A wire has a resistance R. What will be its resistace if (a) it is double on itself and (b) it is stretched so that (i) its length is doubled.

Explanation

In case of a resistance wire,
`R= rho(L)/(S) = rho(L)/(pir^2) [as S = pir^2]`
and in changing dimesions without any chage in mass volume
remsins constant `i.e., V = V' or SL = S'L'`
a. when the wire is doubled on itself (as shown in fig. S6 1(b),
its length will be halved, `i.e.,L' = (L//2)`.
`SxxL = S'xx(L//2), i.e., S' = 2S`
`:. R' = rho [(L')/(S')] = rho ((L//2))/(2S) = (1)/(4) rho(L)/(S)=(1)/(4)R`
,
b. When the length of the wire is doubled, i.e., `L' = 2L
SL = 2LS', i.e., S' = S//2`
`R' = rho[(L)/(S)] = rho ((2L)/(S//2)) = 4rho[(L)/(S)] = 4R`
ii. when the radis of the wire is halved, i.e., `r' = r//2`,
`S' = pir^2 =pi(r//2)^2 = pir^2 //4 = S//4`
And hence from `SL = S'L', i.e., SL = (S//4)L'`,
we have `L' = 4L, R' = rho(L)/(S) = rho(4L)/(S//4) = 16 rho(L)/(S) = 16R`