Further for a balance of torque about \(CG\) we get,
\(\mathrm{W} / 4 \times \mathrm{L} / 2=3 \mathrm{W} / 4 \times \mathrm{x}\) therefore \(\mathrm{x}=1 / 6\)
Thus distance from the left end of the rod
\(=\mathrm{L} / 2+\mathrm{L} / 6=4 \mathrm{L} / 6=2 \mathrm{L} / 3\)