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} $

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Answer:

Correct Answer: D

Solution:

  • 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} $



NCERT Chapter Video Solution

Dual Pane