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NEET generalGeneralHard

Question

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

A

a=gcos⁡θa = g \cos \theta

B

a=gsin⁡θa = \frac{g}{\sin \theta}

C

a=gcosec θa = \frac{g}{\text{cosec } \theta}

D

a=gtan⁡θa = g \tan \theta

Step-by-Step Solution

In non-inertial frame,

Nsin⁡θ=ma…(i)N \sin \theta = ma \quad \dots(i)

Ncos⁡θ=mg…(ii)N \cos \theta = mg \quad \dots(ii)

Dividing equation (i) by (ii), we get: tan⁡θ=ag\tan \theta = \frac{a}{g}

a=gtan⁡θa = g \tan \theta

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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