1.

A uniform steel wire of density 7800 kg//m^(3) is 2.5 m long and weighs15.6xx10^(-3)kg.2.5 m long and weighs 15.6xx10^(-3) kg . If extends by 1.25 mm whenloaded by 8 kg, then the value of Young's modulus for steel will be

Answer»

`1.96xx10^(11)N//m^(2)`
`1.096xx10^(11)N//m^(2)`
`0.196xx10^(11)N//m^(2)`
`10.96xx10^(11)N//m^(2)`

SOLUTION :Volume = MASS / DENSITY
Area of cross section `=("Volume")/("Length")`
`=("Mass")/("Density" xx "Length") =(15.6 xx 10^(-3))/(7800xx2.5)`
`therefore A=8XX10^(-7)m^(2)`
`therefore Y=(FL)/(Al)=(8xx9.8xx2.5)/((8xx10^(-7))xx1.25xx10^(-3))`
`=1.96 xx 10^(11) N//m^(2)`


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