Question

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 :

• A. a = b
• B. a = 2b
• C. 2a = b
• D. 3a = 2b

Answer 1

Answer: (C)

$x_{ CM }=\frac{1}{M} \int x d m ; M=\int d m$
$=\int_{0}^{L}\left(a+\frac{b x}{L}\right) d x$
$=\left(a L+\frac{b L}{2}\right)$
$\Rightarrow \int x d m=\int_{0}^{L}\left(a x+\frac{b x^{2}}{L}\right) d x$
$=\left(\frac{a L^{2}}{2}+\frac{b L^{2}}{3}\right)$
Given, $x_{ CM }=\frac{7 L}{12}$
Therefore,
$\frac{7 \not L}{12}=\frac{\frac{a \not L^{2}}{2}+\frac{b \not L^{2}}{3}}{a \not L+\frac{\not b L}{2}}=\frac{3 a+2 b}{6\left(a+\frac{b}{2}\right)}$
$\Rightarrow 6 a+4 b=7 a+\frac{7 b}{2}$
$\Rightarrow a=\frac{b}{2}$
$\Rightarrow b=2 a$