Question 62 of 130

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

A$2a=b$B$a=2b$C$a=b$D$3a=2b$

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Answer: (A) $2a=b$

Consider an infinitesimal small element of length $dx$ and mars density $\lambda$

${x}_{cm}=\dfrac{\int \left(dm\right)x}{\int dm}=\dfrac{\int \left(\lambda dx\right)x}{\int \lambda dx}$

$=\dfrac{{\int }_{0}^{L}\left(a+\dfrac{b}{L}x\right)xdx}{{\int }_{0}^{L}\left(a+\dfrac{b}{L}x\right)dx}$

$=\dfrac{{\int }_{0}^{L}axdx+{\int }_{0}^{L}\dfrac{b}{L}{x}^{2}dx}{{\int }_{0}^{L}adx+{\int }_{0}^{L}\dfrac{b}{L}xdx}$

$=\dfrac{a{\left[\dfrac{{x}^{2}}{2}\right]}_{0}^{L}+\dfrac{b}{L}{\left[\dfrac{{x}^{3}}{3}\right]}_{0}^{L}}{a{\left[x\right]}_{0}^{L}+\dfrac{b}{L}{\left[\dfrac{{x}^{2}}{2}\right]}_{0}^{L}}$

$=\dfrac{\dfrac{a}{2}\left({L}^{2}-{0}^{2}\right)+\dfrac{b}{3L}\left({L}^{3}-{0}^{3}\right)}{a\left(L-0\right)+\dfrac{b}{2L}\left({L}^{2}-{0}^{2}\right)}$

$=\dfrac{\dfrac{a}{2}{L}^{2}+\dfrac{b{L}^{2}}{3}}{aL+\dfrac{bL}{2}}=\dfrac{\dfrac{a}{2}L+\dfrac{bL}{3}}{a+\dfrac{b}{2}}$

$∵{x}_{cm}=\dfrac{7L}{12}=\dfrac{\left(\dfrac{a}{2}+\dfrac{b}{3}\right)L}{a+\dfrac{b}{2}}\Rightarrow \dfrac{7}{12}=\dfrac{\dfrac{3a+2b}{6}}{\dfrac{2a+b}{2}}$

$\Rightarrow \dfrac{7}{12}=\dfrac{3a+2b}{3}\times \dfrac{1}{\left(2a+b\right)}\Rightarrow 7\left(2a+b\right)=4\left(3a+2b\right)$

$\Rightarrow 14a+7b=12a+8b\Rightarrow 2a=b$