The upper half of an inclined plane of inclination \(\theta\) is perfectly smooth while the lower half is rough A body starting from the rest at top come back to rest at the bottom, then the coefficient of friction for the lower half is given by (A) \(\mu=\sin \theta\) (B) \(\mu=\cot \theta\) (C) \(\mu=2 \cos \theta\) (D) \(\mu=2 \tan \theta\)

As the body slides down the smooth half of the inclined plane, it only experiences acceleration due to gravity. The motion is governed by the following equation: \(v^2 = u^2 + 2as\) where, \(v\) is the final velocity reached at the bottom of the smooth half, \(u\) is the initial velocity (0, since the body starts from rest), \(a\) is the acceleration due to gravity along the inclined plane (\(g\sin\theta\)), \(s\) is the distance traveled (half the inclined plane length, let's denote it by \(L/2\)). Substituting these values, we get: \(v^2 = 0^2 + 2(g\sin\theta)(L/2)\) Simplifying the equation: \(v^2 = gL\sin\theta\)

On the lower half of the inclined plane, the body experiences both gravitational force and frictional force. We can write the equation of motion for this segment: \(v'^2 = v^2 - 2\mu a's'\) where, \(v'\) is the final velocity at the bottom of the rough half (0, since the body comes to rest), \(v\) is the initial velocity reached at the end of the smooth half, \(\mu\) is the coefficient of friction between the body and the rough surface, \(a'\) is the acceleration due to gravity along the inclined plane (\(g\sin\theta\)) with the negative sign because it opposes the body's motion, \(s'\) is the distance traveled (the other half of the inclined plane's length, denoted by \(L/2\)). Substituting these values, we get: \(v'^2 = v^2 - 2\mu (g\sin\theta)(L/2)\) Simplifying the equation: \(v'^2 = v^2 - \mu gL\sin\theta\)

Since the final velocity \(v'\) at the end of the rough half is 0 (the body comes to rest), we can equate the mechanical energy during smooth segment and rough segment using conservation of mechanical energy: \(gL\sin\theta = \mu gL\sin\theta\) Divide both sides by \(g L\sin\theta\): \(1 = \mu\) Here, we find that the coefficient of friction (\(\mu\)) for the lower half of the inclined plane is equal to 1. Checking the given options, we find that: Option (A) \(\mu=\sin \theta\) Option (B) \(\mu=\cot \theta\) Option (C) \(\mu=2 \cos \theta\) Option (D) \(\mu=2 \tan \theta\) None of the given options match our result (\(\mu = 1\)). Therefore, the correct answer is not listed among the given options.