Question

A rod of length is $3m$ 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 $dx$ at a distance $x$ from one end.

It's mass $=k\cdot x\cdot dx$
[ $k=$ proportionality constant $]$
Then centre of gravity of the $\text{rod}{x}_c$ is given by
$x_c=\frac{\underset{0}{\overset{3}{\int }}kxdx\cdot x}{\underset{0}{\overset{3}{\int }}kxdx}=\frac{\underset{0}{\overset{3}{\int }}{x}^{2}dx}{\underset{0}{\overset{3}{\int }}xdx}=\frac{{\frac{x^3}{3}|}_0^3}{{\frac{x^2}{2}|}_0^3}$
or, $x_c=\frac{27/3}{9/2}=2$.
$\therefore$ Centre of gravity of the rod will be at distance of $2m$ from one end.