• # question_answer A pipe of diameter D conveying a discharge Q is to be replaced by parallel pipes of smaller diameter d to discharge the same quantity. What will be the ratio of $\frac{D}{d}\,?$($f$Is same for all pipes) A) $\frac{D}{d}=2$                     B) $\frac{D}{d}=\sqrt{2}$C) $\frac{D}{d}={{4}^{1/5}}$               D) $\frac{D}{d}={{4}^{1/3}}$

Correct Answer: C

Solution :

$Q={{Q}_{1}}+{{Q}_{2}}$ Since ${{d}_{1}}={{d}_{2}}=d$ $\therefore$  ${{Q}_{1}}={{Q}_{2}}$ $\therefore$  $Q=2{{Q}_{1}}$ ${{D}^{2}}v=2{{d}^{2}}{{v}_{1}}$                                               ?1 Also $\frac{fL{{v}^{2}}}{2gD}=\frac{fl{{v}^{2}}}{2g{{d}_{1}}}$ $\frac{{{v}^{2}}}{D}=\frac{v_{1}^{2}}{{{d}^{11}}}$ ${{\left( \frac{v}{{{v}^{1}}} \right)}^{2}}=\frac{D}{{{d}_{1}}}=\frac{D}{d}$ From eq. $1.,\frac{v}{{{v}_{1}}}=2\,{{\left( \frac{d}{D} \right)}^{2}}$ ${{\left( \frac{v}{{{v}_{1}}} \right)}^{2}}=4\,{{\left( \frac{d}{D} \right)}^{4}}$ $\therefore$  $\frac{D}{d}=4\,{{\left( \frac{d}{D} \right)}^{4}}$ $\frac{D}{d}\times 4\,{{\left( \frac{d}{D} \right)}^{4}}=4$ ${{\left( \frac{D}{d} \right)}^{5}}=4$ $\frac{D}{d}={{4}^{1/5}}$

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