Question

A copper wire of cross-sectional area 0.01 cm2 is under a tension of 20N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 1011 N m−2 and Poisson ratio = 0.32.

Answer 1

Given:

Cross-sectional area of copper wire A = 0.01 cm2 = 10−6 m2

Applied tension T = 20 N

Young modulus of copper Y = 1.1 × 1011 N/m2

Poisson ratio σ = 0.32

We know that:

Y=FLA∆L

⇒∆LL=FAY            =2010−6×1.1×1011=18.18×10−5       

Poisson's ratio,σ=∆dd∆LL=0.32Where d is the transverse lengthSo, ∆dd=(0.32)×∆LL              =0.32×(18.18)×10−5=5.81×10−5Again, ∆AA=2∆rr=2∆dd⇒∆A=2∆ddA⇒∆A=2×(5.8×10−5)×(0

Answer 2

The decrease in the cross-sectional area $1.164\times10^{-6}\ cm^2$

Explanation:

Given that,

Cross-sectional area = 0.01 cm²

Tension = 20 N

Young modulus of copper $Y= 1.1\times10^{11}\ N$

Poisson ratio = 0.32

We need to calculate the longitudinal strain

Using formula of longitudinal strain

$strain = \dfrac{stress}{Y}$

Put the value into the formula

$strain=\dfrac{\dfrac{20}{1\times10^{-6}}}{1.1\times10^{11}}$

$strain=\dfrac{2\times10^{7}}{1.1\times10^{11}}$

$strain=1.82\times10^{-4}$

We need to calculate the lateral strain

Using formula of lateral strain

$lateral\ strain =0.32\times1.82\times10^{-4}$

$\dfrac{\Delta D}{D}=5.82\times10^{-5}$

We need to calculate the decrease in the cross-sectional area

Using formula of cross sectional area

$A=\dfrac{\pi D^2}{4}$

On differentiating

$dA=2\piD\dfrac{dD}{4}$.....(I)

Divided by area of equation (I)

$\dfrac{dA}{A}=\dfrac{2\pi D\timesdD}{4\times\pi\times\dfrac{D^2}{4}}$

$\dfrac{dA}{A}=2\dfrac{dD}{D}$

$dA=2A\dfrac{dD}{D}$

Put the value into the formula

$dA=2\times0.01\times(5.82\times10^{-5})$

$dA=1.164\times10^{-6}\ cm^2$

Hence, The decrease in the cross-sectional area $1.164\times10^{-6}\ cm^2$

Learn more :

Topic : young modulus

https://brainly.in/question/6669970