# What will be moment of inertia of a solid sphere of mass 'm' and radius 'r' about an axis passing through the center?

51 views
in Physics
closed
What will be moment of inertia of a solid sphere of mass 'm' and radius 'r' about an axis passing through the center?
1. $\frac{M{{r}^{2}}}{4}$
2. $\frac{M{{r}^{2}}}{2}$
3. $\frac{2M{{r}^{2}}}{3}$
4. $\frac{2}{5}~M{{r}^{2}}$

by (30.0k points)
selected

Correct Answer - Option 4 : $\frac{2}{5}~M{{r}^{2}}$

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 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:

$I = {2 \over 5}mr^2$.

• So the correct answer is option 4.