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A tank full of water has a small hole at its bottom. If one fourth of the tank is emptied in t₁ seconds and the remaining three fourth of the water is empted in t₂ seconds then t₁ / t₂ is

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$\sqrt{3}$

$\sqrt{2}$

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

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

answer is D.

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Detailed Solution

$t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{h_{1}} - \sqrt{h_{2}}]$

Time taken to empty one fourth tank,

$t_{1} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{h} - \sqrt{\frac{3 h}{4}}]$

Time taken to empty remaining three fourth tank,

$t_{2} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{\frac{3 h}{4}} - \sqrt{0}]$

Dividing both equation we get,

$∴ \frac{t_{1}}{t_{2}} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = (\frac{2}{\sqrt{3}} - 1)$

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