Answer
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Hint: Let’s find out the slope of point A by making a tangent through the point as we already know that the slope of a point is equal to the tangent drawn through it. We will end up realising that the shape made by the water is that of a parabola. Refer to the figure as well at the solution below.
Formula used: $F = mr{\omega ^2}$, $\tan = \dfrac{{perpendicular}}{{base}}$, $F = mg$
Complete Step-by-Step solution:
Let a water molecule be there on point A with mass dm.
As we know that the formula for centrifugal force is $F = mr{\omega ^2}$. Where m is mass, r is the radius and $\omega $ is the angular velocity.
Now, the mass of the water droplet is dm.
The radius is x.
The centrifugal force being applied on the water droplet will be-
$ \Rightarrow F = dmx{\omega ^2}$
The force due to gravity is calculated by $F = mg$. Putting the value of the mass of the droplet, we get the value of the force being applied downwards. So-
$ \Rightarrow F = dmg$
Now, for the value of the resultant force, take reference from the figure.
Now, as we can see from the figure, the angle between the gravitational force and the resultant force is $\theta $ because the angle between the tangent and the gravitational force is $90^\circ - \theta $ which came from the fact that the angle between the resultant force and the tangent is 90 degrees.
Applying the formula of $\tan = \dfrac{{perpendicular}}{{base}}$, we get-
$
\Rightarrow \tan \theta = \dfrac{{dmx{\omega ^2}}}{{dmg}} \\
\\
\Rightarrow \tan \theta = \dfrac{{x{\omega ^2}}}{g} \\
$
As we know that $\tan \theta $ also represents the slope i.e. $\tan \theta = \dfrac{{dy}}{{dx}}$.
From this, we can say that-
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{\omega ^2}}}{g}x \\
\\
\Rightarrow dy = \dfrac{{{\omega ^2}}}{g}xdx \\
$
Integrating both sides-
$ \Rightarrow \int {dy} = \dfrac{{{\omega ^2}}}{g}\int {xdx} $ (as $\dfrac{{{\omega ^2}}}{g}$ is constant)
$ \Rightarrow y = \dfrac{{{\omega ^2}{x^2}}}{{2g}}$
The above equation is the equation of parabola defining that the shape made by the water while rotating is that of a parabola. Hence, $y = \dfrac{{{\omega ^2}{r^2}}}{{2g}}$ has its origin at O.
The value of the radius is given r by the question. So, when the value of x is r, the y is-
When $x = r$,
$ \Rightarrow y = \dfrac{{{\omega ^2}{r^2}}}{{2g}}$
Thus, the difference between the label is $y = \dfrac{{{\omega ^2}{r^2}}}{{2g}}$.
Hence, option B is the correct option.
Note: In physics, angular velocity refers to how rapidly an object spins or rotates in relation to another point, i.e. however quickly an object's angular position or orientation varies over time. There are two types of angular velocity: the angular orbital velocity and the angular spin velocity.
Formula used: $F = mr{\omega ^2}$, $\tan = \dfrac{{perpendicular}}{{base}}$, $F = mg$
Complete Step-by-Step solution:
Let a water molecule be there on point A with mass dm.
As we know that the formula for centrifugal force is $F = mr{\omega ^2}$. Where m is mass, r is the radius and $\omega $ is the angular velocity.
Now, the mass of the water droplet is dm.
The radius is x.
The centrifugal force being applied on the water droplet will be-
$ \Rightarrow F = dmx{\omega ^2}$
The force due to gravity is calculated by $F = mg$. Putting the value of the mass of the droplet, we get the value of the force being applied downwards. So-
$ \Rightarrow F = dmg$
Now, for the value of the resultant force, take reference from the figure.
Now, as we can see from the figure, the angle between the gravitational force and the resultant force is $\theta $ because the angle between the tangent and the gravitational force is $90^\circ - \theta $ which came from the fact that the angle between the resultant force and the tangent is 90 degrees.
Applying the formula of $\tan = \dfrac{{perpendicular}}{{base}}$, we get-
$
\Rightarrow \tan \theta = \dfrac{{dmx{\omega ^2}}}{{dmg}} \\
\\
\Rightarrow \tan \theta = \dfrac{{x{\omega ^2}}}{g} \\
$
As we know that $\tan \theta $ also represents the slope i.e. $\tan \theta = \dfrac{{dy}}{{dx}}$.
From this, we can say that-
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{\omega ^2}}}{g}x \\
\\
\Rightarrow dy = \dfrac{{{\omega ^2}}}{g}xdx \\
$
Integrating both sides-
$ \Rightarrow \int {dy} = \dfrac{{{\omega ^2}}}{g}\int {xdx} $ (as $\dfrac{{{\omega ^2}}}{g}$ is constant)
$ \Rightarrow y = \dfrac{{{\omega ^2}{x^2}}}{{2g}}$
The above equation is the equation of parabola defining that the shape made by the water while rotating is that of a parabola. Hence, $y = \dfrac{{{\omega ^2}{r^2}}}{{2g}}$ has its origin at O.
The value of the radius is given r by the question. So, when the value of x is r, the y is-
When $x = r$,
$ \Rightarrow y = \dfrac{{{\omega ^2}{r^2}}}{{2g}}$
Thus, the difference between the label is $y = \dfrac{{{\omega ^2}{r^2}}}{{2g}}$.
Hence, option B is the correct option.
Note: In physics, angular velocity refers to how rapidly an object spins or rotates in relation to another point, i.e. however quickly an object's angular position or orientation varies over time. There are two types of angular velocity: the angular orbital velocity and the angular spin velocity.
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