Q.

Moment of inertia of a thin ring of mass m and radius r about an axis passing through its centre and perpendicular to its plane is I. Three such identical rings are joined together as shown in the figure. Then moment of inertia of the system about axis AB is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

534I

b

8+233I

c

3+432I

d

15+432I

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

I=mR2 (R=Radius)M.O.I. of a ring about a diameter =12mR2=I2d = Distance of centre of ring 3 from axis AB =R+2Rsin60o=3+1R∴IAB=mR22+mR2 x 2+mR22+m3+12R2=72mR2+22+3 x mR2=15+432 x I
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon
Moment of inertia of a thin ring of mass m and radius r about an axis passing through its centre and perpendicular to its plane is I. Three such identical rings are joined together as shown in the figure. Then moment of inertia of the system about axis AB is