Final answer:
Using the parallel axis theorem, the radius of gyration about a parallel axis at a distance 2R from the first axis for a solid sphere is found to be D. √(12/5)R.
Step-by-step explanation:
The question pertains to calculating the radius of gyration for a solid sphere when its axis of rotation is moved a certain distance from its centre of gravity. We can use the parallel axis theorem, which states that the moment of inertia I about any axis parallel to and a distance d from an axis through the center of mass is given by I = I_cm + Md², where I_cm is the moment of inertia through the center of mass, and M is the mass of the object.
In this case, given that the moment of inertia of the sphere about its center of gravity (I_cm) is ⅓ MR², and we know that the parallel axis is 2R away (d = 2R), we can apply the theorem:
I = I_cm + Md² = ⅓ MR² + M(2R)² = ⅓ MR² + 4MR² = ⅔ MR².
To find the radius of gyration k, we must solve the equation I = Mk² for k, yielding:
k = √(I/M) = √(⅔ MR²/M) = √(⅔ R²) = √(⅜ R²) = √(⅜)R = √(⅞u215e/5)R = √(12/5)R.
Therefore, the correct answer is D. √(12/5)R.
Applying the parallel-axis theorem to the moment of inertia of the solid sphere about the first axis, we have:
Iparallel-axis = 2/5MR² + M(2R)²
Iparallel-axis = 2/5MR² + 4MR²
Iparallel-axis = (2/5 + 4)MR²
Iparallel-axis = 22/5MR²
Therefore, the radius of gyration about the parallel axis is given by √(Iparallel-axis/M), which simplifies to √(22/5)R. Hence, the correct answer is option B, √(22/5)R.