JEE Main 2017 · 9 Apr · Shift 1 · Physics Q9

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Two tubes of radii $r_1$ and $r_2$ , and lengths $l_1$ and $l_2$ , respectively, are connected in series and a liquid flows through each of them in stream line conditions. $P_1$ and $P_2$ are pressure differences across the two tubes. If $P_2$ is $4P_1$ and $l_2$ is $\dfrac{l_1}4$ , then the radius $r_2$ will be equal to :

r_1

2r_1

4r_1

\dfrac{r_1}2

Answer and worked solution

Official answer: D

In series the flow rate is the same. By Poiseuille's law $Q\propto\dfrac{Pr^4}l$ :
$\frac{P_1r_1^4}{l_1}=\frac{4P_1r_2^4}{l_1/4}\implies r_2^4=\frac{r_1^4}{16},\quad r_2=\frac{r_1}2.$

Topics:Viscosity and Stokes' law

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 9