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Laws of Motion question

2013 · Q145
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Laws of Motion question

2013 · Q145

NEETPhysicsLaws of MotionMCQ+4 / −1
The upper half of an inclined plane of inclination θ\thetaθ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by
  1. A
    μ\muμ = 2 tanθ\thetaθ
  2. B
    μ\muμ = tanθ\thetaθ
  3. C
    μ\muμ = 1tan⁡θ{1 \over {\tan \theta }}tanθ1​
  4. D
    μ=2tan⁡θ\mu = {2 \over {\tan \theta }}μ=tanθ2​
View written solutionFree

Correct answer: A

NEET 2013 Physics - Laws of Motion Question 45 English Explanation

For upper half of inclined plane

v2 = u2 + 2a S/2 = 2 (g sin θ\theta θ) S/2 = gS sin θ\theta θ

For lower half of inclined plane

0 = u2 + 2 g (sin θ\theta θ – μ\mu μ cos θ\theta θ) S/2

⇒−gSsin⁡θ=gS(sin⁡θ−μcos⁡θ) \Rightarrow - gS\sin \theta = gS\left( {\sin \theta - \mu \cos \theta } \right)⇒−gSsinθ=gS(sinθ−μcosθ)

⇒2sin⁡θ=μcos⁡θ\Rightarrow 2\sin \theta = \mu \cos \theta⇒2sinθ=μcosθ

⇒μ=2sin⁡θcos⁡θ=2tan⁡θ\Rightarrow \mu = {{2\sin \theta } \over {\cos \theta }} = 2\tan \theta⇒μ=cosθ2sinθ​=2tanθ

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