Answer
64.8k+ views
Hint: In this solution, we will use Poiseuille’s Law. This law helps us in relating the flow rate in a tube with a pressure difference and resistance and with a resistance $R$ dependent on the viscosity of the liquid in the tube
Formula used: In this solution, we will use the following formula:
Resistance in a tube: $R = \dfrac{{8\eta l}}{{\pi {r^4}}}$ where $\eta $ is the viscosity of the liquid, $l$ is the length of the tube, $r$ is the radius of the tube.
Complete step by step answer:
We’ve been given that two capillary tubes of bore diameter but of lengths ${l_1}$ and ${l_2}$ are fitted side by side to the bottom of a vessel containing water and we want to find the length of a single tube of the same radius such that the rate of flow remains constant.
The resistance experienced by water when flowing in a tube is calculated by Poiseuille’s Law as
$R = \dfrac{{8\eta l}}{{\pi {r^4}}}$
To experience the same amount of resistance, the net resistance of the two tubes must be equal to the resistance of the new tube. Since the two tubes are placed side by side, the net resistance of the tubes can be calculated as
$\dfrac{1}{{{R_{net}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
$ \Rightarrow \dfrac{1}{{{R_{net}}}} = \dfrac{1}{{\dfrac{{8\eta {l_1}}}{{\pi {r^4}}}}} + \dfrac{1}{{\dfrac{{8\eta {l_2}}}{{\pi {r^4}}}}}$
This net resistance must be equal to the resistance of the newer tube and we can calculate it as
${R_{new}} = {R_{net}}$
$ \Rightarrow \dfrac{1}{{{R_{new}}}} = \dfrac{1}{{{R_{net}}}}$
So now we can write
$\dfrac{1}{{\dfrac{{8\eta {l_{new}}}}{{\pi {r^4}}}}} = \dfrac{1}{{\dfrac{{8\eta {l_1}}}{{\pi {r^4}}}}} + \dfrac{1}{{\dfrac{{8\eta {l_2}}}{{\pi {r^4}}}}}$
Which can be simplified to
\[\dfrac{1}{{{l_{new}}}} = \dfrac{1}{{{l_1}}} + \dfrac{1}{{{l_2}}}\]
$\therefore {l_{new}} = \dfrac{{{l_1}{l_2}}}{{{l_1} + {l_2}}}$
Note: Here we have assumed that the pressure difference across the two ends of the two tubes will remain constant in the longer new tube as well. The combination of two tubes can also be remembered as the combination of two electrical resistances in parallel since they have the same relations and the resistance in the flow can be determined using Poiseuille’s Law.
Formula used: In this solution, we will use the following formula:
Resistance in a tube: $R = \dfrac{{8\eta l}}{{\pi {r^4}}}$ where $\eta $ is the viscosity of the liquid, $l$ is the length of the tube, $r$ is the radius of the tube.
Complete step by step answer:
We’ve been given that two capillary tubes of bore diameter but of lengths ${l_1}$ and ${l_2}$ are fitted side by side to the bottom of a vessel containing water and we want to find the length of a single tube of the same radius such that the rate of flow remains constant.
The resistance experienced by water when flowing in a tube is calculated by Poiseuille’s Law as
$R = \dfrac{{8\eta l}}{{\pi {r^4}}}$
To experience the same amount of resistance, the net resistance of the two tubes must be equal to the resistance of the new tube. Since the two tubes are placed side by side, the net resistance of the tubes can be calculated as
$\dfrac{1}{{{R_{net}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
$ \Rightarrow \dfrac{1}{{{R_{net}}}} = \dfrac{1}{{\dfrac{{8\eta {l_1}}}{{\pi {r^4}}}}} + \dfrac{1}{{\dfrac{{8\eta {l_2}}}{{\pi {r^4}}}}}$
This net resistance must be equal to the resistance of the newer tube and we can calculate it as
${R_{new}} = {R_{net}}$
$ \Rightarrow \dfrac{1}{{{R_{new}}}} = \dfrac{1}{{{R_{net}}}}$
So now we can write
$\dfrac{1}{{\dfrac{{8\eta {l_{new}}}}{{\pi {r^4}}}}} = \dfrac{1}{{\dfrac{{8\eta {l_1}}}{{\pi {r^4}}}}} + \dfrac{1}{{\dfrac{{8\eta {l_2}}}{{\pi {r^4}}}}}$
Which can be simplified to
\[\dfrac{1}{{{l_{new}}}} = \dfrac{1}{{{l_1}}} + \dfrac{1}{{{l_2}}}\]
$\therefore {l_{new}} = \dfrac{{{l_1}{l_2}}}{{{l_1} + {l_2}}}$
Note: Here we have assumed that the pressure difference across the two ends of the two tubes will remain constant in the longer new tube as well. The combination of two tubes can also be remembered as the combination of two electrical resistances in parallel since they have the same relations and the resistance in the flow can be determined using Poiseuille’s Law.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)