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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2017 · 9 April · Online · Physics · Mechanical Properties of Fluids

Two tubes of radii r _ 1 and r _ 2 and lengths l _ 1 and l _ 2,...

Two tubes of radii r1{r}_{1} and r2{r}_{2} and lengths l1{l}_{1} and l2,{l}_{2,} respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1{P}_{1} and P2{P}_{2} are pressure differences across the two tubes. If P2{P}_{2} is 4P14{P}_{1} and l2{l}_{2} is l14\dfrac{{l}_{1}}{4} then the radius r2{r}_{2} will be equal to :

Options

  1. A4r14{r}_{1}
  2. Br1{r}_{1}
  3. C2r12{r}_{1}
  4. Dr12\dfrac{{r}_{1}}{2}

Correct answer

D. r12\dfrac{{r}_{1}}{2}

Step-by-step solution

dVdt=π8pr4ηL\dfrac{dV}{dt} = \dfrac{\pi}{8}\dfrac{pr^{4}}{ηL}

p1r14L1=p2r24L2\dfrac{{p}_{1}{r}_{1}^{4}}{{L}_{1}} = \dfrac{{p}_{2}{r}_{2}^{4}}{{L}_{2}}

p1r14l1=4p1r24(l14)<br/><br/>\dfrac{{p}_{1}{r}_{1}^{4}}{l1} = 4{p}_{1}{r}_{2}^{4} \left(\dfrac{l1}{4}\right)<br /><br />⇒{r}_{2}^{4} = r1416<br/><br/>\dfrac{{r}_{1}^{4}}{16}<br /><br />⇒{r}_{2} = r12\dfrac{{r}_{1}}{2}
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