question_answer

   We have two (narrow) capillary tubes \[{{T}_{1}}\] and \[{{T}_{2}}\]. Then lengths are \[{{l}_{1}}\] and \[{{l}_{2}}\] and radii of cross-sections are \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively. The rate of flow of water under a pressure difference \[p\] through the tube \[{{T}_{1}}\] is\[8\,\,c{{m}^{3}}/s\]. If \[{{l}_{1}}=2{{l}_{2}}\] and \[{{r}_{1}}={{r}_{2}}\]. What will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (=p)?

A)  \[4\,c{{m}^{3}}/s\]                        

B)  \[16/3\,\,c{{m}^{3}}/s\]

C)  \[8/17\,c{{m}^{3}}/s\]

D)  None of these