A copper wire of cross -sectional area 0.01 `cm^2` is under a tension of 22N. Find the percentage change in the cross- sectional area.
(Young's modulus of copper `= 1.1 xx 10^(11) N//m^2` and Poisson's ratio = 0.32 )

Options

A. `12.6 xx10^(-3)`
B. `8.6 xx10^(-3)`
C. `6.4 xx10^(-3)`
D. `2.8 xx 10^(-3)`

Answer

A

Explanation

Given cross - section area of copper wire, `A = 0.01 cm ^2 = 10 ^(-6) m^2` and tension force, F = 22 N,
poisson's ratio `sigma = 0.32` , and young's modulus `Y = 11 xx 10^(11) N//m^2`
since Poisson's ratio , `sigma = ("Lateral strain")/("Longitudinal strain")`
`sigma =- ((DeltaD)/(D))/((DeltaL)/L)`.......(i)
`because ` Area , A =` piD^2`
`rArr (DeltaA ) /A =2 (DeltaD)/D` ........(ii)
From Eqs. (i) and (ii) we get
`rArr (DeltaA )/A = 2sigma (DeltaL)/L` ..........(iii)
Young's modulus, `y = (FL)/(A DeltaL)` .........(iv)
From Eqs. (iii) and (iv) we get
`rArr (DeltaA )/A = (2 sigma F) /(YA) `
Now, putting the given values,
`(DeltaA)/A = (2xx 0.32 xx22)/(1.1 xx 10^(11) xx 10^(-6)) = 12.8 xx 10^(-5)`
`rArr (DeltaA) /A % = 12.8 xx 10^(-3) = 12. xx 10^(-3)`