System of Particles and Rotational Motion - Result Question 60
63. Four identical thin rods each of mass $M$ and length $l$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is :
(a) $\frac{2}{3} Ml^{2}$
(b) $\frac{13}{3} Ml^{2}$
(c) $\frac{1}{3} Ml^{2}$
(d) $\frac{4}{3} Ml^{2}$
[2009]
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Answer:
Correct Answer: 63. (d)
Solution:
- (d) Moment of inertia of a thin rod of length $l$ about an axis passing through centre and perpendicular to the $rod=\frac{1}{12} Ml^{2}$.
Thus moment of inertia of the frame.
$\frac{ml^{2}}{12}+\frac{ml^{2}}{4}=\frac{4 ml^{2}}{12}=\frac{ml^{2}}{3}$
Total M.I. $=4 \times \frac{ml^{2}}{3}$
