
A cane filled with water is revolved in a vertical circle of radius $0.5{\text{m}}$ and the water does not fall. The maximum period of revolution must be
A. $1.45$
B. $2.45$
C. $14.15$
D. $4.25$
Answer
511.5k+ views
Hint: This question involves laws of motion. For the water to fall from the cane when revolved, the forces acting on the water should be balanced out so that the water remains in an equilibrium position.
Complete step by step answer:
The radius of the vertical circle in which the water is revolved is given.
$r = 0.5{\text{m}}$
Let the velocity of the vertical circle be $v$. As the bucket rotates, the water experiences a centrifugal force. Now, this centrifugal force must be greater than the gravitational force acting on the water in a downward direction so that the water does not fall.
${F_c} = F - - - - (1)$
We know that the centrifugal force is defined by,
${F_c} = \dfrac{{m{v^2}}}{r}$
And gravitational force is given by
$F = mg$
Where, $m$ is the mass of the water, $g$ is the acceleration due to gravity and $r$ is the radius.
Substituting the values in equation (1), we get
$\dfrac{{m{v^2}}}{r} = mg$
$ \Rightarrow v = \sqrt {rg} $
Now the circular velocity or the angular velocity is given by
$v = \dfrac{{2\pi r}}{T}$
$T$ is the time period of revolution.
Rearranging and substituting the values we get,
$T = \dfrac{{2\pi \times 0.5}}{{\sqrt {0.5 \times 9.8} }}$
$ \Rightarrow T = \dfrac{{3.14}}{{2.23}}$
$ \therefore T = 1.45\sec $.......(approximately)
Therefore, option A is correct.
Note: When the cane of water is raised and turned upside down, the water experiences a strong pull downwards which is the pull of gravity of the earth’s surface and therefore it falls. When the water is revolved in a circular motion vertically it experiences a force that keeps the water in motion and it acts outwardly which is the centrifugal force. If this force balances the gravitational force the water does not fall. But when the centrifugal force overcomes the gravitational force it pushes the water to the end of the cane so that not even a drop of water falls from the opening of the cane.
Complete step by step answer:
The radius of the vertical circle in which the water is revolved is given.
$r = 0.5{\text{m}}$
Let the velocity of the vertical circle be $v$. As the bucket rotates, the water experiences a centrifugal force. Now, this centrifugal force must be greater than the gravitational force acting on the water in a downward direction so that the water does not fall.
${F_c} = F - - - - (1)$
We know that the centrifugal force is defined by,
${F_c} = \dfrac{{m{v^2}}}{r}$
And gravitational force is given by
$F = mg$
Where, $m$ is the mass of the water, $g$ is the acceleration due to gravity and $r$ is the radius.
Substituting the values in equation (1), we get
$\dfrac{{m{v^2}}}{r} = mg$
$ \Rightarrow v = \sqrt {rg} $
Now the circular velocity or the angular velocity is given by
$v = \dfrac{{2\pi r}}{T}$
$T$ is the time period of revolution.
Rearranging and substituting the values we get,
$T = \dfrac{{2\pi \times 0.5}}{{\sqrt {0.5 \times 9.8} }}$
$ \Rightarrow T = \dfrac{{3.14}}{{2.23}}$
$ \therefore T = 1.45\sec $.......(approximately)
Therefore, option A is correct.
Note: When the cane of water is raised and turned upside down, the water experiences a strong pull downwards which is the pull of gravity of the earth’s surface and therefore it falls. When the water is revolved in a circular motion vertically it experiences a force that keeps the water in motion and it acts outwardly which is the centrifugal force. If this force balances the gravitational force the water does not fall. But when the centrifugal force overcomes the gravitational force it pushes the water to the end of the cane so that not even a drop of water falls from the opening of the cane.
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