Answer

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**Hint:**The angular momentum is a vector quantity which varies with the point under consideration. Here, the angular momentum of the same body as observed from two different points the centre of rotation O and the fixed-point P are considered.

Formula Used: The angular momentum is calculated using the formula:

\[\begin{align}

& \overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p} \\

& \text{or} \\

& \overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v}) \\

\end{align}\]

Where, \[\overrightarrow{L}\] is the angular momentum

\[\overrightarrow{r}\] is the radius of the circle in which the mass undergoes circular motion

\[\overrightarrow{p}\] is the linear momentum

\[v\] is the linear velocity

**Complete answer:**

Angular momentum is the measure of linear momentum in a circular motion. It is a vector quantity like the linear momentum.

Let us consider the given situation. An object of mass m attached to a fixed string at point P is rotating around a fixed-point O with a constant angular velocity \[\omega \]. We have to determine the angular momenta experienced by the body with respect to point O and P.

We can begin with the point O, the center of rotation.

\[\overrightarrow{{{L}_{0}}}=\overrightarrow{r}\times \overrightarrow{p}\]

The linear momentum ‘mv’ being substituted gives, \[\overrightarrow{{{L}_{o}}}=m(\overrightarrow{r}\times \overrightarrow{v})\]

We know that the angular velocity is a constant. Also, the radius of the rotation is a constant.

\[\begin{align}

& \Rightarrow \overrightarrow{{{L}_{o}}}=m(\overrightarrow{OR}\times \overrightarrow{v)} \\

& \Rightarrow \left| \overline{{{L}_{o}}} \right|=mrv \\

& \\

\end{align}\]

The magnitude of the angular momentum at O remains constant throughout.

And, the direction of the angular momentum at O is always along the Z-direction.

Now, let us consider the angular momentum at point P due to the mass.

\[\begin{align}

& \Rightarrow \overrightarrow{{{L}_{p}}}=m(\overrightarrow{PR}\times \overrightarrow{v}) \\

& \Rightarrow \overrightarrow{{{L}_{p}}}=m((\overrightarrow{PO}+\overrightarrow{OR})\times \overrightarrow{v}) \\

\end{align}\]

It is clear from the above equation that \[\overrightarrow{PR}\]changes its direction at every instant. As a result, the direction of the angular momentum at P keeps on varying with time.

And the magnitude is given by,

\[\begin{align}

& \overrightarrow{{{L}_{p}}}=m\left| \overrightarrow{PR} \right|\overrightarrow{v}\widehat{k} \\

& \left| \overrightarrow{{{L}_{p}}} \right|=m\left| \overrightarrow{PR} \right|\overrightarrow{v} \\

\end{align}\]

Which remains constant with time.

From the above, we can conclude that \[\overrightarrow{{{L}_{o}}}\]remains constant both in direction and magnitude, whereas, \[\overrightarrow{{{L}_{p}}}\]has a constant magnitude, but the direction changes.

Therefore, \[\overrightarrow{{{L}_{o}}}\] remains constant and \[\overrightarrow{{{L}_{p}}}\]varies with time.

**The right answer is given by option C.**

**Additional Information:**

We can calculate the angular momentum by using the angular velocity, \[\overrightarrow{\left| L \right|}=m{{\omega }^{2}}r\].

**Note:**

The angular momentum in a closed system remains constant. It is a conserved quantity. As the linear momentum is conserved as long as no external force is applied, the angular momentum is conserved as long as the torque is zero.

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