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Answer 1

Question:- The linear mass density(λ) of a rod of length L kept along  x - axis varies as λ = α + βx , where α and β are positive constants. The  centre of  mass of  the rod is ?

Answer:  

$r_{cm}=\frac{(3\alpha+2\beta L)L}{3(2\alpha+\beta L)}$

Explanation:

Given,

Linear mass density of Rod,

λ = α + βx

We know that, linear mass density of a Rod is given by,

λ = M/L

Since we do not know the total mass of rod. We would use calculas method by taking a small mass dm, whose  length is dm and situated at distance x from the x-axis. Thus we have,

$\lambda=\frac{dm}{dx}\\\;\\dm=\lambda.dx$

$dm=(\alpha+\beta x)dx$

$\textbf{Integrating above equation from x = 0 to x = L}$

$\int\limits^M_0{dm}=\int\limits^L_0{(\alpha+\beta x)dx}$

$M=[\alpha x+\beta\frac{x^2}{2}]\limits^L_0\\\;\\M=\alpha L+\frac{\beta L^2}{2}$

This is the total mass of Rod.

Now, We know that  

$\text{Centre of Mass}(r_{cm})=\frac{1}{M}\int{x}dm$

$r_{cm}=\frac{1}{M}\int{x(\alpha+\beta x)dx}\;\;\;\;\;\;\;\;\;\;\;\{\because dm=(\alpha+\beta x)dx}\}$

$r_{cm}=\frac{1}{M}\int{(\alpha x+\beta x^2)dx}$

$\textbf{Integrating above equation from x = 0 to x = L}$

$r_{cm}=\frac{1}{M}\int\limits^L_0{(\alpha x+\beta x^2)}dx$

$r_{cm}=\frac{1}{M}[\alpha\frac{x^2}{2}+\beta\frac{x^3}{3}]\limits^L_0$

$r_{cm}=\frac{1}{M}(\frac{\alpha L^2}{2}+\frac{\beta L^3}{3})$

$r_{cm}=\frac{1}{M}(\frac{3\alpha L^2+2\beta L^3}{6})\\\;\\\text{Putting vaule of M,}\\\;\\r_{cm}=\frac{2}{2\alpha L+\beta L^2}(\frac{3\alpha L^2+2\beta L^3}{6})\\\;\\r_{cm}=\frac{(3\alpha+2\beta L)L}{3(2\alpha+\beta L)}$

Hence Option 2) is correct

Answer 2

Center of mass is at $\frac{(3 \alpha + 2 \beta L) L}{3(2 \alpha + \beta L)}$

Given Linear mass density of the rod varies with x as,

$\lambda = \alpha + \beta x$

If the rod was uniform, then the Centre of mass would be at L/2.

To make the rod uniform,

$\beta = 0$

Now, Only Option 2 Satisfies the Centre of mass of Rod of uniform mass is L/2.

Option 2 :

$\frac{3 \alpha + 2 \beta L}{3(2 \alpha + \beta L)} \times L \\ \\ = \frac{3 \alpha + 0}{2(3 \alpha + 0)} \times L\\ \\ = \frac{3 \alpha }{6 \alpha } \times L\\ \\ = \frac{L}{2}$

Therefore, The Correct Option is Option B.