In the above question the ratio of the elongation produced in the copper wire and steel wire are

Options

A. `7 : 2`
B. `5 : 2`
C. `2 : 5`
D. `5 : 7`

Answer

B

Explanation

For copper wire, `l_(1)=2.2m` ,
`r_(1)=1.5 mm=1.5xx10^(-3)m`,
`Y_(1)=1.1xx10^(11) Nm^(2) , Delta l_(1)=0.5xx10^(-3)m`
For steel wire, `l_(2)=1.6` ,
`r_(2)=1.5 mm=1.5xx10^(-3) m` ,
`Y_(2)=2.0xx10^(11) Nm^(-2) , Delta l_(2)= ?`
Let F be the streching force in both the wires, then for copper wire, `Y_(1)=(F)/(pip r_(1)^(2))xx(l_(1))/(Delta l_(1))`
or `F=(Y pi r_(1)^(2)xxDelta l_(1))/(l_(1))`
`=(1.1xx10^(11))/(2.2)xx(2)/(7)xx(1.5xx10^(-3))^(2)xx0.5xx10^(-3) =1.8xx10^(2) N`
Since both the wires have same tension and same area of cross-section, hence have same tensile stress S. Thus S=Y x longitudinal strain
`:. (F)/(A) =Y_(1)xx(Delta l_(1))/(l_(1))=Y_(2)xx(Delta l_(2))/(l_(2))`
or `Delta l_(2)=(Y_(1))/(Y_(2))xx(l_(2))/(l_(1))xxDelta l_(1)`
`=(1.1xx10^(11))/(2.0xx10^(11))xx(1.6)/(2.2)xx0.5xx10^(-3)`
`=0.2xx10^(-3)m=0.2 mm`
`:. (Delta l_(1))/(Delta l_(2))=(0.5)/(0.2) =(5)/(2)`