A) \[\text{4}.\text{31}\times \text{1}{{0}^{-1}}\text{ N}\]
B) \[\text{4}.\text{32}\times \text{1}{{0}^{-2}}\text{ N}\]
C) \[\text{4}.\text{32}\times \text{1}{{0}^{-3}}\text{ N}\]
D) 43.2 N
Correct Answer: D
Solution :
From Newtons second law Rate of change of momentum is equal to force \[\therefore \] \[F=\frac{dp}{dt}\] ... (i) where p = mv ... (ii) From Eqs. (i) and (ii), we get \[F=\frac{mv\,\cos \theta -(-mv\,\cos \theta )}{\Delta t}\] \[\therefore \] \[F=\frac{2\,mv\,\cos \theta }{\Delta t}\] \[\Rightarrow \] \[F=2\,A\,v\,\rho \,v\,\cos \theta \] \[\Rightarrow \] \[F=2\,A\,\rho \,{{v}^{2}}\times \frac{1}{2}\] \[\Rightarrow \] \[F=A\,\rho \,{{v}^{2}}\] Given, \[A=3\,c{{m}^{2}}=3\times {{({{10}^{-2}})}^{2}}{{m}^{2}}\], \[\rho ={{10}^{3}}kg/{{m}^{3}}\], \[v=12\,m/s\] \[\therefore \] \[F=3\times {{10}^{-4}}\times {{10}^{3}}\times {{(12)}^{2}}\] = 43.2 N
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