JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2022
- Medium
- Numerical
Four identical discs each of mass ' \(M\) ' and diameter ' \(a\) ' are arranged in a small plane as shown in figure. If the moment of inertia of the system about \(OO ^{\prime}\) is \(\frac{ x }{4} Ma ^{2}\). Then, the value of \(x\) will be \(......\)

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Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(I _{1}= I _{3}=\frac{ MR ^{2}}{4}\)
\(I _{2}=\frac{ MR ^{2}}{4}+ MR ^{2}=\frac{5}{4} MR ^{2}= I _{4}\)
So \(I = I _{1}+ I _{2}+ I _{3}+ I _{4}\)
\(=\frac{ MR ^{2}}{2}+\frac{5}{2} MR ^{2}\)
\(=3 MR ^{2}, \text { Putting } R =\frac{ a }{2}\)
\(I =\frac{3 Ma ^{2}}{4}, \text { So } x =3\)
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