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The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?
1. 1.00 mm
2. 1.16 mm
3. 0.90 mm
4. 1.36 mm

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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.

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