Question

The breaking stress for the wire of the radius $r$ of the given material is $FN/{m^2}$. The breaking stress for the wire, of the same material of radius $2r$ is,
$(A). \dfrac{F}{4}$
$(B). \dfrac{F}{2}$
$(C). F$
$(D). 2F$

Answer 1

Hint: The breaking stress will always remain the same. We can change the cross-section of the wire to change the force that is exerted on the wire. The breaking stress is not dependent on the physical properties and hence it remains the same.

Complete step by step answer:
One of the intensive properties of the materials is the ultimate tensile strength or breaking stress. These tensile values do not depend on the size of the specimen that is being tested.
The tensile strength can be defined as the stress, which can be measured as the force per unit area.
Some materials break very easily and this force is known as the breaking force. That is also again called the breaking stress. It can also be defined as the maximum stress that a material can withstand before the breakage of the material.
As mentioned in the hint, the breaking stress will always remain the same. We can change the cross-section of the wire to change the force that is exerted on the wire. The breaking stress is not dependent on the physical properties and hence it remains the same.

Hence, the correct answer is option (C). That is the breaking stress for the wire, of the same material of radius $2r$ will be $F$ same as the breaking stress for the wire when the radius $r$ of the given material is $FN/{m^2}$.

Note: The tensile strengths are rarely used in the members of the ductile family. But this property is important to the members of the brittle family. The highest point of the strain-stress curve is the ultimate tensile strength and it has the unit of stress.