JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(......\)
- A \(2\)
- B \(1\)
- C \(4\)
- D \(3\)
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 \(I = I _{1}+ I _{2}+ I _{3}+ I _{4}\) \(=\frac{ MR ^{2}}{2}+\frac{5}{2} MR ^{2}\) \(=3 MR ^{2}, \text { Putting } R =\frac{ a }{2}\)…
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