Two wires A and B are of same material. Their lengths are in the ratio 1:2 and diameters are in the ratio 2:1 when stretched by force `F_A` and `F_B` respectively they get equal increase in their lengths. Then the ratio `(F_A)/(F_B)` should be

Options

A. `1:2`
B. `1:2`
C. `2:1`
D. `8:1`

Answer

B

Explanation

Force `F=(YA Deltal)/l`
Substituting area `A=pi((d)/(2))^(2)=(pid^(2))/(4)`
So, we have `F=(Y pid^(2)Deltal)/(4l)`
`Fprop(d^(2)Deltal)/(l)[because(Ypi)/(4)="constant"]`
Hence, using Eq. (i), we have
`A_(A)prop(d_(A)^(2)Deltal_(A))/(4l)" "...(i)`
`F_(B)prop(d_(B)^(2)Deltal_(B))/(l_(B))" "...(ii)`
From Eqs. (i) and (ii), we get
`therefore (F_(A))/(F_(B))=(d_(A)^(2)Deltal_(A))/(l_(A))xx(l_(B))/(d_(B)^(2)Deltal_(B))`
as `Deltal_(A)=Deltal_(B)=Deltal`
Therefore `(F_(A))/(F_(B))=(d_(A)^(2))/(d_(B)^(2))xx(l_(B))/(l_(A))`
Putting `(d_(A))/(d_(B))=2/1`
and`(l_(B))/(l_(A))=2/1`
we get `F_(A):F_(B)=8:1`