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. `0.128xx10^(-6) cm^(2) `
B. `128 xx 10^(-6) cm^(2)`
C. `12.8xx0^(-6) cm^(2)`
D. `128 xx 10^(-6) cm^(2)`

Answer

D

Explanation

Young.s modulus
`Y = (F//A)/(Deltal//l)`
`(Deltal)/(l)= (F)/(YA)`
where `(Deltal)/(l) ` = longitudinal strain Given F = 22 N , Y= `1.1 xx 10^(11) N-m^(2)` ,
A= 0.01 `cm^(2)= 10^(-6) m^(2)`
`(Deltal)/(l)= (22)/(1.1xx10^(11)xx10^(-6))= 2 xx 10^(-4)`
Now poisson ratio
`sigma = ("Lateral strain")/("Longitudinal strain")= (Deltad//d)/(Deltal//l)`
`(Deltad)/(d)= sigma . (Deltal)/(l) = 0.32xx2xx10^(-4)`
Change in diameter ,` (Deltad)/(d) = 6.4 xx10^(-5)`
Area , `" " A = pir^(2)`
Fractional change in area ,
`(DeltaA)/(A) = 2 . (Deltar)/(r)`
`(DeltaA)/(A) = 2xx6.4 xx 10^(-5)`
`Delta= (12.8 xx 10^(-5))A `
Decrease in area `DeltaA = (12.8 xx 10^(-5))xx (0.01) cm^(2)`
`Delta A = 1.28 xx 10^(-6) cm^(2)`