A rod of length  has non-uniform linear mass density given by by $\mathrm{p}\left(\mathrm{x}\right)=\mathrm{a}+\mathrm{b}{\left[\frac{\mathrm{x}}{\mathrm{L}}\right]}^2$, where a and b are constants and $0\le \mathrm{x}\le \mathrm{L}$. The value of x for the centre of mass of the rod is at :

Given,

$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{a}+\mathrm{b}{\left[\frac{\mathrm{x}}{\mathrm{L}}\right]}^2=\mathrm{\lambda }$

$\frac{\mathrm{dm}}{\mathrm{dx}}=\mathrm{\lambda }$

$\mathrm{dm}=\mathrm{\lambda dx}=\left(\mathrm{a}+\mathrm{b}{\left[\frac{\mathrm{x}}{\mathrm{L}}\right]}^2\right)\mathrm{dx}$

${\mathrm{x}}_{\mathrm{com}}=\frac{\int \mathrm{xdm}}{\int \mathrm{dm}}=\frac{\int \mathrm{x\lambda dx}}{\int \mathrm{\lambda dx}}$

${\mathrm{x}}_{\mathrm{com}}=\frac{{\int }_0^1\mathrm{x}\left(\mathrm{a}+\left({\displaystyle \frac{{\mathrm{bx}}^2}{{\mathrm{L}}^2}}\right)\right)\mathrm{dx}}{{\int }_0^1\left(\mathrm{a}+\left(\frac{{\mathrm{bx}}^2}{{\mathrm{L}}^2}\right)\right)\mathrm{dx}}=\frac{{\int }_0^1\mathrm{a}+\left(\frac{{\mathrm{bx}}^3}{{\mathrm{L}}^2}\right)\mathrm{dx}}{{\int }_0^1\mathrm{a}+\left(\frac{{\mathrm{bx}}^2}{{\mathrm{L}}^2}\right)\mathrm{dx}}$

${\mathrm{x}}_{\mathrm{com}}=\frac{{{\left[{\displaystyle \frac{{\mathrm{ax}}^2}{2}}\right]}^1}_0+{\displaystyle \frac{\mathrm{b}}{{\mathrm{L}}^2}}{{\left[\frac{{\mathrm{x}}^2}{4}\right]}^1}_0}{\mathrm{a}{{\left[\mathrm{x}\right]}^1}_0+{\displaystyle \frac{\mathrm{b}}{\mathrm{L}}}{{\left[\frac{{\mathrm{x}}^3}{3}\right]}^1}_0}=\frac{{\displaystyle \frac{{\mathrm{a}}^{{\mathrm{l}}^2}}{2}+\frac{{\mathrm{b}}^{{\mathrm{l}}^2}}{4}}}{\mathrm{al}+{\displaystyle \frac{\mathrm{bl}}{3}}}$

${\mathrm{x}}_{\mathrm{com}}=\frac{(2\mathrm{a}+\mathrm{b})\mathrm{L}}{(3\mathrm{a}+\mathrm{b})4}\times 3=\frac{3}{4}\left[\frac{(2\mathrm{a}+\mathrm{b})}{(3\mathrm{a}+3\mathrm{b})}\right]\mathrm{L}$

Hence the correct answer $\frac{3}{4}\left[\frac{(2\mathrm{a}+\mathrm{b})}{(3\mathrm{a}+3\mathrm{b})}\right]\mathrm{L}.$