d
(d)Given, \({l_1} = {l_2} = 1,\)and \(\frac{{{r_1}}}{{{r_2}}} = \frac{1}{2}\) \(V = \frac{{\pi {P_1}r_1^4}}{{8\eta l}} = \frac{{\pi {P_2}r_2^4}}{{8\eta l}}\)==> \(\frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4} = 16\) ==> \({P_1} = 16{P_2}\) Since both tubes are connected in series, hence pressure difference across combination, \(P = {P_1} + {P_2}\)==> 1 = \({P_1} + \frac{{{P_1}}}{{16}}\) ==> \({P_1} = \frac{{16}}{{17}} = 0.94m\)