A conducting wire has length `L_(1)` and diameter `d_(1)`. After stretching the same wire length becomes `L_(2)` and diameter `d_(2)`, The ratio of resistances before and after stretching is

Options

A. `d_(2)^(4):d_(1)^(4)`
B. `d_(1)^(4):d_(2)^(4)`
C. `d_(2)^(2):d_(1)^(2)`
D. `d_(1)^(2):d_(2)^(2)`

Answer

A

Explanation

Resistance of wire is given by `R=rho(l)/(A)`
where, l is the length of conductor and A is the cross-sectional area.
According to the question,
`R_(1)=rho(l_(1))/(A_(1))` . . .(i)
In the second condition,
`R_(2)=rho(l_(2))/(A_(2))` . .. (ii)
On dividing Eq. (i) by Eq. (ii), we get
`(R_(1))/(R_(2))=(l_(1))/(A_(1))xx(A_(2))/(l_(2))`
`(R_(1))/(R_(2))=(A_(2))/(A_(1))xx(l_(1))/(l_(2))`
`(R_(1))/(R_(2))=(pi((d_(2))/(2))^(2))/(pi((d_(1))/(2))^(2))xx(l_(1))/(l_(2))`
`(R_(1))/(R_(2))=((d_(2))^(2))/((d_(1))^(2))xx(l_(1))/(l_(2))` . . .(iii)
Volume of wire remains constant. Hence, `V_(1)=V_(2)`
`rArrpi((d_(1))/(2))^(2)l_(1)=pi((d_(2))/(2))^(2)l_(2)`
`rArr(pi d_(1)^(2))/(4)l_(1)=(pi d_(2)^(2))/(4)l_(2)`
`(l_(1))/(l_(2))=((d_(2))^(2))/((d_(1))^(2))` . . . (iv)
Substituting the value of Eq. (iv) in Eq (iii), we get
`(R_(1))/(R_(2))=((d_(2))^(2))/((d_(1))^(2))xx((d_(2))^(2))/((d_(1))^(2))`
`(R_(1))/(R_(2))=((d_(2))^(4))/((d_(1))^(4))`