d
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$