wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is:
984925_c5f1c7473c69464991e58f048ebce6e2.png

A
215MR2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
404MR2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
10MR2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
379MR2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 215MR2
Initial moment of inertia of the disc =12(9M)R2
I=92MR2
Moment of inertia of the removed disc about their own centre and mass of the removed disc.
In πR29M
In unit area 9MπR2
In π(R3)29MπR2×π(R3)2
m=M
So,
Iremoved(c)=12M(R3)2
Now, moment of inertia of removed portion about the centre of the bigger disc.
Iremoved=118MR2+12M(2R3)2
Iremoveed=5MR218
The net moment of inertia
Ifinel=92MR2(518)MR2
Ifinel=215MR2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integrating Solids into the Picture
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon