Correct Answer - Option 2 : 1.16 mm

**Concept:**

**Stress:**

Stress is defined as the force per unit area of a material.

i.e. \({\rm{Stress}} = \frac{{{\rm{force}}}}{{{\rm{cross\;sectional\;area}}}}{\rm{\;}}\)

\(\sigma = \frac{F}{A}\)

where, σ = stress,

F = force applied, and

A = cross sectional area of the object.

Units of stress is Nm^{-2} or Pa.

**Calculation:**

Given,

Elastic limit of brass, σ = 379 MPa

Load, P = 400 N

We know that,

P = σA

Where, Area of cross section, \({\rm{A}} = \frac{1}{4}{\rm{\pi }}{{\rm{r}}^2}\)

\( \Rightarrow {\rm{P}} = {\rm{\sigma }}\frac{1}{4}{\rm{\pi }}{{\rm{r}}^2}\)

\(400 = 379 \times \left( {\frac{1}{4} \times 3.14 \times {{\rm{r}}^2}} \right)\)

\({{\rm{r}}^2} = \frac{{400}}{{379}} \times \frac{4}{{3.14}}\)

r^{2 }= 1.34

r = 1.15 mm

Therefore, the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit is 1.16 mm.