A body of mass 2 kg is placed on a rough inclined plane of...

Let $P$ be the magnitude of the horizontally applied push. The forces acting on the block are its weight $mg$, the normal reaction $N$, the horizontal push $P$, and the frictional force $f$.

Resolving the forces perpendicular to the inclined plane, the normal reaction is:
N = mg \cos 30^{}^{{}^{\circ}} + P \sin 30^{}^{{}^{\circ}}

To initiate movement up the incline, the component of the push along the incline must overcome the downward component of the weight and the limiting static friction:
P \cos 30^{}^{{}^{\circ}} = mg \sin 30^{}^{{}^{\circ}} + f

Since $f = \mu N$, we have:
P \cos 30^{}^{{}^{\circ}} = mg \sin 30^{}^{{}^{\circ}} + \mu (mg \cos 30^{}^{{}^{\circ}} + P \sin 30^{}^{{}^{\circ}})

Rearranging to solve for $P$:
P (\cos 30^{}^{{}^{\circ}} - \mu \sin 30^{}^{{}^{\circ}}) = mg (\sin 30^{}^{{}^{\circ}} + \mu \cos 30^{}^{{}^{\circ}})

P = \dfrac{mg (\sin 30^{}^{{}^{\circ}} + \mu \cos 30^{}^{{}^{\circ}})}{\cos 30^{}^{{}^{\circ}} - \mu \sin 30^{}^{{}^{\circ}}}

Substituting the given values ($m = 2 \text{ kg}$, $g = 10 \text{ m/s}^2$, $\mu = 0.2$):
$P = \dfrac{20 \left(0.5 + 0.2 \times \dfrac{\sqrt{3}}{2}\right)}{\dfrac{\sqrt{3}}{2} - 0.2 \times 0.5}$

$P = \dfrac{20 (0.5 + 0.1732)}{0.866 - 0.1}$

$P = \dfrac{20 (0.6732)}{0.766} \approx 17.57 \text{ N}$

The closest value among the given options is $17.5 \text{ N}$.

Answer: $17.5 \text{ N}$