Q162 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
Consider a uniform square plate of side $'a'$ and mass $'m'$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
- A$\dfrac{5ma^2}{6}$
- Correct answerB$\dfrac{2ma^2}{3}$
- C$\dfrac{7ma^2}{12}$
- D$\dfrac{ma^2}{12}$
Solution
Let the square plate lie in the $xy$-plane with one of its corners at the origin and its adjacent sides along the $x$ and $y$ axes.
The moment of inertia of the plate about the $x$-axis (which lies along one edge) is $I_x = \dfrac{ma^2}{3}$.
The moment of inertia of the plate about the $y$-axis (which lies along the adjacent edge) is $I_y = \dfrac{ma^2}{3}$.
By the perpendicular axis theorem, the moment of inertia about the $z$-axis (which is perpendicular to the plane and passes through the corner) is given by:
$I_z = I_x + I_y$
$I_z = \dfrac{ma^2}{3} + \dfrac{ma^2}{3} = \dfrac{2ma^2}{3}$
Answer: $\dfrac{2ma^2}{3}$
The moment of inertia of the plate about the $x$-axis (which lies along one edge) is $I_x = \dfrac{ma^2}{3}$.
The moment of inertia of the plate about the $y$-axis (which lies along the adjacent edge) is $I_y = \dfrac{ma^2}{3}$.
By the perpendicular axis theorem, the moment of inertia about the $z$-axis (which is perpendicular to the plane and passes through the corner) is given by:
$I_z = I_x + I_y$
$I_z = \dfrac{ma^2}{3} + \dfrac{ma^2}{3} = \dfrac{2ma^2}{3}$
Answer: $\dfrac{2ma^2}{3}$
Hence, the correct answer is option: 'B'.