Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block of mass $m$ is placed on a smooth inclined wedge ABC of inclination $\theta$ as shown in the figure. The wedge is given an acceleration ' $a$ ' towards the right. The relation between $a$ and $\theta$ for the block to remain stationary on the wedge is:

$a = g \csc \theta$

$a = g \sin \theta$

$a = g \cos \theta$

$a = g \tan \theta$

Step-by-Step Solution

Frame of Reference: Analyze the problem from the non-inertial frame of reference of the wedge, which is accelerating with acceleration $a$ towards the right.
Pseudo Force: In this non-inertial frame, a pseudo force of magnitude $ma$ acts on the block of mass $m$ . Since the wedge accelerates to the right, the pseudo force acts to the left.
Forces on the Block:

Weight ( $mg$ ) acting vertically downwards.
Normal Reaction ( $N$ ) acting perpendicular to the incline.
Pseudo Force ( $ma$ ) acting horizontally towards the left.

Equilibrium Condition: For the block to remain stationary relative to the wedge, the net force along the incline must be zero.

Component of weight down the incline: $mg \sin \theta$ .
Component of pseudo force up the incline: $ma \cos \theta$ .

Calculation: Equating the components along the incline: $ma \cos \theta = mg \sin \theta$ $a = g \frac{\sin \theta}{\cos \theta}$ $a = g \tan \theta$ (Reference: NCERT Class 11, Physics Part I, Chapter 5, Laws of Motion).

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