A) \[\rho {{v}^{2}}\]
B) \[\frac{1}{4}\rho {{v}^{2}}\]
C) \[\frac{3}{4}\rho {{v}^{2}}\]
D) \[\frac{1}{2}\rho {{v}^{2}}\]
Correct Answer: C
Solution :
The momentum per second carried by liquid per second is \[\rho {{v}^{2}}A.\] [A is area of cross section of pipe] The force exerted due to reflected back molecules \[2\left( \frac{1}{4}\rho a{{v}^{2}} \right)\] So, the resultant pressure is \[\frac{\left( \frac{1}{2}\rho A{{v}^{2}}+\frac{1}{4}\rho A{{v}^{2}} \right)}{A}=\frac{3}{4}\rho {{v}^{2}}\]
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