Question

The moment of inertia of a ring of mass 10g and radius 1cm about a tangent to the ring and normal to its plane is:
A. $2 \times {10^{ - 6}}kg{m^2}$
B. $5 \times {10^{ - 7}}kg{m^2}$
C. $5 \times {10^{ - 8}}kg{m^2}$
D. $4 \times {10^{ - 6}}kg{m^2}$

Answer 1

Hint Moment of inertia can be calculated by using perpendicular axis theorem or by parallel axis theorem. The moment of inertia of a ring about a tangent to the ring and normal to the plane is 2MR$^2$ where, M is the mass of the ring and R is the radius of the ring.

Complete step-by-step answer:
Here, it is given that
Mass of the ring is $M = 10g = \dfrac{{10}}{{1000}}kg = {10^{ - 2}}kg$
Radius of the ring is $R = 1cm = \dfrac{1}{{100}}m = {10^{ - 2}}m$
The moment of inertia can be calculated by using the parallel or perpendicular axis theorem.
Now, the moment of inertia of a ring about a tangent to the ring and normal to the plane is
$ \Rightarrow I = 2M{R^2}$…………………….. (1)
Now, substitute the values of m and R in equation (1), we get
$
   \Rightarrow I = 2 \times {10^{ - 2}} \times {\left( {{{10}^{ - 2}}} \right)^2} \\
   \Rightarrow I = 2 \times {10^{ - 2}} \times {10^{ - 4}} \\
   \Rightarrow I = 2 \times {10^{ - 6}}kg{m^2} \\
 $
Thus, the moment of inertia of a ring about the tangent to the ring and normal to the plane is $2 \times {10^{ - 6}}kg{m^2}$

Hence, option A is the correct answer

Additional information The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. In general it can be expressed as $I = M{R^2}$, M is the mass of an object and R is the radius of the object.

Note the moment of inertia of planar objects are guided by the perpendicular axis theorem i.e. ( these are the moment of inertia along Z, X, Y axis respectively). The moment of inertia of the rigid body is guided by parallel axis theorem. Care should be taken in calculating the distance of separation from the centre of mass to the tangent placed.