A tank full of water has a small hole at its bottom. Let $t_{1}$ be the time taken to empty first one thbird of the tank and $t_{2}$ be the time taken to empty second one third of the tank and $t_{3}$ be the time taken to empty rest of the tank then

t_{1} = t_{2} = t_{3}

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t_{1} > t_{2} > t_{3}

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t_{1} < t_{2} < t_{3}

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t_{1} > t_{2} < t_{3}

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Solution

The correct option is B $t_{1} < t_{2} < t_{3}$
As the water level is decreasing the pressure at the bottom due to water (pgh) is decreasing. Because of this, inner pressure and o'cater pressure at the small hole are having decreasing difference. Due to this lowering pressure difference, the velocity of efflux will decrease. Thus it will take more time to sap out same amount of water at lower water level than that at higher.
So, $t_{3} > t_{2} > t_{1}$ Option - C is correct.

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