Question

Moment of inertia of a disc about an axis which is tangent and parallel to its plane is $I$. Then the moment of inertia of disc about a tangent, but perpendicular to its plane will be
(A) $\dfrac{3}{4}I$
(B) $\dfrac{5}{6}I$
(C) $\dfrac{3}{2}I$
(D) $\dfrac{6}{5}I$

Answer 1

Hint
A quantity that demonstrates an object's propensity to withstand angular acceleration, that is the addition of the products of the mass of every other particle in the object with the square of the distance from the axis of its own rotation.
The moment of inertia of the disc about an axis parallel to its plane is given as;
$\Rightarrow {I_t} = {I_d} + M\,{R^2}$
Where, $R$ denotes the radius of the disc, $M$ denotes the mass of the disc, ${I_t}$ denotes the moment of inertia of disc about a tangent.

Complete step by step answer
In order to explain how to calculate the moment of inertia of a disk, we will take the example of a uniform thin disk which is rotating about an axis through its centre. In the figure, we can see a uniform thin disk with radius r rotating about an axis passing through the centre.
The moment of inertia of the disc about an axis parallel to its plane is given as;
$\Rightarrow {I_t} = {I_d} + M\,{R^2} $
$\Rightarrow I = \dfrac{1}{4} \times M{R^2} + M{R^2} $
$\Rightarrow I = \dfrac{5}{4} \times M{R^2} $
Now, moment of inertia about a tangent perpendicular to its plane is;
$I' = \dfrac{3}{2} \times M{R^2}$
Substitute the values of the moment of inertia in the above equation or formula;
$\Rightarrow I' = \dfrac{3}{2} \times \dfrac{4}{5} \times I $
$\Rightarrow I' = \dfrac{6}{5}I $
Therefore, the moment of inertia of disc about a tangent, but perpendicular to its plane will be $I' = \dfrac{6}{5}I$
Hence the option (D) $I' = \dfrac{6}{5}I$ is the correct answer.

Note
A property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.