The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ρ = ρ₀(1 - (x²)/(L²)) kg/m and length L (in meter) is (3L)/(α) m. The value of α is ___________. (where x is the distance from end A) Numerical answer type. Work it out, then open the answer.

Answer: 8. M = ∫₀L ρ₀(1 - (x²)/(L²))dx = (2ρ₀ L)/(3) ∫₀L xρ dx = ρ₀(L²/2 - L²/4) = (ρ₀ L²)/(4) xcm = (L²/4)/(2L/3) = 3L/8 So α = 8.

The distance of centre of mass from end A of a one dimensional rod… | JEE Main 2022 (28 Jul, Shift 2)

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Continuous Bodies: Standard Results & Variable Density

The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $\rho = \rho_0\left(1 - \frac{x^2}{L^2}\right)\text{ kg/m}$ and length L (in meter) is $\frac{3L}{\alpha}\text{ m}$ . The value of $\alpha$ is ___________. (where x is the distance from end A)

Numerical answer type. Work it out, then open the answer.

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Answer: 8

$M = \int_0^L \rho_0\left(1 - \frac{x^2}{L^2}\right)dx = \frac{2\rho_0 L}{3}$
$\int_0^L x\rho\,dx = \rho_0\left(\frac{L^2}{2} - \frac{L^2}{4}\right) = \frac{\rho_0 L^2}{4}$
$x_{cm} = \frac{L^2/4}{2L/3} = \frac{3L}{8}$
So $\alpha = 8$ .

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