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A Steel Wire Having Cross Sectional Area 1.5 Mm2 When Stretched by a Load Produces a Lateral Strain 1.5 X 10^-5. Calculate the Mass Attached to the Wire. - Physics

A steel wire having cross sectional area 1.5 mm2 when stretched by a load produces a lateral strain 1.5 x 10-5. Calculate the mass attached to the wire.

(Ysteel = 2 x 1011 N/m2, Poisson’s ratio σ = 0.291,g = 9.8 m/s2)

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Solution

Given that A = 1.5 mm2, lateral strain = 1.5 *10−5,

Ysteel = 2 * 1011 N/m2, σ = 0.291 and g = 9.8 m/s2

Poisson's ratio, σ =`"Lateral strain"/"Longitudinal strain"`

∴Longitudinal strain = `(1.5*10^-5)/0.291=5.14*10^-5`

Longitudinal stress = Y * Longitudinal strain

                                =2*1011*5.14*10-5

∴Longitudinal stress =10.28*106

But longitudinal stress =`(mg)/A`

→10.28*106 = `(M*9.8)/(1.5*10^-6)`

∴M =`(10.28*10^6*1.5*10^-6)/9.8=15.42/9.8`

∴M = 1.58kg

 

 

 

 

 

 

Concept: Definition of Stress and Strain
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