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A tank full of water drains through a small hole at its bottom. The upper one-fourth of the water (by height) takes t1 seconds to drain, and the remaining three-fourths takes t2 seconds. Find the ratio t1/t2.

1. sqrt(3)
2. sqrt(2)
3. 1/sqrt(2)
4. 2/sqrt(3) - 1

Correct answer: 2/sqrt(3) - 1

Solution

For a draining tank the time to fall from height h_i to h_f is proportional to (sqrt(h_i) - sqrt(h_f)). Draining the top quarter: from H to 3H/4 gives t1 ~ sqrt(H) - sqrt(3H/4). Draining the rest: from 3H/4 to 0 gives t2 ~ sqrt(3H/4) - 0. The ratio t1/t2 = (1 - sqrt(3)/2)/(sqrt(3)/2) = (2 - sqrt(3))/sqrt(3) = 2/sqrt(3) - 1.

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