Two Brass rods of same length but with different diameters are heated by equal amounts of heat. The expansion is

If Q is the heat supplied and Δt is the rise in temperature. 

Then Q = m.s.Δt

$\Rightarrow \mathrm{\Delta t}=\frac{\mathrm{Q}}{\mathrm{m}.\mathrm{s}}$

$\therefore \mathrm{\Delta t}\propto \frac{1}{\mathrm{m}}$

Since mass of thick rod is more , rise in temperature in thick rod is less.

Also , increase in length = l $\propto$ Δt

$\therefore \mathrm{\Delta }l\propto \mathrm{\Delta t}$