JEE Main 2022 · 28 Jul · Shift 2 · Physics Q22

← JEE Main 2022 · 28 Jul · Shift 2

PhysicsInteger+4 / -1Question 22 of 90

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

Your answer

Answer and worked solution

Official answer: 8

Centre of mass. $\int\rho x\,dx=\frac{\rho_0L^2}4$ and $\int\rho\,dx=\frac{2\rho_0L}3$ , so $x_{cm}=\frac{3L}8$ .

Answer: 8

Topics:Centre of mass

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JEE Main 2022 · 28 Jul · Shift 2, Physics question 22