Question

Let $I$ be the moment of inertia of a uniform square plate about an axis $AB$ that passes through its centre and is parallel to two of its sides. $CD$ is a line in the plane of the plate that passes through the centre of the plate and makes an angle $\theta$ with $AB$. The moment of inertia of the plate about the axis $CD$ is then equal to

• A. $I$
• B. $I \, sin^2 \theta$
• C. $I \, cos^2 \theta$
• D. $I \,cos^2 (\theta/2)$

Answer 1

The Correct Option is A

In the figure, $A'B'$ is perpendicular to $AB$
$ CD'$ is perpendicular to $CD$
By symmetry, $I_{AB} = I_{A'B'}$
$I_{CD} = I_{C'D'}$
By theorem of perpendicular axes,
$I_{z} = I_{AB} +I_{A'B'} = 2I_{AB} \quad.....\left(i\right)$
Again$ I_{z} = I_{CD} +I_{C'D'} = 2I_{CD} \quad.....\left(ii\right)$
$\therefore$ From $\left(i\right)$ and $\left(ii\right)$,
$ 2I_{AB} = 2I_{CD}$ or $I_{AB} = I_{CD}$ or $I= I_{CD}$
Hence moment of inertia about $CD$ axis $= I$