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The ratio of moment of inertia of solid sphere and hollow sphere is _______. (Given that both have the same mass and the same radius)
1. \({3\over 5}\)
2. \( {5\over 3}\)
3. \(1 \over 3\)
4. \(2 \over 5\)

1 Answer

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Best answer
Correct Answer - Option 1 : \({3\over 5}\)

CONCEPT:  

  • Moment of Inertia: It is a quantity that expresses a body's tendency to resist angular acceleration, it is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation, is called the moment of Inertia.
  • The moment of inertia is simply the mass times the square of the perpendicular distance of central to the axis of rotation.


I = m × r​2 

where I is the Moment of Inertia, m is point mass, r is the perpendicular distance from the axis of rotation.

The moment of inertia of different bodies is given in the below table:

Shape Axis of rotation Moment of inertia
Ring axis passing through the center perpendicular to the plane of the ring \(I = mr^2\)
Ring axis passing through the diameter of the ring \(I = {1 \over 2}mr^2\)
Solid Cylinder axis passing through the center perpendicular to the plane of the ring \(I = {1 \over 2}mr^2\)
Solid sphere through center \(I = {2 \over 5}mr^2\)
Hollow sphere through center \(I = {2 \over 3}mr^2\)
Rod  through midpoint perpendicular to the rod \(I = {1 \over 12}ml^2\)


EXPLANATION:

From the above table, it is clear that the moment of inertia of a solid sphere of mass 'm' and radius 'R' about an axis passing through the center is:

\(I_1 = {2 \over 5}mr^2\).

Moment of inertia of a solid sphere of mass 'm' and radius 'R' about an axis passing through the center is:

\(I_2 = {2 \over 3}mr^2\).

Their ratio \({I_1 \over I_2} = {{2 \over 5}mr^2 \over {2 \over 3}mr^2}\)

\({I_1 \over I_2} = {3 \over 5}\)

So the correct answer is option 1.

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