A particle is placed at rest inside a hollow hemisphere of radius 'R'. The co-efficient of friction between the particle and the hemisphere is $\mathrm{\mu }=\frac{1}{\sqrt{3}}$. The maximum height upto which the particle can remain stationary is

Particle in Hemisphere – Maximum Height Calculation

Let's clarify the problem and the solution, as there is a discrepancy in the coefficient of friction in your query versus the solutions found: you wrote μ = 1/3, but most reliable sources for this classic problem use μ = 1/√3. Below, I provide solutions for both cases.

For μ = 1/√3

Step-by-step solution:

1. Forces on the particle:
   • Gravitational force:mg downward.
   • Normal force:N = mg cosθ (perpendicular to surface).
   • Frictional force:f = μN = μ mg cosθ (along the surface).
2. Condition for equilibrium:The component of gravity along the surface must be balanced by friction:mg sinθ ≤ μ mg cosθtanθ ≤ μθ ≤ tan⁻¹(μ)
3. Given μ = 1/√3:tanθ ≤ 1/√3θ ≤ 30°
4. Maximum height calculation:The height above the base is:h = R - R cosθFor θ = 30°:h = R - R cos(30°) = R(1 - √3/2)

Correct option:

h = (1 - √3/2)R

For μ = 1/3

If the coefficient of friction is indeed μ = 1/3 (as you wrote), the solution changes:

1. Condition for equilibrium:tanθ ≤ 1/3θ ≤ tan⁻¹(1/3) ≈ 18.43°
2. Maximum height calculation:h = R - R cos(tan⁻¹(1/3))cos(tan⁻¹(1/3)) = 3/√10h = R(1 - 3/√10)Numerically:h ≈ R(1 - 0.9487) ≈ 0.0513R(This is not among the standard options and is much lower than the previous result.)