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, \({\rm{A}} = \frac{1}{4}{\rm{\pi }}{{\rm{r}}^2}\)
\( \Rightarrow {\rm{P}} = {\rm{\sigma }}\frac{1}{4}{\rm{\pi }}{{\rm{r}}^2}\)
\(400 = 379 \times \left( {\frac{1}{4} \times 3.14 \times {{\rm{r}}^2}} \right)\)
\({{\rm{r}}^2} = \frac{{400}}{{379}} \times \frac{4}{{3.14}}\)
r2 = 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.