Question

The elastic limit of brass is 379 M Pa. What should be the minimum diameter of a brass rod if it is to support a 400N load without exceeding its elastic limit?
$A.{\text{ }}1.16{\text{ }}mm$
$B.{\text{ }}0.90{\text{ }}mm$
$C.{\text{ }}1.36{\text{ }}mm$
$D.{\text{ }}1.00{\text{ }}mm$

Answer 1

Hint: To approach the solution of this question knowledge of elastic limit is must also remember to use the formula $S = \dfrac{F}{A}$where S is the stress, F is the force acting on the body and A is the area of the body.

Formula used: $S = \dfrac{F}{A}$here S is stress, F is the force experienced by the material and A is the area of the material
Area of brass: $\pi {r^2}$ here r is the radius of the brass

Complete step-by-step solution:
Before we move forward towards the solution to the problem let’s discuss about the concept of stress which says that when a material is experiencing any force per unit area of material it named as stress, a body which can experience the force before breaking into small pieces this maximum force which can be hold by the material is called breaking stress. By the definition of stress, the formula of stress of any material is given as $S = \dfrac{F}{A}$ here F is the maximum force can be experienced by the material used and A is the area of the material used
According to the given information the elastic limit of material brass is 379 M Pa = $379 \times {10^{6}}$ and the maximum force the brass can hold without exceeding its elastic limit is 400N
We know that area of brass = $\pi {r^2}$or $\pi \dfrac{{{d^2}}}{4}$ here r is the radius of brass and d is the diameter of material brass
Since we know that the calculation of stress is done by formula $S = \dfrac{F}{A}$
Substituting the given values in the above equation, we get
$379 \times {10^{6}} = \dfrac{{400}}{{\pi \dfrac{{{d^2}}}{4}}}$
$ \Rightarrow $${d^2} = \dfrac{{400 \times 4 \times 7}}{{379 \times 22}}$
$ \Rightarrow $${d^2} = 1.34 \times {10^{ - 6}}$
$ \Rightarrow $$d = \sqrt {1.34 \times {{10}^{ - 6}}} $
$ \Rightarrow $d = $1.157 \times {10^{ - 3}}$m $ \approx $ $1.16 \times {10^{ - 3}}$m
So, the diameter of brass rod which can support 400N force without exceeding its elastic limit is $1.16 \times {10^{ - 3}}$ m or 1.16 mm
Hence option A is the correct option.

Note: The term elastic limit that is mentioned in the above question explains the maximum stress limit of any material which occurs when a solid material experiences any force acting per unit area or stress, when we remove the stress or force acting per unit area from any material it regains its particles regains the original position and the property of any material to maintain its configurations of particles under the effect of force are called the elastic property of any material, the elastic property of any material depends upon the magnitude of force under which it is holding its original configuration of particles for example steel is the most elastic material.