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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