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# Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in the level of water in the tank in half hours.

Last updated date: 10th Sep 2024
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Hint: In order to solve this problem we need to find the volume of water flowing through the pipe in one second and then multiply 30 x 60 to get the volume of pipe in 30 minutes of half an hour. Then we need to find the volume of the cylinder taking height as variable and then we need to equate the volume we found earlier through the pipe to that of the filled cylinder’s volume to get the value of height. Doing this will solve your problem and will give you the right answer.

The figure of the system can be drawn as:

Internal diameter of cylindrical pipe is equal to 2cm
So the volume of water flowing in the pipe = $\pi {r^2}H$
$\Rightarrow \pi {(1)^2} \times 0.4 \times 100c{m^3}$
$\Rightarrow \pi {(1)^2} \times 0.4 \times 100 \times 30 \times 60$…………………………. (1)
Let H be the height of cylindrical tank so its volume is given as $\pi {R^2}H$where R=40 cm
So the volume of water in the cylindrical tank after 30 minutes = $\pi {(40)^2}H$…………………. (2)
$\Rightarrow \pi {(1)^2} \times 0.4 \times 100 \times 30 \times 60$=$\pi {(40)^2}H$