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. $Isi{n}^2\theta$
• C. $Ico{s}^2\theta$
• D. $Ico{s}^2(\theta /2)$

Answer 1

Answer: (A)

In the figure, $A^{\prime }{B}^{\prime }$ is perpendicular to $AB$
$C{D}^{\prime }$ is perpendicular to $CD$
By symmetry, $I_{AB}=I_{A^{\prime }{B}^{\prime }}$
$I_{CD}=I_{C^{\prime }{D}^{\prime }}$
By theorem of perpendicular axes,
$I_z=I_{AB}+I_{A^{\prime }{B}^{\prime }}=2I_{AB}.....\left(i\right)$
Again$I_z=I_{CD}+I_{C^{\prime }{D}^{\prime }}=2I_{CD}.....\left(ii\right)$
$\therefore$ 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 $=I$