A) 3
B) 4
C) 5
D) 6
Correct Answer: B
Solution :
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 \[{{Q}_{1}}={{Q}_{2}}\] \[\sqrt{\frac{h}{3}}=\frac{\sqrt{5h}}{{{\left[ {{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}-1 \right]}^{1/2}}}\] \[\sqrt{{{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}}-1=\sqrt{5\times 3}\] \[{{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}-1=15\] \[{{\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)}^{2}}=16\] \[\frac{{{A}_{1}}}{{{A}_{2}}}=4\]
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