A) 9.4 m
B) 4.9 m
C) 0.49 m
D) 0.94 m
Correct Answer: D
Solution :
Given, \[{{l}_{1}}={{l}_{2}}=1,\]and \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{1}{2}\] \[V=\frac{\pi {{P}_{1}}r_{1}^{4}}{8\eta l}=\frac{\pi {{P}_{2}}r_{2}^{4}}{8\eta l}\]Þ \[\frac{{{P}_{1}}}{{{P}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{4}}=16\] Þ \[{{P}_{1}}=16{{P}_{2}}\] Since both tubes are connected in series, hence pressure difference across combination, \[P={{P}_{1}}+{{P}_{2}}\]Þ 1 = \[{{P}_{1}}+\frac{{{P}_{1}}}{16}\] Þ \[{{P}_{1}}=\frac{16}{17}=0.94m\]
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