When a wire of length 10m is subjected t...

When a wire of length `10m` is subjected to a force of `100N` along its length, the lateral strain produced is `0.01xx10^(-3)`. The poisson's ratio was found to be `0.4`. If area of cross section of wire is `0.25 m^(2)`, its young's modulus is

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Poisson's ratio
`sigma = ("lateral of strain")/("longitudinal strian") = (-(Delta r)/(r))/((Delta l)/(l))`
`therefore = (Delta l)/(l) = (-(Delta r)/(r))/(sigma) = (10^(-5))/(0.4) = 25xx10^(-6)`
From
`Y = (Fl)/(A Delta l) = (F)/(A((Delta l)/(l))) = (100)/(25xx10^(-3)xx25xx10^(-6))`
`= 1.6xx10^(8) Nm^(2)`

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When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is 0.01 xx 10^(-3) . The Poisson's ratio was found to be 0.4. If the area of cross-section of wire is 0.025 m^2 , its Young's modulus is

`1.6xx10^8 N m^(-2)`

`2.5xx10^10 N m^(-2)`

`1.25 xx 10^11 N m^(-2)`

`16xx10^9 N m^(-2)`

When a wire is subjected to a force along its length, its length increases by 0.4% andits radius decreaes by 0.2%. Then, the poisson's ratio of the material of the wire is

0.8

0.5

0.2

0.1

An elongation of 0.2% in a wire of cross section 10^(-4)m^(2) causes tension 1000 N. Then its young's modulus is

`5xx10^(8)N//m^(2)`

`5xx10^(9)N//m^(2)`

`10^(8)N//m^(2)`

`10^(7)N//m^(2)`