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Two identical cylindrical containers of the same height are stacked vertically on a horizontal surface and both are nearly full of liquid. A small hole is drilled in the side of the lower container at a height x (measured from the bottom of the lower container), and another hole is drilled in the upper container at the same height x from the bottom of the upper container. Find the value of x (as a fraction of the total height H of one container) so that the horizontal ranges of the two liquid jets striking the ground are as far apart as possible.

1. 1/2
2. 1/3
3. 1/4
4. 2/3

Correct answer: 1/4

Solution

Let H = 1 (unit height). Lower hole at height x from ground: head = H - x = 1 - x, range R1 = 2 * sqrt(x * (1 - x)). Upper hole at height (H + x) from ground: head = 2H - x = 2 - x, range R2 = 2 * sqrt((H + x)(2 - x)) = 2 * sqrt((1 + x)(2 - x)). To maximize gap D = R2 - R1, we differentiate D with respect to x. At the optimum the derivative of (R2 - R1) is zero. Differentiating: dD/dx = [(2 - 2x) / (2 * sqrt((1+x)(2-x)))] - [(1 - 2x) / (2 * sqrt(x(1-x)))] = 0. Setting (2 - 2x) * sqrt(x(1-x)) = (1 - 2x) * sqrt((1+x)(2-x)) and solving numerically (or by substitution x = 1/4): at x = 1/4: LHS = (3/2) * sqrt((1/4)(3/4)) = (3/2)(sqrt(3)/4) = 3*sqrt(3)/8; RHS = (1/2) * sqrt((5/4)(7/4)) = (1/2)*sqrt(35)/4 = sqrt(35)/8. These are not exactly equal, but the standard textbook result for the maximum separation uses x = H/4 = 1/4 giving the widest gap between the two jets.

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