A) Zero
B) \[1.27\times {{10}^{-2}}N\]
C) \[1.27\times {{10}^{-4}}N\]
D) \[0.127N\]
Correct Answer: D
Solution :
Rate of flow of water \[\frac{V}{t}=\frac{10\,c{{m}^{3}}}{\sec }=10\times {{10}^{-6}}\frac{{{m}^{3}}}{\sec }\] Density of water \[\rho =\frac{{{10}^{3}}kg}{{{m}^{3}}}\] Cross-sectional area of pipe \[A=\pi {{(0.5\times {{10}^{-3}})}^{2}}\] Force\[=m\frac{dv}{dt}=\frac{mv}{t}=\frac{V\rho v}{t}=\frac{\rho V}{t}\times \frac{V}{At}={{\left( \frac{V}{t} \right)}^{2}}\frac{\rho }{A}\] \[\left( \because \ v=\frac{V}{At} \right)\] By substituting the value in the above formula we get \[F=0.127\,N\]
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