Question

A thin ring has mass 0.25 kg and radius 0.5 m it's moment of inertia about an axis passing through its center and perpendicular to its plane is...
A)0.0625 kg msq.
B)0.625 kg msq.
C)6.25kg msq.
D)62.5kg msq.

Answer 1

= M × Area
= 0.25 × 0.5 × 0.5
= 625/1000 = 0.625
So, option b is correct.

Answer 2

Dear Student,

◆ Answer -

(A) 0.0625 kg/m^2

● Explanation -

# Given -

m = 0.25 kg

r = 0.5 m

# Solution -

Moment of inertia about an axis passing through center of the ring and perpendicular to its plane is given by formula -

I = mr^2

I = 0.25 × 0.5^2

I = 0.0625 kg/m^2

Hence, moment of inertia about an axis passing through center of the ring and perpendicular to its plane is 0.0625 kg/m^2.

Thanks dear. Hope this helps you...