A plane is inclined at an angle θ with the horizontal. A body of mass m rests on it. If the coefficient of friction is μ, then the minimum force that has to be applied parallel to the inclined plane to make the body just move up the inclined plane is

The correct option is μ mg cosθ + mg sinθ.

To find the minimum force required to make the body just start moving up the inclined plane, we need to analyze all the forces acting on the body along and perpendicular to the inclined surface.

First, let us identify the forces acting on the body of mass m resting on a plane inclined at an angle θ:
1. The gravitational force (weight) acting vertically downwards is:
$W=mg$
2. The normal force (N) exerted by the inclined plane on the body, perpendicular to the surface.

We can resolve the gravitational force into two perpendicular components:
- A component perpendicular to the inclined plane:
$W_{\perp }=mgcos\theta$
- A component parallel to the inclined plane (directed down the incline):
$W_{\parallel }=mgsin\theta$

Since there is no acceleration perpendicular to the inclined plane, the forces in this direction must balance each other:
$N=mgcos\theta$

Now, if we want to move the body up the inclined plane, the applied force (F) must be directed up the incline. Consequently, the force of friction (f) will oppose this impending motion and act downwards along the incline. Since we want to just make the body move, we must overcome the maximum static friction:
$f_s=\mu N=\mu mgcos\theta$

Therefore, the total force acting down the incline that the applied force F must overcome is the sum of the component of gravity down the incline and the maximum frictional force down the incline:
$F=f_s+mgsin\theta$

Substituting the expression for the frictional force, we get the minimum parallel force:
$F=\mu mgcos\theta +mgsin\theta$