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 A uniform thin rod AB of length L has linear mass density \mu (x)=a+\frac{bx}{L}      , where x is measured from A. If the CM of the   rod lies at adistance of   \left ( \frac{7}{12} L\right )from A, then a and b are related as :

 

  • Option 1)

    a=b

  • Option 2)

    a=2b

  • Option 3)

    2a=b

  • Option 4)

    3a=2b

 

Answers (1)

best_answer

As we learnt

Centre of Mass of a continuous Distribution -

x_{cm}=\frac {\int xdm}{\int dm}, \; \;y_{cm}=\frac{\int ydm}{\int dm}, \;z_{cm}=\frac{\int zdm}{\int dm}

- wherein

dm is mass of small element. x, y, z are coordinates of dm part.

 

 As we know that 

\Rightarrow\ \;x_{C}=\frac{\int_{0}^{L}(\frac{a+bx}{L})xdx}{\int_{0}^{L}(\frac{a+bx}{L})dx}

x_{C}=\frac{\int_{0}^{L}(\frac{ax+bx^{2}}{L})dx}{\int_{0}^{L}(\frac{a+bx}{L})dx}=\frac{\frac{al^{2}}{2}+\frac{bL^{2}}{3}}{aL+\frac{bL}{2}}=\frac{L^{2}(\frac{a}{2}+\frac{b}{3})}{L(a+\frac{b}{2})}

Given x_{C}=\frac{7L}{12}

    \frac{7L}{12}=\frac{(\frac{a}{2}+\frac{b}{3})L}{(a+\frac{b}{2})}  After solving we get b = 2a.


Option 1)

a=b

This is an incorrect option.

Option 2)

a=2b

This is an incorrect option.

Option 3)

2a=b

This is the correct option.

Option 4)

3a=2b

This is an incorrect option.

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