A) \[18\,\,c{{m}^{3}}\]
B) \[1\,\,c{{m}^{3}}\]
C) \[16\,\,c{{m}^{3}}\]
D) \[4\,\,c{{m}^{3}}\]
Correct Answer: B
Solution :
From Poiseuille's formula, if \[l\]and \[r\]be the length and radius of the tube and P the pressure difference across its ends, then the volume of the liquid flowing per second through the tube is given by \[Q=\frac{\pi \,{{\Pr }^{4}}}{8\,\eta \,l}\] Where \[\eta \] is the coefficient of viscosity of the liquid Given, \[{{Q}_{1}}=16\,\,\,c{{m}^{3}}/s,\,r=a\] \[\therefore \] \[16=\frac{\pi P{{a}^{4}}}{8\eta l}\] ?(1) When \[{{l}_{2}}=l,\,\,{{r}_{2}}=\frac{a}{2}\] \[\therefore \] \[{{Q}_{2}}=\frac{\pi P{{\left( \frac{a}{2} \right)}^{4}}}{8\,\eta \,l}\] ?(2) Dividing \[Eq\]. (1) by Eq. (2), we get \[\frac{16}{{{Q}_{2}}}=\frac{\pi P{{a}^{4}}}{8\eta \,l}\times \frac{8\eta l}{\pi P{{\left( \frac{a}{2} \right)}^{4}}}=16\] \[\Rightarrow \] \[{{Q}_{2}}=1\,c{{m}^{3}}/s.\] Note: While giving the formula \[Q=\frac{\pi {{\Pr }^{4}}}{8\eta l},\]Poiseuille assumed that the flow of liquid in the tube is stream lined and parallel to the axis of the tube and the pressure difference across the tube is just enough to overcome the viscous force.
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