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Substituting the values, \( V = \pi (3)^2 (5) = 45\pi \) cubic meters.
At a drainage rate of 2 cubic meters per hour, the time \( t \) required is \( t = \frac{45\pi}{2} \approx \frac{141.37}{2} = 70.685 \) hours.
#### 70.685

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