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JIPMER 2017
Exam
JIPMER
Year
2017
Subject
Physics
Chapter
Laws of Motion

JIPMER 2017 · Physics · Laws of Motion

The upper half of an inclined plane of inclination θ is perfectly...

The upper half of an inclined plane of inclination θ\theta is perfectly smooth while the lower half 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 the lower half of the plane is given byThe upper half of an inclined plane of inclination θ is perfectly smooth while the lower half rough. A block starting

Options

  1. Aμ=2tan⁡θ\mu=2 \tan \theta
  2. Bμ=tan⁡θ\mu=\tan \theta
  3. Cμ=2/(tan⁡θ)\mu=2 /(\tan \theta)
  4. Dμ=1/tan⁡θ\mu=1 / \tan \theta

Correct answer

A. μ=2tan⁡θ\mu=2 \tan \theta

Step-by-step solution

Suppose length of plane is LL. When block descends the plane, rise in kinetic energy = fall in potential energy =mgLsin⁡θi=m g L \sin \theta i The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half rough. A block starting Work done by friction =μ×=\mu \times (reaction )×) \times distance 0+μ(mgcos⁡θ)×(L/2)=μ(mgcos⁡θ)×(L/2)\begin{aligned} 0+\mu(m g \cos \theta) & \times(L / 2) \\ & =\mu(m g \cos \theta) \times(L / 2)\end{aligned} Now, work done = change in KEmgLsin⁡θ=μ(mgcos⁡θ)×(L/2)⇒tan⁡θ=μ/2∴μ=2tan⁡θ\begin{array}{rlrl} & & m g L \sin \theta=\mu(m g \cos \theta) \times(L / 2) \\ \Rightarrow & \tan \theta=\mu / 2 \\ \therefore & \mu=2 \tan \theta\end{array}
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