Moment of inertia of a ring of mass m = 3 gm and radius r = 1 cm about an axis passing through its edge and parallel to its natural axis is:

Question

Moment of inertia of a ring of mass m = 3 gm and radius r = 1 cm about an axis passing through its edge and parallel to its natural axis is: (A)  1 text gm text cm2  (B)  6 text gm text cm2  (C)  100 text gm text cm2  (D)  10 text gm text cm2

Tardigrade Admin · Accepted Answer

Applying theorem of parallel axis I=2mR2 I=2× 3× (1)2=6gm cm2

Q. Moment of inertia of a ring of mass m = 3 gm and radius r = 1 cm about an axis passing through its edge and parallel to its natural axis is:

$1 g m c m^{2}$

$6 g m c m^{2}$

$100 g m c m^{2}$

$10 g m c m^{2}$

Applying theorem of parallel axis $I = 2 m R^{2}$ $I = 2 \times 3 \times (1)^{2} = 6 g m c m^{2}$

Q. Moment of inertia of a ring of mass m = 3 gm and radius r = 1 cm about an axis passing through its edge and parallel to its natural axis is:

1gmcm2

6gmcm2

100gmcm2

10gmcm2

Applying theorem of parallel axis I=2mR2 I=2×3×(1)2=6gmcm2

$1 g m c m^{2}$

$6 g m c m^{2}$

$100 g m c m^{2}$

$10 g m c m^{2}$

Applying theorem of parallel axis $I = 2 m R^{2}$ $I = 2 \times 3 \times (1)^{2} = 6 g m c m^{2}$