Rotational Motion Question 189

Question:If $I _{xy}$ is the moment of inertia of a ring about a tangent in the plane of the ring and ${{I} _{x’y’}}$ is the moment of inertia of a ring about a tangent perpendicular to the plane of the ring then

Options:

A) $I _{xy}={{I} _{x’y’}}$

B) $I _{xy}=\frac{1}{2}{{I} _{x’y’}}$

C) ${{I} _{x’y’}}=\frac{3}{4}I _{xy}$

D) $I _{xy}=\frac{3}{4}{{I} _{x’y’}}$

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Answer:

Correct Answer: D

Solution:

[d]

$ I _{x y}$, moment of inertia of a ring about its tangent in the plane of ring

$I _{x^{\prime} y}=\frac{3}{2} M R^2$ .

Moment of inertia about a tangent perpendicular to the plane of ring

$I _{x y}=2 M R^2$

$ \therefore I _{x y}=\frac{3}{4}\left(2 M R^2\right) or I _{x y}=\frac{3}{4} I _{x^1 y^1} $



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