A given object takes n times the time to slide down 45° rough inclined plane as it takes the time to slide down an identical perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :

Solution

Physics sir · 4 steps

Same length on both slopes, so t² goes as 1/a. Compare the two accelerations and μ drops out.

1. Two accelerations. On the smooth slope, a₁ = g sin 45° = g/√2. On the rough slope, friction μ mg cos 45° acts up the slope, so a₂ = g(sin 45° - μ cos 45°) = g(1-μ)/√2.
2. Time and acceleration. Both slopes have the same length L and the object starts from rest, so L = 1/2at². That gives t² = 2L/a, so t² is inversely proportional to a.

   t₂²/t₁² = a₁/a₂ ⇒ n² = 1/(1-μ)

3. Solve for μ. Rearranging gives 1 - μ = 1/n², so μ = 1 - 1/n².
4. So. The coefficient of kinetic friction is 1 - 1/n², which is option (a).

The idea to keep. Same distance from rest means t ∝ 1/√a. At 45°, sin θ = cos θ, so the rough-slope acceleration is just the smooth one times (1-μ).

Answer from the NTA final answer key. Solution by PrepWiser.