A disc has mass $M$ and radius $R$. How much tangential force should be applied to the rim of the disc so as to rotate with angular velocity $\omega $ in time $t$?
A. $\dfrac{{MR\omega }}{{4t}}$
B. $\dfrac{{MR\omega }}{{2t}}$
C. $\dfrac{{MR\omega }}{t}$
D. $MR\omega t$

Hint: We know that torque is the product of force, $F$, and the perpendicular distance between the force and the axis of rotation $R$. Torque is also the product of the moment of inertia and angular acceleration.
Equating the two equations for torque and substituting the given values we can find the equation for tangential force.

Formulas used:
$\tau = F \times R$
Where $\tau $ is the torque, $F$ is the force and $R$ is the perpendicular distance between force and the axis of rotation.
$\tau = I \times \alpha $
Where, $I$ is the moment of inertia and $\alpha $ is the angular acceleration.
Moment of inertia of a disc is
$I = \dfrac{1}{2}M{R^2}$
Where, $I$ is the moment of inertia of disc, $M$ is the mass and $R$ is the radius of the disc.
$\alpha = \dfrac{\omega }{t}$
Where $\omega $ is the angular velocity and $t$ is the time.

Complete step by step answer:
Given,
Mass of the disc = $M$
Radius of the disc = $R$
Angular velocity = $\omega $
Time = $t$
We need to find the tangential force.
Since angular velocity is given we can calculate angular acceleration.
Angular acceleration is the rate of change of angular velocity,
$\alpha = \dfrac{\omega }{t}$
Where $\omega $ is the angular velocity and $t$ is the time.
Moment of inertia of a disc is given as,
$\Rightarrow I = \dfrac{1}{2}M{R^2}$
Torque is the rotational analogue of force.
It is the product of the moment of inertia and angular acceleration.
$\tau = I \times \alpha $
Where $I$ is the moment of inertia and $\alpha $ is the angular acceleration.
Therefore, substituting for $I$ and $\alpha $ we get
$\Rightarrow \tau = \dfrac{1}{2}M{R^2} \times \dfrac{\omega }{t}$
Torque is also the product of force, $F$ and the perpendicular distance between the force and the axis of rotation $R$.
Therefore
$\tau = F \times R$
$ \Rightarrow F = \dfrac{\tau }{R}$
Substituting for torque, we get
$\Rightarrow F = \dfrac{{\dfrac{1}{2}M{R^2} \times \dfrac{\omega }{t}}}{R} = \dfrac{{MR\omega }}{{2t}}$
This is the value of tangential force.

Therefore, the correct answer is option (B).

Note:
Here, remember that we considered the disc rotates about an axis perpendicular to the plane of disc passing through its centre of mass. That is why we took the perpendicular distance between the tangential force and the axis of rotation to be the radius of the disc and the moment of inertia as $\dfrac{1}{2}M{R^2}$. If the axis of rotation is not the axis at the centre of mass perpendicular to the plane of the disc then the moment of inertia will be different.