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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The linear mass density of a thin rod \(A B\) of length \(L\) varies from \(A\) to \(B\) as \(\lambda( x )=\lambda_{0}\left(1+\frac{ x }{ L }\right),\) where \(x\) is the distance from A. If \(M\) is the mass of the rod then its moment of inertia about an axis passing through \(A\) and perpendicular to the rod is\(......ML^{2}\)
- A \(0.416\)
- B \(0.428\)
- C \(0.4\)
- D \(0.388\)
Answer & Solution
Correct Answer
(D) \(0.388\)
Step-by-step Solution
Detailed explanation
\(I =\int r ^{2} dm =\int x ^{2} \lambda dx\) \(I =\int_{0}^{ L } x ^{2} \lambda_{0}\left(1+\frac{ x }{ L }\right) dx\) \(I=\lambda_{0} \int_{0}^{L}\left(x^{2}+\frac{x^{3}}{L}\right) d x\) \(I=\lambda\left[\frac{L^{3}}{3}+\frac{L^{3}}{4}\right]\)…
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