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The linear mass density of a thin rod AB of length L varies from A to B as $λ (x) = λ_{0} (1 + \frac{x}{L}),$ 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:

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detailed solution

Correct option is B

M=∫oLλo1+xLdx=32λoL I=∫dmx2M= ∫oLλo1+xLx2dx32λoL =718ML2

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The linear mass density of a thin rod AB of length L varies from A to B as λ(x) =λ01+xL, 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:

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The linear mass density of a thin rod AB of length L varies from A to B as $λ (x) = λ_{0} (1 + \frac{x}{L}),$ 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: