A) \[{{a}_{1}}{{a}_{2}}\sqrt{\frac{2gh}{a_{1}^{2}-a_{2}^{2}}}\]
B) \[{{a}_{1}}{{a}_{2}}\sqrt{\frac{2g}{h\left( a_{1}^{2}-a_{2}^{2} \right)}}\]
C) \[{{a}_{1}}{{a}_{2}}\sqrt{\frac{(a_{1}^{2}+a_{2}^{2})h}{2g\left( a_{1}^{2}-a_{2}^{2} \right)}}\]
D) \[\frac{2{{a}_{1}}{{a}_{2}}gh}{\sqrt{a_{1}^{2}-a_{2}^{2}}}\]
Correct Answer: A
Solution :
[a] Let \[{{a}_{1}}\] and \[{{a}_{2}}\]be cross sectional areas and\[{{v}_{1}}\] and \[{{v}_{2}}\] the velocities of liquid flow at A and B. If \[{{p}_{1}}\]and \[{{p}_{2}}\]are pressures of liquid recorded by manometer, then \[{{P}_{1}}+\frac{1}{2}pv_{1}^{2}={{p}_{2}}+\frac{1}{2}\rho v_{2}^{2}\] \[\Rightarrow {{p}_{1}}-{{p}_{2}}=\frac{1}{2}\rho v_{1}^{2}\left( \frac{v_{2}^{2}}{v_{1}^{2}}-1 \right)\] By equation of continuity, \[{{a}_{1}}{{v}_{1}}={{a}_{2}}{{v}_{2}}\] Also, \[{{p}_{1}}-{{p}_{2}}=h\rho g\] Substituting these values, we have \[{{v}_{1}}=\sqrt{\frac{2gh}{\frac{a_{1}^{2}}{a_{2}^{2}}-1}}\] Rate of flow of liquid is\[{{a}_{1}}{{v}_{1}}={{a}_{1}}{{a}_{2}}\sqrt{\frac{2gh}{a_{1}^{2}-a_{2}^{2}}}\]
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