If a solid cylinder, thin hollow cylinder and a sphere, all with same material and are having same outside diameter are allowed to roll down in same inclination without sliding. All surfaces are having same coefficient of friction. What is the order of the bodies reaching bottom 

Correct option is C

​$$\begin{aligned}&\textbf{1. Rolling Without Slipping Condition:} \\&\text{The acceleration } a \text{ of a rolling object is given by:} \\&a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \\[1em]&\text{where:} \\&g: \text{Acceleration due to gravity,} \\&\theta: \text{Inclination angle,} \\&I: \text{Moment of inertia,} \\&m: \text{Mass,} \\&r: \text{Radius (same for all objects)} \\[1.5em]&\textbf{2. Moment of Inertia } (I) \textbf{ for Each Object:} \\&\text{Solid cylinder: } I = \frac{1}{2} m r^2 \\&\text{Thin hollow cylinder: } I = m r^2 \\&\text{Solid sphere: } I = \frac{2}{5} m r^2 \\[1.5em]&\textbf{3. Substitute } I \text{ into Acceleration Formula:} \\&\text{Solid cylinder:} \\&a_{\text{solid cyl}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{1.5} = \frac{2}{3} g \sin \theta \\[1em]&\text{Thin hollow cylinder:} \\&a_{\text{hollow cyl}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} = \frac{1}{2} g \sin \theta \\[1em]&\text{Solid sphere:} \\&a_{\text{sphere}} = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{1.4} = \frac{5}{7} g \sin \theta \\[1.5em]&\textbf{Comparison of Accelerations:} \\&\quad a_{\text{sphere}} \left( \frac{5}{7} g \sin \theta \right) > a_{\text{solid cyl}} \left( \frac{2}{3} g \sin \theta \right) > a_{\text{hollow cyl}} \left( \frac{1}{2} g \sin \theta \right)\end{aligned}$$​​