Rotation 1 Question 17
20. Let $I$ be the moment of inertia of a uniform square plate about an axis $A B$ that passes through its centre and is parallel to two of its sides. $C D$ is a line in the plane of the plate that passes through the centre of the plate and makes an angle $\theta$ with $A B$. The moment of inertia of the plate about the axis $C D$ is then equal to
(a) $I$
(b) $I \sin ^{2} \theta$
(c) $I \cos ^{2} \theta$
(d) $I \cos ^{2}(\theta / 2)$
$(1998,2 M)$
Objective Question II (One or more correct option)
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Answer:
Correct Answer: 20. (a)
Solution:
- $A^{\prime} B^{\prime} \perp A B$ and $C^{\prime} D^{\prime} \perp C D$
From symmetry $\quad I _{A B}=I _A{ }^{\prime} B^{\prime}$
and
$$ I _{C D}=I _{C^{\prime} D^{\prime}} $$
From theorem of perpendicular axes,
$$ \begin{aligned} I _{Z Z}=I _{A B}+I _{A^{\prime} B^{\prime}} & =I _{C D}+I _{C^{\prime} D^{\prime}} \\ & =2 I _{A B}=2 I _{C D} \\ I _{A B} & =I _{C D} \end{aligned} $$
Alternate
The relation between $I _{A B}$ and $I _{C D}$ should be true for all values of $\theta$
At
$$ \theta=0, I _{C D}=I _{A B} $$
Similarly, at $\theta=\pi / 2, I _{C D}=I _{A B}$ (by symmetry)
Keeping these things in mind, only option (a) is correct.
