Q115 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
Moment of inertia of a disc of mass $M$ and radius 'R' about any of its diameter is $MR^2/4$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $(x/2) MR^2$. The value of $x$ is
- A$1$
- Correct answerB$3$
- C$5$
- D$7$
Solution
Given the moment of inertia of the disc about its diameter is $I_{d} = \dfrac{MR^2}{4}$.
By the perpendicular axis theorem, the moment of inertia of the disc about an axis passing through its center and perpendicular to its plane is $I_{c} = I_{d} + I_{d} = \dfrac{MR^2}{4} + \dfrac{MR^2}{4} = \dfrac{MR^2}{2}$.
By the parallel axis theorem, the moment of inertia of the disc about an axis perpendicular to its plane and passing through a point on its edge is $I = I_{c} + MR^2$.
Substituting the value of $I_{c}$, we get $I = \dfrac{MR^2}{2} + MR^2 = \dfrac{3}{2}MR^2$.
Comparing this with the given expression $\dfrac{x}{2}MR^2$, we get $x = 3$.
Answer: $3$
By the perpendicular axis theorem, the moment of inertia of the disc about an axis passing through its center and perpendicular to its plane is $I_{c} = I_{d} + I_{d} = \dfrac{MR^2}{4} + \dfrac{MR^2}{4} = \dfrac{MR^2}{2}$.
By the parallel axis theorem, the moment of inertia of the disc about an axis perpendicular to its plane and passing through a point on its edge is $I = I_{c} + MR^2$.
Substituting the value of $I_{c}$, we get $I = \dfrac{MR^2}{2} + MR^2 = \dfrac{3}{2}MR^2$.
Comparing this with the given expression $\dfrac{x}{2}MR^2$, we get $x = 3$.
Answer: $3$
Hence, the correct answer is option: 'B'.