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Question

A steel rod is having a radius of 10 mm and length 1.0 m. A force of 100 kN is used to stretch it along its length. If the Young's modulus of steel is 2×1011 Nm2, then the strain in the rod will be

A
0.002
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B
0.003
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C
0.008
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D
0.0016
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Solution

The correct option is A 0.0016
Given :
Applied force (F)=100 kN or 105 N
Young's modulus of elasticity (Y)=2×1011 Nm2
Length of the steel rod (L)=1.0 m
Area of the steel rod (A)=π×r2=π×(10×103)2=π× 104 m2

We know that, Y=Longitudinal StressLongitudinal Strain
Longitudinal Strain(ΔLL)=FA×Y
From the given data,
ΔLL=105(π×104)×(2×1011)=5×103π
ΔLL=0.0016
Thus, option (d) is the correct answer.

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