System Of Particles And Rotational Motion Ques 68
The moment of inertia of a disc of mass $M$ and radius $R$ about an axis, which is tangential to the circumference of the disc and perpendicular to its diameter, is
[1999]
(a) $\frac{3}{2} M R^{2}$
(b) $\frac{2}{3} M R^{2}$
(c) $\frac{5}{4} M R^{2}$
(d) $\frac{4}{5} M R^{2}$
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Answer:
Correct Answer: 68.(c)
Solution:
- (c) Moment of inertia of disc about its diameter is $I_d=\frac{1}{4} M R^{2}$
MI of disc about a tangent passing through rim and in the plane of disc is
$I=I_d+M R^{2}=\frac{1}{4} M R^{2}+M R^{2}=\frac{5}{4} M R^{2}$
BETA
