Rotational Motion Question 40
Question: The moment of inertia of a disc of mass M and radius R about a tangent to its rim in its plane is:
[AIPMT 1999]
Options:
A) $ \frac{2}{3}MR^{2} $
B) $ \frac{3}{2}MR^{2} $
C) $ \frac{4}{5}MR^{2} $
D) $ \frac{5}{4}MR^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
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Moment of inertia of a disc about its diameter is
$ I _{d}=\frac{1}{4}MR^{2} $
Now, according to perpendicular axis theorem moment of inertia of disc about a tangent passing through rim and in the plane of discs
$ I=I _{d}+MR^{2} $
$ =\frac{1}{4}MR^{2}+MR^{2} $
$ =\frac{5}{4}MR^{2} $
BETA
