A given object takes n times more time to slide down \(45^{\circ}\) rough inclined plane as it takes to slide down a perfectly smooth \(45^{\circ}\) incline. The coefficient of kinetic friction between the object and the incline is (A) \(\left[1 /\left(2-\mathrm{n}^{2}\right)\right]\) (B) \(\left[1-\left(1 / \mathrm{n}^{2}\right)\right]\)

First, let's analyze the forces acting on the object in both scenarios. We have two main forces acting on the object: gravity and friction (in the case of a rough incline). The force of gravity has two components: one is parallel to the incline, and the other is perpendicular to the incline. The force of friction is opposite to the motion of the object and is directly proportional to the force pressing the object and the plane together (normal force).

In the case of a smooth inclined plane, there is no friction force acting on the object. The only force causing the object to slide down the incline is the gravitational force parallel to the incline. This parallel force can be found using the equation: \(F_{g \parallel} = mg \sin \theta\), where m is the mass of the object, g is the gravitational acceleration, and \(\theta\) is the angle of inclination. Since the angle of inclination is 45 degrees, this equation simplifies to: \(F_{g \parallel} = mg \sin 45^{\circ} = mg/\sqrt{2}\) Using Newton's Second Law of Motion, the acceleration of the object on the smooth incline can be found: \(F_{g \parallel} = m a_{smooth}\), which simplifies to: \(a_{smooth} = g/\sqrt{2}\)

In the case of a rough inclined plane, there is friction force acting on the object. The friction force can be found using the equation: \(F_{friction} = \mu F_{normal}\), where \(\mu\) is the coefficient of kinetic friction and \(F_{normal}\) is the normal force between the object and the incline. The force acting perpendicular to the plane is equal to the normal force in this case: \(F_{normal} = mg \cos \theta\), which simplifies to: \(F_{normal} = mg/\sqrt{2}\) Combining both equations and substituting back the force due to gravity component, \(F_{friction} = \mu (mg/\sqrt{2})\) Now, the net force acting on the object along the incline is the difference between the gravitational force parallel to the incline and the friction force. Applying Newton's Second Law of Motion: \(F_{g \parallel} - F_{friction} = m a_{rough}\) Substituting both forces, we get: \((mg/\sqrt{2}) - (\mu (mg/\sqrt{2})) = ma_{rough}\) This simplifies to: \((1-\mu)(g/\sqrt{2}) = a_{rough}\)

According to the problem, the object takes n times more time to slide down the rough incline than it takes to slide down the smooth incline. Therefore, the time taken in both cases can be represented as: \(t_{rough} = n t_{smooth}\) The distance traveled down the incline in both cases is the same. Using the equation of motion, we have: \(d = \frac{1}{2} a t^2\) Since distance d is the same, we can equate the smooth and rough situations: \(\frac{1}{2} a_{smooth} t_{smooth}^2 = \frac{1}{2} a_{rough} t_{rough}^2\)

Now, substitute the expressions of time and accelerations from Step 3 and Step 4 into the equation above: \(\frac{1}{2} (g/\sqrt{2}) t_{smooth}^2 = \frac{1}{2} ((1-\mu)(g/\sqrt{2})) (n t_{smooth})^2\) Simplifying and canceling terms, we can solve for the coefficient of kinetic friction: \(\mu = 1 - \frac{1}{n^2}\) Therefore, the answer is (B) \(\left[1-\left(1 / \mathrm{n}^{2}\right)\right]\).