Class 11 NEET Physics System of Particles and Rotational Motion

Let us consider an elementary length dx at a distance x from one end.
It's mass =k.x.dx
[k= proportionality constant ]
Then centre of gravity of the rod‌xc is given by
xc=

3 ∫ 0 kxdx.x

3 ∫ 0 kxdx

=

3 ∫ 0 x2dx

3 ∫ 0 xdx

=

x3 3 |03

x3 3 |03

or xc=

27∕3

9∕2

=2
∴ Centre of gravity of the rod will be at distance of 2m from one end.