A ring solid sphere and a disc are rolling down from the top of the same height, then the sequence to reach on surface is

The correct option is Sphere, disc, ring.

To understand the order in which the objects reach the bottom of the incline, we need to analyze their acceleration down the slope. For any solid body of mass ^M, radius ^R, and moment of inertia ^I rolling down an inclined plane of angle $\theta$ without slipping, its acceleration ^a is given by the formula:

$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}$

where ^g is the acceleration due to gravity. The object with the greatest acceleration will reach the bottom first (taking the least time), and the one with the smallest acceleration will reach last.

Let us write down the moment of inertia ^I for each object:
1. For a solid sphere: $I=\frac{2}{5}M{R}^2$, which gives the ratio $\frac{I}{M{R}^2}=0.4$.
2. For a disc: $I=\frac{1}{2}M{R}^2$, which gives the ratio $\frac{I}{M{R}^2}=0.5$.
3. For a ring: $I=M{R}^2$, which gives the ratio $\frac{I}{M{R}^2}=1.0$.

Now, let us calculate the acceleration for each object:

- Acceleration of the solid sphere:
$a_{\text{sphere}}=\frac{g\sin \theta }{1+0.4}=\frac{g\sin \theta }{1.4}\approx 0.714g\sin \theta$

- Acceleration of the disc:
$a_{\text{disc}}=\frac{g\sin \theta }{1+0.5}=\frac{g\sin \theta }{1.5}\approx 0.667g\sin \theta$

- Acceleration of the ring:
$a_{\text{ring}}=\frac{g\sin \theta }{1+1.0}=\frac{g\sin \theta }{2.0}=0.5g\sin \theta$

Comparing the accelerations, we have:

$a_{\text{sphere}}>a_{\text{disc}}>a_{\text{ring}}$

Since the sphere has the highest acceleration, it will reach the bottom first. The disc will reach next, and the ring, having the lowest acceleration, will reach last. Therefore, the sequence to reach the surface is: Sphere, disc, ring.