A) 0.5 m
B) 1 m
C) 1.5 m
D) 2.2 m
Correct Answer: B
Solution :
The equation of continuity \[{{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}\] \[A\times 4=\frac{2}{3}A\times {{v}_{2}}\] \[{{v}_{2}}=6\,m{{s}^{-1}}\] From Bernoullis theorem, \[p+\rho g{{h}_{1}}+\frac{1}{2}\rho v_{1}^{2}=p+\rho g{{h}_{2}}+\frac{1}{2}\rho v_{2}^{2}\] Or \[g({{h}_{1}}-{{h}_{2}})=\frac{1}{2}(v_{1}^{2}-v_{1}^{2})\] \[g\times h=\frac{1}{2}[{{(6)}^{2}}-{{(4)}^{2}}]\]\[[\because {{h}_{1}}-{{h}_{2}}=h]\] \[10\times h=\frac{1}{2}[36-16]\] Or \[h=\frac{20}{20}\] \[=1\,m\]You need to login to perform this action.
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