Question

A steel wire ($Y = 200GPa$) of cross-sectional $area = 1m{m^2}$ is stretched by a $force = 1000N$. The percentage increase in length of wire is
A. 1%
B. 2%
C. 0.5%
D. 1.5%

Answer 1

Young’s modulus of elasticity: Within the elastic limit the ratio of longitudinal stress to the longitudinal strain is called young`s modulus of material of the wire. As we know by the law of elasticity that stress and strain is proportional to each other and the proportionality constant is known as young modulus.

Complete step by step answer:
As we all know that :
Young’s modulus = Longitudinal stress divides by longitudinal strain
So mathematically,
$Y = \dfrac{{F/A}}{{\Delta l/l}}$
Where Y is young's modulus
F is force applied on the wire
A is cross sectional area of the wire
l is original length of the wire
$\Delta l$is the change in length by applying the force
So according to given question we have:
$F = 1000$N
$Y = 200 \times {10^9}$ pa
$A = 1m{m^2}$
And we have to find value of$\dfrac{{\Delta l}}{l} \times 100$
So apply elasticity equation:

$200 \times {10^9} = \dfrac{{1000/{{10}^{ - 6}}}}{{\Delta l/l}}$
$\dfrac{{\Delta l}}{l} = \dfrac{{1000/{{10}^{ - 6}}}}{{200 \times {{10}^{ - 9}}}}$
On further calculating the above equation
$\dfrac{{\Delta l}}{l} = \dfrac{1}{{200}}$
On multiplying 100 on both side
$\dfrac{{\Delta l}}{l} \times 100 = 0.5$
So we can say that on applying the 1000 n force change in length of the wire will be 0.5% of its length.
Option number C will be the correct answer.

Note: According to the Robert Hooke,
The extension produced in the wire is directly proportional to the load applied i.e. with in the elastic limit, the stress is directly proportional to the strain.
The constant of proportionality is called modulus of elasticity or coefficient of elasticity.