Forum

From a circular rin...
 
Notifications
Clear all

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre

  

0
Topic starter

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times 'MR2'. Then the value of 'K' is :

(1) \(\frac{3}{4}\)

(2) \(\frac{7}{8}\)

(3) \(\frac{1}{4}\)

(4) \(\frac{1}{8}\)

Topic Tags
1 Answer
0

Correct answer: (1) \(\frac{3}{4}\)

Explanation:

Mremain = \(\frac{3}{4}\)M

I = Mremain R2

= \(\frac{3}{4}MR^2\)

Share:

How Can We Help?