Ratio of time to empty a tank in two stages
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A large tank filled with water to a height 'h' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from $h$ to $\frac{h}{2}$ and from $\frac{h}{2}$ to bottom is (a) $\sqrt{2}$ (b) $1/2$ (c) $\sqrt{2}-1$ (d) $\frac{1}{\sqrt{2}-1}$
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Hi Mohammad, let's solve this fluid dynamics problem together.
Draining a Tank
To solve this, we use the principle of continuity and Torricelli's law. For a tank with cross-sectional area A and a small hole of area a, the velocity of the exiting water is given by the square root of two g h.
Rearranging this equation to solve for the time interval dt, we get this expression.
Integrating this, the time taken to drop from an initial height H1 to a final height H2 is proportional to the difference of the square roots of those heights.
Now, let's find the time t1 to fall from h to h over two.
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