![]() Integral Calculus, Centroids and Moments of Inertia, Vector Calculus. Using the parallel axis theorem keeps in mind that the reference axis must pass through the center of mass of the object. integral is to be evaluated over a sphere for the given steady vector field. Note- The parallel axis theorem is used for finding the moment of inertia of the area of a rigid body whose axis is parallel to the axis of the known moment body and it is through the center of gravity of the object. Therefore, the moment of inertia of a solid sphere about its diameter (axis) is expressed as follows ![]() By parallel axis theorem, the moment of inertia at 2R is I ( 2 5 M R 2) + M ( 2 R) 2 I ( 22 5 M R 2) The radius of gyration is M k 2 ( 2 5 M R 2) Therefore k 22 5 R Hence the correct option is B. The moment of inertia is the reluctance to change the state of motion (rotation) of an object rotating around a given axis m mass of the solid sphere r - Radius of the sphere. Radius of gyration of a body about an axis of rotation is defined as the radial distance to a point which would have a moment of inertia the same as the body's actual distribution of mass, if the total mass of the body were concentrated. Parallel axis theorem states that the moment of inertia of a body about its parallel axis is equal to the sum of its moment inertia about its axis and the product of the square of distance between them and its mass. And then we will use the parallel axis theorem formula to get the answer. Moment of inertia of a body about a given axis is 1.5 kg m2. ![]() represents the variation of I(x) with x correctly. ![]() Hint- For solving this numerical firstly we will see the terms like radius of gyration and moment of inertia of the sphere. The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a. ![]()
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