Question

A rod of length is $3\, m$ and its mass acting per unit length is directly proportional to distance $x $ from one of its end then its centre of gravity from that end will be at :-

• A. 1.5 m
• B. 2 m
• C. 2.5 m
• D. 3.0 m

Answer 1

Answer: (B)

Let us consider an elementary length $d x$ at a distance $x$ from one end.

It's mass $=k \cdot x \cdot d x$
[ $k=$ proportionality constant $]$
Then centre of gravity of the $\text{rod} x_{c}$ is given by
$x_{c}=\frac{\int\limits_{0}^{3} k x d x \cdot x}{\int\limits_{0}^{3} k x d x}=\frac{\int\limits_{0}^{3} x^{2} d x}{\int\limits_{0}^{3} x d x}=\frac{\left.\frac{x^{3}}{3}\right|_{0} ^{3}}{\left.\frac{x^{2}}{2}\right|_{0} ^{3}}$
or, $x_{c}=\frac{27 / 3}{9 / 2}=2$.
$\therefore$ Centre of gravity of the rod will be at distance of $2\, m$ from one end.