What is the expression for the moment of inertia of a solid cylinder about its axis of symmetry and an axis perpendicular to its length?

Hint The moment of inertia of a solid cylinder about its hub or axis will be equivalent to the moment of inertia about a hub or axis going through its focal point of gravity and opposite to its length. So with the help of the above statement, we can make the relation.

Complete Step By Step Solution
As we know the moment of inertia about the axis of a cylinder can be represented as
$ \Rightarrow \dfrac{1}{2}M{R^L}$
Also through the center of the gravity and perpendicular to its length the moment of inertia about the axis passing will be given by
$ \Rightarrow m\left( {\dfrac{{{L^2}}}{{12}} + \dfrac{{{R^2}}}{4}} \right)$
So it can also be written as
$ \Rightarrow \dfrac{m}{{12}}\left( {{L^2} + 3{R^2}} \right)$
Since we know both will be equal to each other.
Therefore on equating the equation, we get
$ \Rightarrow \dfrac{1}{2}M{R^L} = \dfrac{m}{{12}}\left( {{L^2} + 3{R^2}} \right)$
On solving the above equation, we get
$ \Rightarrow 6{R^2} = {L^2} + 3{R^2}$
Again solving the above equation and we will get
$ \Rightarrow 3{R^2} = {L^2}$
And hence we can also write it as
$ \Rightarrow \sqrt 3 R = L$, this is our required relation.

Note Moment of inertia is the property of a body to resist the angular acceleration due to external torque. For better understanding the concept of moment of inertia, we must have a good conceptual hold on linear kinetics. Linear kinetics Ian's hugely based on Newton's laws of motion.
Newton's ${1^{st}}$ law of motion states that if a body is at rest or is moving, continues to do so without any change unless an external force is applied. This property is known as inertia. Roughly inertia of a body is determined by its mass. As from our daily experience, we know that it is difficult to move a heavy object but it is not so difficult when the object is lighter in mass. This we are quite familiar with the property of inertia of a body.