A) 4Q
B) Q
C) \[\frac{Q}{4}\]
D) \[\frac{Q}{8}\]
Correct Answer: D
Solution :
\[V=\frac{\pi {{\Pr }^{4}}}{8\eta l}\] \[\therefore \] \[V\propto p{{r}^{4}}\] (\[\eta \] and I are contants) \[\therefore \] \[\frac{{{V}_{2}}}{{{V}_{1}}}=\left( \frac{{{p}_{2}}}{{{p}_{1}}} \right){{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{4}}\] \[=2\times {{\left( \frac{1}{2} \right)}^{4}}=\frac{1}{8}\] \[{{V}_{2}}=\frac{Q}{8}\]
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