A steel wire of length 4.5 m and cross-sectional area `3 xx 10^(-5) m^(2)` stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of `1 xx 10^(-5) m^(2)` under a given load. The ratio of the Young's modulus of steel to that of copper is

Options

A. `1.3`
B. `1.5`
C. `1.7`
D. `1.9`

Answer

C

Explanation

For copper wire., `L_(C) = 3.2m, A_(C)= 4xx10^(-5) m^(2)`
For steel wire, `L_(S) = 4.5 m, A_(S) = 3xx10^(-5)m^(2)`
As, Young's modulus, `Y = ((F//A))/((DeltaL//L))`
As applied force F and extesnion `DeltaL` are same for steel and copper wire
`therefore (F)/(DeltaL)= (Y_(S)A_(S))/(L_(S)) = (Y_(C)A_(C))/(L_(C))`
Where the subscripts C and S refers to copper and steel respectively.
`therefore (Y_(S))/(Y_(C)) =(L_(S))/(L_(C)) xx(A_(C))/(A_(S)) =((4.5m) xx (4xx10^(-5)m^(2)))/((3.5m)xx(3xx10^(-5)m^(2)))=1.7`