
A steel rod has a radius of 10mm and a length of 1.0m. A force stretches it along its length and produces a strain of 0.16%. Young’s modulus of the steel is $2.0 \times {10^{11}}N/{m^2}$. What is the magnitude of the force stretching the rod?
A. 100kN
B. 314kN
C. 31.4kN
D. 200kN
Answer
568.5k+ views
Hint: The given problem is from elastic properties of material. According to hooke's law, the stress is directly proportional to strain of object due to applied force. Here we use this relation to find out the magnitude of force.
Complete step by step answer:
As we know that according to hooke's law, the stress in an object is directly proportional to strain developed in the object.
Hooke's law- stress ∝ strain
After reducing the proportionality sign, the given relation can be represented as
$\sigma = \in \times Y$
Where, $\sigma = \dfrac{F}{A}$= Stress
$ \in = \dfrac{{\Delta l}}{l}$= Strain
Y= Young’s modulus of elasticity
$\Delta l$= change in length
F= Applied force
A= Cross sectional area
Given data in question as following
radius $(r)= 10×10-3m$
${\Delta}l= {\dfrac{0.16}{100}}l$
Y= $2.0 \times {10^{11}}N/{m^2}$
The equation of hooke's law can be written as
$\dfrac{F}{A} = \dfrac{{Y\Delta l}}{l}$
After putting the values in this equation, we get
$
F = 2 \times {10^{11}} \times 3.14 \times {10^{ - 4}} \times 0.16 \times {10^{ - 2}} \\
F = 100kN \\
$
Hence, the correct answer is option (A).
Stress is defined as the restoring force per unit area of the material. It is a tensor quantity. It is measured in pascal. There are several types of stress but mainly it is categorized into two forms that are normal stress and tangential stress or shearing stress.
Strain is defined as the amount of deformation experienced by the body in the direction of force applied, divided by the initial dimensions of the body. it just defines the relative change in shape.
Note: An elastic modulus is a quantity that measures an object or substance’s resistance to being deformed elastically, when a stress is applied on it. The elastic modulus of an object can also be defined as the slope of its stress strain curve in the elastic deformation region.
Complete step by step answer:
As we know that according to hooke's law, the stress in an object is directly proportional to strain developed in the object.
Hooke's law- stress ∝ strain
After reducing the proportionality sign, the given relation can be represented as
$\sigma = \in \times Y$
Where, $\sigma = \dfrac{F}{A}$= Stress
$ \in = \dfrac{{\Delta l}}{l}$= Strain
Y= Young’s modulus of elasticity
$\Delta l$= change in length
F= Applied force
A= Cross sectional area
Given data in question as following
radius $(r)= 10×10-3m$
${\Delta}l= {\dfrac{0.16}{100}}l$
Y= $2.0 \times {10^{11}}N/{m^2}$
The equation of hooke's law can be written as
$\dfrac{F}{A} = \dfrac{{Y\Delta l}}{l}$
After putting the values in this equation, we get
$
F = 2 \times {10^{11}} \times 3.14 \times {10^{ - 4}} \times 0.16 \times {10^{ - 2}} \\
F = 100kN \\
$
Hence, the correct answer is option (A).
Stress is defined as the restoring force per unit area of the material. It is a tensor quantity. It is measured in pascal. There are several types of stress but mainly it is categorized into two forms that are normal stress and tangential stress or shearing stress.
Strain is defined as the amount of deformation experienced by the body in the direction of force applied, divided by the initial dimensions of the body. it just defines the relative change in shape.
Note: An elastic modulus is a quantity that measures an object or substance’s resistance to being deformed elastically, when a stress is applied on it. The elastic modulus of an object can also be defined as the slope of its stress strain curve in the elastic deformation region.
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