Question

A tube of radius R and length L is connected in series with another tube of radius \[\dfrac{R}{2}\]and length\[\dfrac{L}{8}\]. If the pressure across the tubes taken together is P, the pressure across the two tubes separately are:
A. \[\dfrac{P}{2}\]and \[\dfrac{P}{2}\]
B. \[\dfrac{P}{3}\]and \[\dfrac{{3P}}{2}\]
C. \[\dfrac{P}{4}\]and \[\dfrac{{3P}}{2}\]
D. \[\dfrac{P}{3}\]and \[\dfrac{{2P}}{3}\]

Answer 1

Hint: In this question, we need to determine the individual pressure across the two tubes. For this we will use the relation between the flow rate formulas and find the pressure in the individual tubes.

Complete step by step answer:Radius of tube 1 = R
Length of tube 1 = L
Radius of tube 2 =\[\dfrac{R}{2}\]
Length of tube 2=\[\dfrac{L}{8}\]
Let,
The pressure drop across tube 1 is \[{P_1}\]
And the pressure drop across tube 2 is \[{P_2}\]

Hence we can say\[{P_1} + {P_2} = P - - (i)\], since the pressure across the tubes taken together is P
Now we know Flow rate\[ = \dfrac{{\pi \Delta {{\Pr }^4}}}{{8\eta L}}\]
Now since the both tube are in series, hence we can say flow rate will be same for both the pipes so we write
\[
  \dfrac{{\pi \Delta {\operatorname{P} _1}{r_A}^4}}{{8\eta {L_A}}} = \dfrac{{\pi \Delta {\operatorname{P} _2}{r_B}^4}}{{8\eta {L_B}}} \\
  \dfrac{{\pi \Delta {\operatorname{P} _1}{R^4}}}{{8\eta L}} = \dfrac{{\pi \Delta {\operatorname{P} _2}{{\left( {\dfrac{R}{2}} \right)}^4}}}{{8\eta \dfrac{L}{8}}} \\
  \dfrac{{\pi \Delta {\operatorname{P} _1}{R^4}}}{{8\eta L}} = \dfrac{1}{2}\dfrac{{\pi \Delta {\operatorname{P} _2}{R^4}}}{{8\eta L}} \\
  2{\operatorname{P} _1} = {\operatorname{P} _2} \\
 \]
Where \[{P_1} + {P_2} = P\]from equation (i), now we substitute \[2{\operatorname{P} _1} = {\operatorname{P} _2}\]in this equation to find the pressure across the two tubes separately
\[
  {P_1} + 2{P_1} = P \\
  3{P_1} = P \\
  \therefore {P_1} = \dfrac{P}{3} \\
 \]
Hence the pressure in pipe B will be equal to
\[
  {\operatorname{P} _2} = 2{\operatorname{P} _1} \\
   = 2 \times \dfrac{P}{3} \\
   = \dfrac{{2P}}{3} \\
 \]
Therefore the pressure in pipe \[{P_1} = \dfrac{P}{3}\] and pipe \[{\operatorname{P} _2} = \dfrac{{2P}}{3}\]respectively
Option D is correct

Note:Flow rate is the volume of fluid flowing through an area each second which is given by the formula
\[Q = \dfrac{{\pi \Delta {{\Pr }^4}}}{{8\eta L}}\], where L is the length of tube, r is the radius of the tube, P is the pressure of fluid through the pipe.