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QNEET-2018-Ph-907: NEET 2018 Physics - Class 11 NEET-2018-Ph-907 Solution: A block of mass m is placed on a smooth inclined wedge ABC of…

A block of mass m is placed on a smooth inclined wedge ABC of inclination θ\theta as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ\theta for the block to remain stationary on the wedge is a=gcos⁡θa = g\cos\theta a=gsin⁡θa = \frac{g}{\sin\theta} a=gtan⁡θa = g\tan\theta a=gcsc⁡θa = g\csc\theta

Multiple Choice Options

  • A a=gcos⁡ecθa = {g \over {\cos ec\theta }} a = cos ec _ g _
  • B a=gcos⁡ecθa = {g \over {\cos ec\theta }} a = cos ec _ g _
  • C a=gcos⁡θa = g\cos \theta a = g cos _
  • D a=gtan⁡θa = g\tan \theta a = g tan _

Step-by-Step Verified Solution

At equilibrium ma cos _ = mg sin _ _ a = g tan _

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