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

\(\begin{array}{l}
{x_{cm}} = \frac{{\int\limits_0^L {(ax + \frac{{b{x^2}}}{L})dx} }}{{\int\limits_0^L {\left( {a + \frac{{bx}}{L}} \right)dx} }}\\
\,\,\,\,\,\,\,\,\, = \frac{{\frac{{a{L^2}}}{2} + \frac{{b{L^2}}}{3}}}{{aL + \frac{{bL}}{2}}} = \frac{{L\left( {\frac{a}{2} + \frac{b}{3}} \right)}}{{a + \frac{b}{2}}}\\
Now\,\frac{7}{{12}} = \frac{{\frac{a}{2} + \frac{b}{3}}}{{a + \frac{b}{2}}}
\end{array}\)

On solving we get, \(b = 2a\)