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Rotational Motion

JEE Main level, single correct, medium

91 of 280

Question 91 of 280

A thin bar of length $\mathrm{L}$ has a mass per unit length $\lambda$ , that increases linearly with distance from one end. If its total mass is $M$ and its mass per unit length at the lighter end is $\lambda_{\mathrm{O}}$ , then the distance of the centre of mass from the lighter end is:

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Answer: (C) $\frac{\mathrm{L}}{3}+\frac{\lambda_0 \mathrm{~L}^2}{4 \mathrm{M}}$

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JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

See it move: Rotational dynamics and rolling simulationDrag the sliders and watch the numbers in this question change.Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Rotational Motion (Moment of inertia of rigid bodies), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A thin bar of length L\mathrm{L}L has a mass per unit length λ\lambdaλ, that increases linearly with distance from one end. If its total mass is MMM and its mass per unit length at the lighter end is λO\lambda_{\mathrm{O}}λO​, then the distance of the centre of mass from the lighter end is:

JEE Main question paper PDF

More Physics practice

A thin bar of length $\mathrm{L}$ has a mass per unit length $\lambda$ , that increases linearly with distance from one end. If its total mass is $M$ and its mass per unit length at the lighter end is $\lambda_{\mathrm{O}}$ , then the distance of the centre of mass from the lighter end is: