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Laws of Motion question

2018 · Q141
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Laws of Motion question

2018 · Q141

NEETPhysicsLaws of MotionMCQ+4 / −1
A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and q for the block to remain stationary on the wedge is NEET 2018 Physics - Laws of Motion Question 12 English
  1. A
    a=gcos⁡ecθa = {g \over {\cos ec\theta }}a=cosecθg​
  2. B
    a=gcos⁡ecθa = {g \over {\cos ec\theta }}a=cosecθg​
  3. C
    a=gcos⁡θa = g\cos \thetaa=gcosθ
  4. D
    a=gtan⁡θa = g\tan \thetaa=gtanθ
View written solutionFree

Correct answer: D

NEET 2018 Physics - Laws of Motion Question 12 English Explanation

At equilibrium

ma cos $\theta $ = mg sin $\theta $

$ \Rightarrow $ a = g tan $\theta $
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