A heavy uniform rod hangs vertically from a fixed support and...

Let the length of the rod be $L$, its total weight be $W$, and its initial cross-sectional area be $A$.

The tension $T$ at a distance $x$ from the bottom of the hanging rod is equal to the weight of the rod below that cross-section, given by $T = \dfrac{W x}{L}$.

The longitudinal stress at this point is $\sigma = \dfrac{T}{A} = \dfrac{W x}{A L}$.

The stress is maximum at the top support ($x = L$) and zero at the free bottom end ($x = 0$).

Longitudinal strain is directly proportional to longitudinal stress. According to Poisson's ratio, a longitudinal extension is accompanied by a lateral contraction, which is proportional to the longitudinal strain.

Since the longitudinal strain is maximum at the top, the lateral contraction is also maximum at the top.

Therefore, the diameter of the rod is smallest at the top and progressively increases downwards.