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A copper wire of cross sectional area `0.01 cm^2` is under a tension of `20 N`. Find the decrease in the cross sectional area. Young modulus of copper `=1.1xx10^11Nm^-2` and Poisson ratio `0.32`.

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The correct Answer is:
A, B, C, D

`A=0.01 cm^2`
`=10^-6m^2`
`=T=20N`
`Y=1.1xx10^11N/m^2`
we know that `Y=(FL)/(A/_\\\\L)`
`=(/_\\\\L)/L=F/(AY)`
`=20/(10^-6xx1.1x10^11)x18.18xx10^-5`
`d=((/_\\\\D)/D)/((/_\\\\L)/L)=0.32`
`So, (/_\\\\D)/D=(0.32)xx(KL)/L`
`=32xx(18.18)xx10^5=5.81xx10^-5`
Again `(/_\\A)/A=(2/_\\\\r)/r`
`rarr /_\\\\A=(2/_\\\\r)/r`
`/_\\\\A=2xx(5.8xx10^-5)xx(0.01)`
`=1.16xx10^9N/m^2`
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