A) zero
B) \[1.27\times {{10}^{-2}}N\]
C) \[1.27\times {{10}^{-4}}N\]
D) \[0.127\,N\]
Correct Answer: D
Solution :
The force is given by \[F=\frac{dp}{dt}=\frac{d}{dt}(m\upsilon )=V\rho \upsilon \] ?(i) where V is the volume of water flowing per second \[\upsilon =\]speed of the flow \[V=\]cross-sectional area \[\times \] velocity \[=A\upsilon \] or \[\upsilon =\frac{V}{A}\] ?(2) So, putting the value of v in equation (1), we get \[F={{V}_{\rho }}\frac{V}{A}\] \[=\frac{{{V}^{2}}\rho }{A}=\frac{{{V}^{2}}\rho }{\pi {{r}^{2}}}\] \[=\frac{{{V}^{2}}\rho }{\pi {{\left( \frac{D}{2} \right)}^{2}}}=\frac{4{{V}^{2}}\rho }{\pi {{D}^{2}}}\] \[=\frac{4\times {{(10\times {{10}^{-6}})}^{2}}\times {{10}^{3}}}{3.14\times {{({{10}^{-3}})}^{2}}}\] \[=0.127\,N\]
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