Question

In the above problem, if the same tubes are connected side by side ( parallel) then the length of the single tube of same radius that can replace the two tubes so that the rate of volume flow through it is equal to total rate of volume flow through the two tubes is :

• A. $3l$
• B. $\frac{2l}{3}$
• C. $1.5l$
• D. $\frac{l}{3}$

Answer 1

Answer: (B)

$\frac{2l}{3}$

$Q$ through smaller pipe= $\frac{\pi P{r}^4}{8\eta L}$-----------(1)

$Q$ through longer pipe= $\frac{\pi P{r}^4}{8\eta 2L}$-----------(2)

$Q$ (through single pipe of Length $L_1$) = $\frac{\pi P{r}^4}{8\eta L_1}$-----------(3)

According to Question,

$\frac{\pi P{r}^4}{8\eta L}+\frac{\pi P{R}^4}{8\eta \left(2L\right)}=\frac{\pi P{r}^4}{8\eta L_1}$

$\Rightarrow \frac{2}{2L}+\frac{1}{2L}=\frac{1}{L_1}$

$\Rightarrow L_1=\frac{2L}{3}$