Answer

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**Hint:**Here, using the angular velocity formula, which is the angle per unit time, the angle through which the rigid body rotates before it comes to rest is computed. The angular velocity is zero at rest, so we will calculate the time and put it in the angle equation.

**Complete Step-by-Step Solution:**

Let us suppose the angle through which it rotates be \[\theta \]

According to the question,

The angular velocity is given as\[\omega = a - bt\]

Now, if the body is at rest, the angular velocity of such a body is zero, i.e., \[\omega = 0\]

We can now equate the value of angular velocity such that

\[a - bt = 0\]

Upon solving, we get

\[t = \dfrac{a}{b}\]

Also, we know that angular velocity is also measured as angle per unit time, therefore,

$\omega = \dfrac{{d\theta }}{{dt}}$

Rearranging the terms, we get

\[d\theta = \omega dt\]

Now, we will integrate both the sides

\[\int\limits_0^\theta {d\theta } = \int\limits_0^{\dfrac{a}{b}} {\omega dt} \]

Upon further solving, we get

\[\theta = \int\limits_0^{\dfrac{a}{b}} {(a - bt) dt} \]

Now, let us integrate the above expression

\[\theta = \left[ {at - \dfrac{{b{t^2}}}{2}} \right]_0^{\dfrac{a}{b}}\]

\[ \Rightarrow \theta = \dfrac{{{a^2}}}{b} - \dfrac{{{a^2}}}{{2b}}\]

Let us simplify and get the final answer

\[\therefore \theta = \dfrac{{{a^2}}}{{2b}}\]

**Hence, the correct option is (B.)**

**Additional Information:**Angular velocity is the quantity of a vector and is described as the angular displacement change rate that specifies the object's angular velocity or rotational velocity and the axis around which the object is rotating. At a given period of time, the amount of change in the angular displacement of the particle is called angular velocity. The angular velocity vector's track is vertical to the rotation plane, in a direction usually indicated by the right-hand rule.

**Note:**In the above question, we have found out the angle by integrating the equation of \[\theta \] between the limits \[0\] to \[\dfrac{a}{b}\]. We could have also landed to the same result by putting the value of $t$ in the $\theta $ equation after integrating it.

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