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 streamline conditions. $P_1$ and $P_2$ are pressure differences across the two tubes. If $P_2$ is $4P_1$ and $l_2$ is $ \frac{l_1}{4}$ , then the radius $r_2$ will be equal to
JEE MAIN 2017, Diffcult
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The volume of liquid flowing through both the tubes i.e., rate of flow of liquid same.

$Therefore,V = {V_1} = {V_2}$

$i.e.,\frac{{\pi {p_1}r_1^4}}{{8\eta {l_1}}} = \frac{{\pi {p_2}r_2^4}}{{8\eta {l_2}}}$

$or\,\,\,\,\,\,\frac{{{p_1}r_1^4}}{{{l_1}}} = \frac{{{p_2}r_2^4}}{{{l_2}}}$

$\,\,\,\,{P_2} = 4\,{P_1}\,\,and\,{l_2} = {l_1}/4$

$\frac{{{p_1}r_1^4}}{{{l_1}}} = \frac{{4{p_1}r_2^4}}{{{l_1}/4}} \Rightarrow r_2^4 = \frac{{r_1^4}}{{16}}$

$ \Rightarrow {r_2} = {r_1}/2$

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