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In screening round of National Anveshika Experimental Skills Test ( NAEST 2017 ), a question was asked on the flow of fluids. The water is flowing through a pipe of circular cross-section. It comes out of the pipe horizontally and travels in a parabolic path hitting the ground at some distance. When end of the pipe is pressed, the water hits the ground farther away.
Choose the correct option(s) from the following:
Solution: When opening of the pipe is pressed, the water travels more distance due to increase in in the flow speed. This can be explained using a result from the geometry and the equation of continuity.
The shape of pipe's cross-section changes from circular to elliptical when its opening is pressed. In this process, the perimeter of the pipe remains same but its cross-section area decreases. It can be shown that area of the circle is maximum for a given perimeter. Thus, cross-section area of the opening decreases from $A_1$ to $A_2$ when it is pressed.
Let $v_0$ and $A_0$ be the flow speed and cross-sectional area at the inlet. The flow speed is $v_1$ when opening is circular and it is $v_2$ when opening is elliptical. Apply continuity equation in two cases to get \begin{align} A_0 v_0 =A_1 v_1 \\ A_0 v_0 = A_2 v_2 \end{align} which gives \begin{align} v_2&=v_1(A_1/A_2) \\ &> v_1\;\; (\text{because}\, A_1>A_2) \end{align} Note that equation of continuity is a statement of mass conservation for incompressible blow.
Let height of the pipe from the ground be $h$. We assume that water stream takes a parabolic path similar to the path taken by a particle projected horizontally. The time taken by a fluid particle to reach the ground is \begin{align} t=\sqrt{2h/g} \end{align}
The horizontal distance travelled by the stream in two cases are \begin{align} x_1&=v_1 t =v_1\sqrt{2h/g}, \\ x_2&=v_2 t =(A_1/A_2) v_1\sqrt{2h/g}, \\ & > x_1 \;\; (\text{because}\, A_1>A_2) \end{align} Thus, stream travels more horizontal distance when pipe's opening is pressed.