A fluid is drawn up through a tube as shown in the figure.If the height difference from the top of the tank to the bottom of the siphon is H = 1.1 m,then what is the outflow speed? Assume the pressure is the same at the top of the tank and at the siphon outlet.

A) 3.3 m/s
B) 21.6 m/s
C) 4.6 m/s
D) 5.4 m/s

Question

A fluid is drawn up through a tube as shown in the figure.If the height difference from the top of the tank to the bottom of the siphon is H = 1.1 m,then what is the outflow speed? Assume the pressure is the same at the top of the tank and at the siphon outlet. 
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A) 3.3 m/s 
B) 21.6 m/s 
C) 4.6 m/s 
D) 5.4 m/s

Quizplus · Accepted Answer

Using Bernoulli's equation, we can equate the pressure difference to the difference in potential energy per unit volume of fluid: 
$$\frac{1}{2}ho v_1^2 + ho gh_1 = \frac{1}{2}ho v_2^2 + ho gh_2$$
where $v_1$ and $v_2$ are the velocities at the top of the tank and at the siphon outlet, respectively. Since we assume the pressures are the same, we can drop the pressure terms on both sides. We want to solve for $v_2$:
$$\frac{1}{2}v_1^2 + gh_1 = \frac{1}{2}v_2^2 + gh_2$$
Rearranging and solving for $v_2$ gives:
$$v_2 = \sqrt{v_1^2 + 2gh} = \sqrt{2gh}$$
Plugging in the given values, we have:
$$v_2 = \sqrt{2(9.81	ext{ m/s}^2)(1.1	ext{ m})} \approx \boxed{	extbf{(C) }4.6	ext{ m/s}}$$

A Fluid Is Drawn Up Through a Tube as Shown