JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A circular disc \(D_1\) of mass \(M\) and radius \(R\) has two identical discs \(D_2\) and \(D_3\) of the same mass \(M\) and radius \(R\) attached rigidly as its opposite ends (see figure). The moment of inertia of the system about the axis \(OO\)’, passing through the centre of \(D_1\) as shown in the figure, will
- A \(MR^2\)
- B \(3MR^2\)
- C \(\frac{4}{5}\,M{R^2}\)
- D \(\frac{2}{3}\,M{R^2}\)
Answer & Solution
Correct Answer
(B) \(3MR^2\)
Step-by-step Solution
Detailed explanation
\(I = \frac{{M{R^2}}}{2} + 2\left( {\frac{{M{R^2}}}{4} + M{R^2}} \right)\) \( = \frac{{M{R^2}}}{2} + \frac{{M{R^2}}}{2} + 2M{R^2} = 3\,M{R^2}\)
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