A) +10 %
B) +46 %
C) -10 %
D) -40 %
Correct Answer: B
Solution :
Volume o liquid flowing through capillary per second is given by \[V=\frac{\pi p{{r}^{4}}}{8\eta l}\] \[\therefore \] \[V\propto {{r}^{4}}\] \[\Rightarrow \] \[\frac{{{V}_{2}}}{{{V}_{1}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{4}}\] \[\therefore \] \[{{V}_{2}}={{V}_{1}}{{\left( \frac{110}{100} \right)}^{4}}={{V}_{1}}{{(1.1)}^{4}}\] \[=1.4641\,V\] \[\therefore \] \[\frac{\Delta V}{V}=\frac{{{V}_{2}}-{{V}_{1}}}{V}\] \[=\frac{1.4641\,V-V}{V}\] \[=0.46\] or \[46%\]
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