A thin rod of length $5$ m has a linear mass density $λ (x) = 2 x kg/m$ , where $x$ is the distance from one end. Find the position of the centre of mass from that end.

hard 4 marks Class 12 · Mathematics · Application of Integrals

Answer

, \frac{10}{3} m

The correct answer is $, \frac{10}{3} m$ — Set up the first moment: $\int_{0}^{5} x l amb d a (x), d x = \int_{0}^{5} 2 x^{2}, d x = \frac{2}{3} x^{3}_{0}^{5} = \frac{250}{3}$ kg·m.

Step-by-step solution

Calculate total mass: $M = in t_{0}^{5} 2 x, d x = 5^{2} = 25$ kg.
Set up the first moment: $\int_{0}^{5} x l amb d a (x), d x = \int_{0}^{5} 2 x^{2}, d x = \frac{2}{3} x^{3}_{0}^{5} = \frac{250}{3}$ kg·m.

Common mistake: Not dividing the first moment by total mass

Concept tested: Centre of mass of a variable density rod

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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A thin rod of length 5 m has a linear mass density λ(x)=2x kg/m, where x is the distance from one end. Find the position of the centre of mass from that end.

Step-by-step solution

A thin rod of length $5$ m has a linear mass density $λ (x) = 2 x kg/m$ , where $x$ is the distance from one end. Find the position of the centre of mass from that end.