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)

P perimeter

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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A uniform thin rod AB of length L has linear mass density , where x is measured from A. If the CM of the rod lies at adistance of from A, then a and b are related as : Option 1) Option 2) Option 3) Option 4)

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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)
$3a=2b$