For the same cross-sectional area and for a given load, the ratio of depressions for the beam of square cross-section and circular cross-section is

The correct option is 3 : π.

Let us understand this step-by-step by deriving the relationship between the depression of a beam and its cross-sectional geometry.

For a beam of length $L$, Young's modulus $Y$, and loaded with a force $W$, the depression $\delta$ is inversely proportional to the geometrical moment of inertia $I_g$ of its cross-section:
$\delta \propto \frac{1}{I_g}$
Since the load, length, and material of the beams are the same, the ratio of the depression of a square cross-section beam (${\delta }_{\text{sq}}$) to that of a circular cross-section beam (${\delta }_{\text{circ}}$) is given by:
$\frac{{\delta }_{\text{sq}}}{{\delta }_{\text{circ}}}=\frac{I_{g\text{,circ}}}{I_{g\text{,sq}}}$

Step 1: Expressing the moment of inertia in terms of cross-sectional area $A$

Let both beams have the same cross-sectional area $A$.

For the square cross-section:
Let the side of the square be $a$. Therefore, the area is:
$A=a^2\Rightarrow a^4=A^2$
The geometrical moment of inertia of a square cross-section is:
$I_{g\text{,sq}}=\frac{a^4}{12}=\frac{A^2}{12}$

For the circular cross-section:
Let the radius of the circular cross-section be $r$. Therefore, the area is:
$A=\pi r^2\Rightarrow r^2=\frac{A}{\pi }$
The geometrical moment of inertia of a circular cross-section is:
$I_{g\text{,circ}}=\frac{\pi r^4}{4}=\frac{\pi }{4}{\left(\frac{A}{\pi }\right)}^2=\frac{A^2}{4\pi }$

Step 2: Calculating the ratio of depressions

Now, we substitute these moments of inertia into our ratio equation:
$\frac{{\delta }_{\text{sq}}}{{\delta }_{\text{circ}}}=\frac{\left(\frac{A^2}{4\pi }\right)}{\left(\frac{A^2}{12}\right)}$
Simplifying the expression:
$\frac{{\delta }_{\text{sq}}}{{\delta }_{\text{circ}}}=\frac{12}{4\pi }=\frac{3}{\pi }$

Thus, the ratio of the depression of the square beam to that of the circular beam is 3 : π.