A metre rod is half made of copper and half made of iron. If the mass of the copper part is 900 g and the mass of iron is 800 g, then calculate the position at which the rod can remain in equilibrium.

Given,
Length of the rod = 1 meter or 100 cm
Mass of copper part = m₁ = 900 g
Mass of iron part = m₂ = 800 g

The copper part is 50 cm long and its center of gravity will be at 25 cm.

The iron part is also 50 cm long and its center of gravity will be at 75 cm.
Then distance between their CG is = (75 - 25) cm = 50 cm.

Let x be the distance of CG of copper part from pivot, then (50 - x) will be the distance of CG of iron part from pivot.

900 x = 800 (50-x)
900 x = 40000−800 x
1700 x = 40000

$\text x=\dfrac{40000}{1700}$

x = 23.53 cm

So, the metre rod will be balanced at a distance (25 + 23.53) cm = 48.53 cm from the end of copper.