question_answer

A jet of water with cross sectional area 'a' is striking against a wall at an angle 6 to the horizontal and rebounds elastically. If the velocity of water jet is v and the density is p, the normal force acting on the wall is:

A) \[2{{\operatorname{av}}^{2}}\rho cos\theta \]               

B) \[{{\operatorname{av}}^{2}}\rho cos\theta \]

C) \[2\operatorname{av}\rho cos\theta \]                  

D) \[\operatorname{av}\,cos\theta \]

Correct Answer: A

Solution :

Newton's second law of motion \[{{\operatorname{F}}_{ext}}=\frac{dp}{dt}\] \[{{\operatorname{F}}_{ext}}={{p}_{2}}-{{p}_{1}}\] \[{{p}_{2}}=\operatorname{mv}\cos \theta \,\,\,\,\,\,\,\,\,\,\,\ell =\frac{mass}{volume}=\frac{m}{av}\] \[{{p}_{2}} = \left( \ell av \right) v Cos\theta  =\ell a{{v}^{2}} Cos\theta \] \[{{\operatorname{p}}_{i}} =-\ell e{{v}^{2}}Cos\theta \] volume flow rate = velocity x area \[{{F}_{e\operatorname{xt}}} = {{P}_{2}}-{{P}_{i}}= a{{v}^{2}}Cos\theta  -\left( -\ell a{{v}^{2}}cos\theta  \right)\] \[{{F}_{e\operatorname{xt}}} =2\ell a{{v}^{2}}\cos \theta \]