Question:
The linear mass density of a thin rod AB of length \(L\) varies from A to B as \(\lambda(x)=\lambda_0\left(1+\dfrac{x}{L}\right)\), 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 :
The linear mass density of a thin rod AB of length \(L\) varies from A to B as \(\lambda(x)=\lambda_0\left(1+\dfrac{x}{L}\right)\), 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 :
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Integrate \(dm=\lambda\,dx\) to get \(M\), and \(\int x^2\,dm\) to get \(I\); then eliminate \(\lambda_0\).
Updated On: Nov 5, 2026
- \(\dfrac{2}{5}ML^2\)
- \(\dfrac{5}{12}ML^2\)
- \(\dfrac{7}{18}ML^2\)
- \(\dfrac{3}{7}ML^2\)
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The Correct Option is C
Solution and Explanation
Mass: \(M=\int_0^L\lambda_0\left(1+\dfrac{x}{L}\right)dx=\lambda_0\left(L+\dfrac{L}{2}\right)=\dfrac{3}{2}\lambda_0L\).
Moment of inertia about A: \(I=\int_0^L x^2\lambda_0\left(1+\dfrac{x}{L}\right)dx=\lambda_0\left(\dfrac{L^3}{3}+\dfrac{L^3}{4}\right)=\dfrac{7}{12}\lambda_0L^3\).
Dividing, \(\dfrac{I}{M}=\dfrac{7/12}{3/2}L^2=\dfrac{7}{18}L^2\).
So \(I=\dfrac{7}{18}ML^2\).
Answer: (3) \(\dfrac{7}{18}ML^2\)
Moment of inertia about A: \(I=\int_0^L x^2\lambda_0\left(1+\dfrac{x}{L}\right)dx=\lambda_0\left(\dfrac{L^3}{3}+\dfrac{L^3}{4}\right)=\dfrac{7}{12}\lambda_0L^3\).
Dividing, \(\dfrac{I}{M}=\dfrac{7/12}{3/2}L^2=\dfrac{7}{18}L^2\).
So \(I=\dfrac{7}{18}ML^2\).
Answer: (3) \(\dfrac{7}{18}ML^2\)
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