Let us consider a bar of length l and density \( \rho \). This bar has a circular cross-sectional area of radius r. So the mass of this bar is \( \pi{r^2}l\rho \).
Now let us consider another bar of the same length and density but it has a square cross-sectional area with each side of length a. So the mass of this bar is \( a^2{l}\rho \).

Since the weight of these two bars is equal, then
\( \pi{r^2}l\rho=a^2{l}\rho\\or,\ \pi{r^2}=a^2\tag{1} \)

Let, \( I_1 \) and \( I_2 \) be the geometrical moment of inertia of the circular bar and the square bar respectively. So we can write,

and, \( I_2=\frac{a^4}{12}\tag{3} \)

Let, \( W_1 \) and \( W_2 \) be the loads of the circular bar and the square bar respectively. So we can write,
the depression of the circular bar is \( y_1=\frac{W_1{l^3}}{3YI_1} \)
and the depression of the square bar is \( y_2=\frac{W_2{l^3}}{3YI_2} \).
[Read In Detail] Where Y is Young’s modulus of the material of the bars.

Using equation (2) we can write, \( y_1=\frac{4W_1{l^3}}{3\pi{Y}r^4} \)
Using equation (3) we can write, \( y_2=\frac{12W_2{l^3}}{3Y{a^4}}=\frac{4W_2{l^3}}{Ya^4} \)

Since the loads produce the equal depression for the two beams,
therefore, \( \displaystyle{\frac{4W_1{l^3}}{3\pi{Y}r^4}=\frac{4W_2{l^3}}{Ya^4}} \)
or, \( \displaystyle{\frac{W_1}{W_2}=\frac{3\pi{r^4}}{a^4}} \)
Using equation (1) we can write
\( \displaystyle{\frac{W_1}{W_2}=\frac{3\pi{r^4}}{{\pi}^2{r^4}}} \)
or, \( \displaystyle{\frac{W_1}{W_2}=\frac{3}{\pi}} \)