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Two capillary of length $L$ and $2L$ and of radius $R$ and $2R$ are connected in series. The net rate of flow of fluid through them will be (given rate to the flow through single capillary, $X=\dfrac{\pi P{{R}^{4}}}{8\eta L}$ )
A.$\dfrac{8}{9}X$
B.$\dfrac{9}{8}X$
C.$\dfrac{5}{7}X$
D.$\dfrac{7}{5}X$

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Last updated date: 13th Jun 2024
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Answer
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Hint:The flow of fluid through any capillary tube depends upon the length and radius of the capillary tube. In order to find the net flow of fluid in the combination of capillary tubes given in the problem, we must find the equivalent flow through the series combination of capillary tubes accordingly. Then we shall compare it with the value of $X$ (flow through single tube) given to get the final solution.

Complete answer:
The flow of fluid, $v$ though any capillary tube is given by Poiseuille equation as:
$v=\dfrac{\pi P{{R}^{4}}}{8\eta L}$
Where,
$P=$ pressure in capillary tube at that point
$R=$ radius of capillary tube
$\eta =$ coefficient of viscosity
$L=$ length of capillary tube
However, the resistance in flow is expressed as:
$r=\dfrac{8\eta L}{\pi {{R}^{4}}}$
In first capillary tube, we have $R=R$ and $L=L$,
Therefore, the flow resistance through first tube is:
${{r}_{1}}=\dfrac{8\eta L}{\pi {{R}^{4}}}$
In second capillary tube, we have $R=2R$ and $L=2L$,
Therefore, the flow resistance through second tube is:
\[\begin{align}
  & {{r}_{2}}=\dfrac{8\eta \left( 2L \right)}{\pi {{\left( 2R \right)}^{4}}} \\
 & \Rightarrow {{r}_{2}}=\dfrac{16\eta L}{16\pi {{R}^{4}}} \\
 & \Rightarrow {{r}_{2}}=\dfrac{\eta L}{\pi {{R}^{4}}} \\
\end{align}\]
The net flow resistance in series is given as:
${{r}_{net}}={{r}_{1}}+{{r}_{2}}$
$\begin{align}
  & \Rightarrow {{r}_{net}}=\dfrac{8\eta L}{\pi {{R}^{4}}}+\dfrac{\eta L}{\pi {{R}^{4}}} \\
 & \Rightarrow {{r}_{net}}=\dfrac{9\eta L}{\pi {{R}^{4}}} \\
\end{align}$
Multiplying and dividing by 8, we can also write is as:
$\Rightarrow {{r}_{net}}=\dfrac{8\eta L}{\pi {{R}^{4}}}\times \dfrac{9}{8}$
Thus, the net flow through tube is, ${{v}_{net}}=\dfrac{P}{{{r}_{net}}}=\dfrac{P}{\dfrac{\pi P{{R}^{4}}}{8\eta L}\times \dfrac{9}{8}}$
$\Rightarrow {{v}_{net}}=\dfrac{\pi P{{R}^{4}}}{8\eta L}\times \dfrac{8}{9}$
Given that, $X=\dfrac{\pi P{{R}^{4}}}{8\eta L}$
Comparing the values of ${{v}_{net}}$ and $X$, we get
 $\Rightarrow {{v}_{net}}=X\times \dfrac{8}{9}$

Therefore, the correct option is (A) $\dfrac{8}{9}X$.

Note:
In physics, we often have a combination of things connected with each other. Just like the electric wires in an electric circuit, the elastic springs, capillary tubes and many such entities are connected with each other in some sort of pattern, that is, either they are joined parallel or in series with each other. Therefore, we apply the same rules as that of wires to simplify these combinations of springs, capillary tubes, etc as well.