A uniform disc of mass '$m$' and radius '$R$' is rolling up a rough inclined plane which makes an angle of $30^\circ$ with the horizontal. If the coefficient of friction is $\mu$ and only forces acting are gravitational and frictional, then the magnitude of the frictional force '$f$' is $\left(\sin 30^\circ = \dfrac{1}{2}\right)$

Let the linear acceleration of the disc down the incline be $a$ and its angular acceleration be $\alpha$.

The forces acting on the disc along the inclined plane are the component of gravity downwards, $mg \sin \theta$, and the frictional force $f$ acting upwards along the incline to oppose the angular deceleration.

The equation for linear motion is:
$mg \sin \theta - f = ma$

The equation for rotational motion about the center of mass is:
$fR = I\alpha$

For pure rolling, the relationship between linear and angular acceleration is:
$a = R\alpha$

The moment of inertia of a uniform disc about its central axis is:
$I = \dfrac{1}{2}mR^2$

Substituting $I$ and $\alpha = \dfrac{a}{R}$ into the torque equation:
$fR = \left(\dfrac{1}{2}mR^2\right)\left(\dfrac{a}{R}\right)$

$f = \dfrac{1}{2}ma \Rightarrow a = \dfrac{2f}{m}$

Substitute this value of $a$ into the linear motion equation:
$mg \sin \theta - f = m\left(\dfrac{2f}{m}\right)$

$mg \sin \theta - f = 2f$

$3f = mg \sin \theta$

$f = \dfrac{mg \sin \theta}{3}$

Given that $\theta = 30^\circ$ and $\sin 30^\circ = \dfrac{1}{2}$:
$f = \dfrac{mg}{3} \times \dfrac{1}{2} = \dfrac{mg}{6}$

Answer: $\dfrac{mg}{6}$