Q 11-09-153JEE MainJEE Main 2017 (9 Apr)Medium
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:
Answer: (D) $\dfrac{r_1}{2}$
In series the same volume flows per second. By Poiseuille's law $Q = \dfrac{\pi Pr^4}{8\eta l}$:
$$\frac{P_1r_1^4}{l_1} = \frac{P_2r_2^4}{l_2} = \frac{4P_1r_2^4}{l_1/4} = \frac{16P_1r_2^4}{l_1}$$
$$r_2^4 = \frac{r_1^4}{16} \;\Rightarrow\; r_2 = \frac{r_1}{2}$$
Solution by Sreeraj P, M.Sc Physics