To understand why the correct answer is option 'A', let's break down the problem into two parts: the rolling motion and the sliding motion of the sphere.

Rolling Motion:
- The sphere is rolling down the inclined plane without slipping, which means that the point of contact between the sphere and the plane is at rest.
- In this case, the sphere's linear speed is related to its angular speed (or rotational speed) by the equation v = ωR, where v is the linear speed, ω is the angular speed, and R is the radius of the sphere.
- Since the sphere is rolling without slipping, the frictional force between the sphere and the plane provides the necessary torque to keep the sphere rolling.
- The moment of inertia of a solid sphere about an axis passing through its center is given by I = (2/5)MR², where M is the mass of the sphere and R is the radius.
- The radius of gyration (k) is defined as the distance from the axis of rotation at which the entire mass of the object can be considered to be concentrated to calculate the moment of inertia.
- For a solid sphere, the radius of gyration is related to the radius by the equation k = R/√5.
- Using the equation v = ωR and the radius of gyration, we can write v = (k√5)ω.

Sliding Motion:
- Now, let's consider the sphere sliding down the same inclined plane without rolling.
- In this case, the sphere is purely sliding, and there is no rotational motion.
- The speed of the sphere at the bottom of the incline is given by v' = √(2gh), where g is the acceleration due to gravity and h is the height of the incline.
- Given that the speed at the bottom is 5v/4, we have v' = 5v/4.
- Substituting the value of v = (k√5)ω from the rolling motion, we can write 5v/4 = √(2gh).

Comparing the speeds:
- From the rolling motion, we have v = (k√5)ω.
- From the sliding motion, we have 5v/4 = √(2gh).
- Equating these two expressions, we get (k√5)ω = √(2gh).
- Simplifying this equation, we find ω = (4√5/5√2)(√h/R).
- Comparing this with the equation v = ωR, we can conclude that v = (4√5/5√2)(√hR/R).
- Simplifying further, we have v = (4√5/5√2)√h.
- Comparing this expression with v' = 5v/4 = √(2gh), we can equate them and solve for h.

Calculating the radius of gyration:
- By comparing the expressions for v and v', we can solve for h.
- After solving for h, we find h = (16/5)(R²/2k²).
- Substituting the value of k = R/√5, we get h = (16/5)(R²/2(R²/5)).
- Simplifying this equation, we