The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is 4 kg-m2 . The diameter of the ring is

Question

The moment of inertia of a circular ring of mass 1 kg about an axis passing through its centre and perpendicular to its plane is 4 kg-m2 . The diameter of the ring is (A) 2 m (B) 4 m (C) 5 m (D) 6 m

Tardigrade Admin · Accepted Answer

Moment of inertia of circular ring about an axis passing through its center of mass and perpendicular to its plane I=M R2 here I=4 k g-m2 m =1 kg =R2=(4/1)=4 R=2 m Therefore, diameter of ring =4 m

Q. The moment of inertia of a circular ring of mass $1 k g$ about an axis passing through its centre and perpendicular to its plane is $4 k g - m^{2}$ . The diameter of the ring is

2 m

4 m

5 m

6 m

Moment of inertia of circular ring about an axis passing through its center of mass and perpendicular to its plane
$I = M R^{2}$
here $I = 4 k g - m^{2}$
$m = 1 k g$
$= R^{2} = \frac{4}{1} = 4 R = 2 m$
Therefore, diameter of ring $= 4 m$

Q. The moment of inertia of a circular ring of mass 1kg about an axis passing through its centre and perpendicular to its plane is 4kg−m2 . The diameter of the ring is

2 m

4 m

5 m

6 m

Moment of inertia of circular ring about an axis passing through its center of mass and perpendicular to its plane I=MR2 here I=4kg−m2 m=1kg =R2=14​=4R=2m Therefore, diameter of ring =4m

Q. The moment of inertia of a circular ring of mass $1 k g$ about an axis passing through its centre and perpendicular to its plane is $4 k g - m^{2}$ . The diameter of the ring is

Moment of inertia of circular ring about an axis passing through its center of mass and perpendicular to its plane
$I = M R^{2}$
here $I = 4 k g - m^{2}$
$m = 1 k g$
$= R^{2} = \frac{4}{1} = 4 R = 2 m$
Therefore, diameter of ring $= 4 m$