Verified Solution • 2025–26 Session moderate

QJEEM17Ph2dea: JEE Main 2017 Physics - Class 11 JEEM17Ph2dea Solution: Two tubes of radii r 1 and r 2 , and lengths l 1 and l 2 ,…

Two tubes of radii r1 and r2, and lengths l1 and l2 , respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1 and P2 are pressure differences across the two tubes.

If P2 is 4P1 and l2 is 114{{{1_1}} \over 4} , then the radius r2 will be equal to :

Multiple Choice Options

  • A r1
  • B 2r1
  • C 4r1
  • D r12{{{r_1}} \over 2}

Step-by-Step Verified Solution

We know,

Rate of flow of liquid through narrow tube,

dvdt{{dv} \over {dt}} = πPr⁡48ηl{{\pi {{\Pr }^4}} \over {8\eta l}}

Both tubes are connected in series so rate of flow of liquid is same.

∴   \therefore\,\,\, πP1r148ηl1{{\pi {P_1}r_1^4} \over {8\eta {l_1}}} = πP2r248ηl2{{\pi {P_2}r_2^4} \over {8\eta {l_2}}}

⇒\Rightarrow    \,\,\, P1r14l1{{{P_1}{r_1}^4} \over {{l_1}}} = P2r24l2{{{P_2}{r_2}^4} \over {{l_2}}}

Given,

P2 = 4P1   and   l2{l_2} = l14{{{l_1}} \over 4}

⇒\Rightarrow    \,\,\, P1r14l1{{{P_1}{r_1}^4} \over {{l_1}}} = 4P1r24l14{{4{P_1}{r_2}^4} \over {{{{l_1}} \over 4}}}

⇒\Rightarrow    \,\,\, r24{r_2}^4 = r1416{{{r_1}^4} \over {16}}

⇒\Rightarrow    \,\,\, r2 = r12{{{r_1}} \over 2}

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