u _{ a }$ $\frac{ V }{ t _{1 b }}= A _{ b } u _{ b }$ $\frac{ t _{ la }}{ t _{ lb }}=\left(\frac{ r _{ b }}{ r _{ a }}\right)^{2}$ (ii) Consider a $dx$ element of tube, since velocity is same,time taken is proportional to length $\frac{ t _{2 a }^{\prime}}{ t _{2 b }^{\prime}}=\left(\frac{ l _{ a }}{ l _{ b }}\right)$ But since area of cross section is different, $dx$ is different in the tubes, $dV$ is same $\begin{array}{c} \Longrightarrow A _{ a } dx _{ a }= A _{ b } dx _{ b } \\\ \frac{ dx _{ a }}{ dx _{ b }}=\frac{ A _{ b }}{ A _{ a }}=4 \\\ \frac{ t _{2 a }}{ t _{2 b }}=\frac{ t _{2 a }^{\prime}}{ t _{2 b }^{\prime}} \times \frac{ dx _{ a }}{ dx _{ b }}=\left(\frac{1_{ a }}{ l _{ b }}\right) \times 4 \\\ \Longrightarrow \frac{ t _{ a }}{ t _{ b }}=\frac{ t _{1 a }}{ t _{ b b }} \times \frac{ t _{2 a }}{ t _{2 b }}=4 \times \frac{4}{2}=8 \\\ t _{ b }=\frac{ t _{ a }}{8}=\frac{1}{8} minute \end{array}$