• # question_answer  Two venturimeters of different area ration are connected at different locations of a pipeline to measure discharge. Similar manometers are used across the two venturimeters to register the head differences. The first venturimeter of area ratio 2 registers a head difference 'h' while the second venturimeter registers '5h?. The area ratio for the second venturimeter is: A) 3                                 B) 4C) 5                                 D) 6

For venturimeter, $Q={{C}_{d}}\left[ \frac{{{A}_{1}}{{A}_{2}}}{A_{1}^{2}-A_{1}^{2}} \right]\times {{\left[ 2g\,\left( \frac{{{s}_{2}}}{{{s}_{1}}}-1 \right)\,h \right]}^{1/2}}$ $\alpha \propto \frac{{{A}_{1}}}{\sqrt{{{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}-1}}\,\sqrt{h}$ For $\frac{{{A}_{1}}}{{{A}_{2}}}=2$ ${{Q}_{1}}\propto \sqrt{\frac{h}{3}}$ ${{Q}_{1}}\propto \frac{\sqrt{5h}}{\sqrt{{{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}-1}}$ For ${{Q}_{1}}={{Q}_{2}}$ $\sqrt{\frac{h}{3}}=\frac{\sqrt{5h}}{{{\left[ {{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}-1 \right]}^{1/2}}}$ $\sqrt{{{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}}-1=\sqrt{5\times 3}$ ${{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}-1=15$ ${{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}=16$ $\frac{{{A}_{1}}}{{{A}_{2}}}=4$