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A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in t₁ seconds and the remaining three-fourths of the tank is emptied in t₂ seconds. Then the ratio $\frac{t_{1}}{t_{2}}$ is:

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$\frac{2}{\sqrt{3}} - 1$

√3

√2

1/√2

answer is D.

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Detailed Solution

$t_{1} \propto (\sqrt{h} - \sqrt{3 h / 4}) t_{2} \propto (\sqrt{3 h / 4})$

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