a
(a)Fluid resistance is given by \(R = \frac{{8\eta l}}{{\pi {r^4}}}.\) When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is \({R_e} = {R_1} + {R_2} = \frac{{8\eta L}}{{\pi {r^4}}} + \frac{{8\eta \times 2L}}{{\pi {{(2R)}^4}}} = \left( {\frac{{8\eta L}}{{\pi {r^4}}}} \right) \times \frac{9}{8}\)Equivalent resistance becomes \(\frac{9}{8}\)times so rate of flow will be \(\frac{8}{9}X\)