Rotational Motion Question 198
Question: 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 comers is
Options:
A) $\frac{5}{6}ma^{2}$
B) $\frac{1}{12}ma^{2}$
C) $\frac{7}{12}ma^{2}$
D) $\frac{2}{3}ma^{2}$
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Answer:
Correct Answer: D
Solution:
[d]
$ I _{y_1}=\frac{M R^2}{4} $
$ \therefore I^{\prime} y_1=\frac{M R^2}{4}+M R^2=\frac{5}{4} M R^2 .$
$ I _{y_2}=\frac{M R^2}{2} $
$ \therefore I _{y_2}^{\prime}=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2 $
$ I^{\prime} y_1=M K_1^2, I^{\prime} y_2=M K_2^2 $
$ \therefore \quad \frac{K_1^2}{K_2^2}=\frac{I_{y_1}^{\prime}}{I_{y_2}^{\prime}} \Rightarrow K_1: K_2=\sqrt{5}: \sqrt{6} $
BETA
