
A large cylindrical tank has a hole of area $A$ at its bottom and water is poured into the tank through a tube of cross-sectional Area $A$ ejecting water at the speed $v$. Which of the following is true?
A. Water level in the tank keeps on rising.
B. No water can be stored in the tank.
C. Water level will rise to a height $\dfrac{{{v^2}}}{{2g}}$ and then stop.
D. The water level will be oscillating.
Answer
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Hint:We need to find the amount of water being ejected and the amount of water that would be added per unit time. This would give a relation between the addition and ejection of water, which would help us understand if the height of the water level would increase or not.
Complete step by step answer:
The initial velocity of the water coming out of the tank will be negligible due to the negligible height of the water in the cylindrical tank. The quantity of the incoming water in the cylindrical tank will be more than the quantity of the outgoing water in the cylindrical tank. Hence the velocity of the water coming out of the cylindrical tank will start increasing as the level of the water in the cylinder tank will begin to rise or increase. Now, the level of the water will rise or increase up to a height of$h$for which the velocity of the water coming from the outlet of the cylindrical tank will be equal to the velocity of the water coming in the inlet of the cylindrical tank. Therefore, we can calculate the height of the water mathematically, by using the following formula:
$v = \sqrt {2gh} $
$\therefore h = \dfrac{{{v^2}}}{{2g}}$
Where,$h \to $is the height of water.
$v \to $is the velocity of water.
$g \to $gravity.
Hence, the correct option is (C) Water level will rise to a height $\dfrac{{{v^2}}}{{2g}}$ and then stop.
Note:The questions may be assigned with a number also. i.e. The same question can be interpreted in the following way: A large cylindrical tank has a hole of $30c{m^2}$ at its bottom and water is proved into the tank through a tube of cross sectional area $30c{m^2}$, ejecting water at a speed V.
Complete step by step answer:
The initial velocity of the water coming out of the tank will be negligible due to the negligible height of the water in the cylindrical tank. The quantity of the incoming water in the cylindrical tank will be more than the quantity of the outgoing water in the cylindrical tank. Hence the velocity of the water coming out of the cylindrical tank will start increasing as the level of the water in the cylinder tank will begin to rise or increase. Now, the level of the water will rise or increase up to a height of$h$for which the velocity of the water coming from the outlet of the cylindrical tank will be equal to the velocity of the water coming in the inlet of the cylindrical tank. Therefore, we can calculate the height of the water mathematically, by using the following formula:
$v = \sqrt {2gh} $
$\therefore h = \dfrac{{{v^2}}}{{2g}}$
Where,$h \to $is the height of water.
$v \to $is the velocity of water.
$g \to $gravity.
Hence, the correct option is (C) Water level will rise to a height $\dfrac{{{v^2}}}{{2g}}$ and then stop.
Note:The questions may be assigned with a number also. i.e. The same question can be interpreted in the following way: A large cylindrical tank has a hole of $30c{m^2}$ at its bottom and water is proved into the tank through a tube of cross sectional area $30c{m^2}$, ejecting water at a speed V.
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