If l xy is the moment of inertia of a ring about a tangent in the plane of the ring and Ix prime y prime is the moment of inertia of a ring about a tangent perpendicular to the plane of the ring then

Question

If l xy is the moment of inertia of a ring about a tangent in the plane of the ring and Ix prime y prime is the moment of inertia of a ring about a tangent perpendicular to the plane of the ring then (A) Ix y=Ix prime y (B) Ix y=(1/2) Ix prime y (C) Ix prime y prime=(3/4) Ix y (D) Ix y=(3/4) Ix prime y

Tardigrade Admin · Accepted Answer

Ix y, moment of inertia of a ring about its tangent in the plane of ring Ix prime y=(3/2) MR 2 Moment of inertia about a tangent perpendicular to the plane of ring Ix y=2 MR 2 ∴ Ix y=(3/4)(2 M2)=(3/2) M R2 text or Ix y=(3/4) Ix1 y 1

Q. If $l_{x y}$ 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

$I_{x y} = I_{x^{'} y}$

$I_{x y} = \frac{1}{2} I_{x^{'} y}$

$I_{x^{'} y^{'}} = \frac{3}{4} I_{x y}$

$I_{x y} = \frac{3}{4} I_{x^{'} y}$

Q. If lxy​ is the moment of inertia of a ring about a tangent in the plane of the ring and Ix′y′​ is the moment of inertia of a ring about a tangent perpendicular to the plane of the ring then

Ixy​=Ix′y​

Ixy​=21​Ix′y​

Ix′y′​=43​Ixy​

Ixy​=43​Ix′y​

Ixy​, moment of inertia of a ring about its tangent in the plane of ring Ix′y​=23​MR2 Moment of inertia about a tangent perpendicular to the plane of ring Ixy​=2MR2 ∴Ixy​=43​(2M2)=23​MR2 or Ixy​=43​Ix1y​1

Q. If $l_{x y}$ 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

$I_{x y} = I_{x^{'} y}$

$I_{x y} = \frac{1}{2} I_{x^{'} y}$

$I_{x^{'} y^{'}} = \frac{3}{4} I_{x y}$

$I_{x y} = \frac{3}{4} I_{x^{'} y}$