A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. Then for the two wires to have the same resistance, the ratio `l_(B)//l_(A)` of their respective lengths must be

Options

A. 2
B. 1
C. `(1)/(2)`
D. `(1)/(4)`

Answer

A

Explanation

Given, `rho_(B)=2rho_(A),R_(A)=R_(B)=R`
Let `r_(A) = r rArr r_(B)=2r`
According to formula,
`R_(A)=rho_(A).(l_(A))/(pi r_(A)^(2))` …(i)
`R_(B)=rho_(B).(l_(B))/(pi r_(B)^(2))` …(ii)
`because " " R_(A)=R_(B) rArr rho_(A)(l_(A))/(pi r_(A)^(2))=rho_(B).(l_(B))/(pi r_(B)^(2))`
or `(l_(B))/(l_(A))=(r_(B)^(2))/(r_(A)^(2)).(rho_(A))/(rho_(B))=((2r)^(2))/(r^(2)).(rho_(A))/(rho_(B))`
`= (4r^(2))/(r^(2)).(rho_(A))/(2rho_(A))=(4)/(2)=2`