AIPMT PRELIMS 2004 Physics Inclined Planes MCQ Question

To understand the problem, we need to analyze the forces acting on the block placed on the inclined wedge that is being accelerated horizontally.

Step-by-step Explanation:

1. Free Body Diagram of the Block:

   • The block of mass $m$ experiences several forces:
     • The gravitational force acting downwards: $\vec{W} = mg$
     • The normal force exerted by the wedge on the block, which acts perpendicular to the surface of the wedge, denoted as $\vec{N}$.
     • If the wedge is being accelerated horizontally, the block also experiences a pseudo force $\vec{F}_{\text{pseudo}}$ acting horizontally in the opposite direction of the acceleration.

2. Components of Forces:

   • The gravitational force can be resolved into two components:
     • Perpendicular to the inclined surface: $mg \cos \theta$
     • Parallel to the inclined surface: $mg \sin \theta$

3. Net Force Analysis:

   • For the block to remain in equilibrium on the wedge (meaning it does not slip), the net acceleration of the block must equal the horizontal acceleration of the wedge. The pseudo force acting on the block due to the horizontal acceleration $a$ of the wedge is given by $F_{\text{pseudo}} = ma$, which acts horizontally.

4. Equations of Motion:

   • We can write the equations of motion for the block along the inclined plane. The normal force $N$ must balance the perpendicular component of the gravitational force and the pseudo force acting parallel to the incline.
   • The force balance in the direction perpendicular to the incline gives: $N = mg \cos \theta$
   • The gravitational force acts downwards and is balanced by the normal force in the perpendicular direction. The pseudo force does not affect the force in this direction, as it acts horizontally.

5. Correct Answer:

   • The normal force $N$, which is the force exerted by the wedge on the block, is given by: $N = mg \cos \theta$
   • Therefore, the correct answer is C) $mg \cos \theta$.

Clarification of Incorrect Options:

• A) $mg \sin \theta$: This option incorrectly represents the force acting parallel to the incline but does not account for the normal force. It is not the force exerted by the wedge.

• B) $mg$: This option suggests that the normal force equals the weight of the block. It is incorrect because the normal force only balances the component of weight acting perpendicular to the incline, which is $mg \cos \theta$.

• D) $mg \cos \theta$: This option is identical to option C, which is correct. Therefore, it should not be presented as a separate option.

Conclusion:

In summary, the force exerted by the wedge on the block is due to the balance of forces acting perpendicular to the incline, resulting in $N = mg \cos \theta$. This analysis shows how forces interact in a non-inertial frame (due to the horizontal acceleration of the wedge) and confirms why option C is the correct answer.