A) \[\frac{1}{2}A\rho {{v}^{3}}\]
B) \[\frac{1}{2}A\rho {{v}^{2}}\]
C) \[\frac{1}{2}A\rho v\]
D) \[A\rho v\]
Correct Answer: A
Solution :
Energy supplied to liquid per second by the pump = \[\frac{1}{2}\frac{m{{v}^{2}}}{t}\]= \[\frac{1}{2}\frac{V\rho {{v}^{2}}}{t}\]= \[\frac{1}{2}A\times \left( \frac{l}{t} \right)\times \rho \times {{v}^{2}}\] \[\left[ \frac{l}{t}=v \right]\] \[=\frac{1}{2}A\times v\times \rho \times {{v}^{2}}\]= \[\frac{1}{2}A\rho {{v}^{3}}\]
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