NEET Solved Paper 2016 Question 1
Question: From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
Options:
A) $ 15,MR^{2}/32 $
B) $ 13,MR^{2}/32 $
C) $ 11,MR^{2}/32 $
D) $ 9,MR^{2}/32 $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {I _{\text{Total disc }}}=\frac{MR^{2}}{2} $ $ M _{Removed}=\frac{M}{4}(Mass\propto area) $ $ {I _{Removed}}(about,same,Perpendicular,axis) $ $ =\frac{M}{4}\frac{{{(R/2)}^{2}}}{2}+\frac{M}{4}{{( \frac{R}{2} )}^{2}}=\frac{3MR^{2}}{32} $ $ {I _{Remaing,disc}}=I _{Total}-{I _{Removed}} $ $ =\frac{MR^{2}}{2}-\frac{3}{32}MR^{2}=\frac{13}{32}MR^{2} $
