A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude:
A) mg
B) mg cos θ
C) mg / cos θ
D) mg tan θ

The correct answer is C. mgcos⁡θ\frac{mg}{\cos \theta}.
Detailed Explanation:
1. System Setup:
o A block of mass mm is placed on a smooth wedge inclined at an angle θ\theta.
o The system is accelerated horizontally, ensuring that the block does not slip on the wedge.
2. Forces Acting on the Block:
o Weight (mgmg): Acts vertically downward.
o Normal force (NN): Acts perpendicular to the surface of the wedge.
o The wedge is accelerated horizontally with an acceleration aa, which generates a pseudo-force on the block in the wedge's reference frame. This pseudo-force acts horizontally in the direction opposite to the acceleration of the system.
3. Condition for No Slipping: For the block to remain stationary relative to the wedge, the net force on the block along the inclined plane must be zero. This means the block must stay at rest relative to the wedge.
4. Resolving Forces:
o The component of the block's weight perpendicular to the inclined plane is mgcos⁡θmg \cos \theta. This is balanced by the normal force NN, i.e., N=mgcos⁡θ.N = \frac{mg}{\cos \theta}.
o The component of the weight along the inclined plane (mgsin⁡θmg \sin \theta) is balanced by the horizontal pseudo-force component acting along the incline.
5. Force Exerted by the Wedge: The force exerted by the wedge on the block is the normal force NN. Since the wedge supports the block against slipping, the magnitude of this force is:
N=mgcos⁡θ.N = \frac{mg}{\cos \theta}.
Conclusion:
The force exerted by the wedge on the block is mgcos⁡θ\frac{mg}{\cos \theta}, making option C the correct answer.