JEE Main On 08 April 2018 Question 16
Question: Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is: [JEE Main Online 08-04-2018]
Options:
A) $ \frac{73}{2}M{R^{2}} $
B) $ \frac{181}{2}M{R^{2}} $
C) $ \frac{19}{2}M{R^{2}} $
D) $ \frac{55}{2}M{R^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=( \frac{MR^{2}}{2} )+6[ \frac{MR^{2}}{2}+M{{(2R)}^{2}} ]=\frac{MR^{2}}{2}+6[ 4.5MR^{2} ] $
$ I=27.5MR^{2}=\frac{55}{2}MR^{2};I _{P}=\frac{55}{2}MR^{2}+(7M){{(3R)}^{2}}=\frac{181MR^{2}}{2} $
