question_answer

A tank is filled with water upto height H. When a hole is made at a distance h below the level of water,  what will be the horizontal range of water jet?                                                                            [BHU M-2003]

A)  \[2\sqrt{h\left( H-h \right)}\]                   

B)  \[4\sqrt{h\left( H+h \right)}\]

C)  \[4\sqrt{h\left( H-h \right)}\]                   

D)  \[2\sqrt{h\left( H+h \right)}\]

Correct Answer: A

Solution :

                 Key Idea: The sum of the pressure and the total energy per unit volume of the liquid must be the same at the surface of liquid and at every point of orifice. Form Bernoulli?s theorem,                 \[P+0+\rho gH=P+\frac{1}{2}\rho {{v}^{2}}+\rho g\left( H-h \right)\] Where P is atmospheric pressure \[\rho \] is density       and \[v\] is velocity of efflux.                 \[\therefore \]                  \[\frac{1}{2}\rho {{v}^{2}}=\rho gh\]                 \[\Rightarrow \]                                               \[v=\sqrt{2gh}\] After emerging from orifice, liquid falls along a parabolic path. \[\therefore \]  \[s=\frac{1}{2}g{{t}^{2}}\]                                 \[H-h=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]               \[t=\sqrt{\frac{2\left( H-h \right)}{g}}\] The horizontal distance is X=horizontal velocity \[\times \] time =v\[\times \]t                                 \[=v\times t\]                                 \[=\sqrt{2gh}\times \sqrt{\frac{2\left( H-h \right)}{g}}\]                                 \[=2\sqrt{h\left( H-h \right)}\] Note: When the orifice is exactly in the middle of the wall of the vessel, the stream of liquid will fall at a maximum distance equal to the height of liquid in the vessel.