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

Options

A. `1.96xx10^(11)N//m^(2)`
B. `1.096xx10^(11)N//m^(2)`
C. `0.196xx10^(11)N//m^(2)`
D. `10.96xx10^(11)N//m^(2)`

Answer

A

Explanation

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)`