A) \[27.8\,m{{s}^{-1}}\]
B) \[41.0\,m{{s}^{-1}}\]
C) \[9.6\,m{{s}^{-1}}\]
D) \[19.7\,m{{s}^{-1}}\]
Correct Answer: B
Solution :
According to Bernoulli's theorem, \[{{P}_{B}}+h\rho g={{P}_{A}}+\frac{1}{2}\rho v_{A}^{2}\] \[(\text{As }{{v}_{A}}>>{{v}_{B}})\] \[3.10P+53\times 660\times 10=P+\frac{1}{2}\times 660v_{A}^{2}\] Þ \[2.1\times 1.01\times {{10}^{5}}+3.498\times {{10}^{5}}\]\[=\frac{1}{2}\times 660\times v_{A}^{2}\] Þ \[5.619\times {{10}^{5}}=\frac{1}{2}\times 660\times v_{A}^{2}\] \ \[{{v}_{A}}=\sqrt{\frac{2\times 5.619\times {{10}^{5}}}{660}}\]= 41 m/sYou need to login to perform this action.
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