A uniform wire of resistance `R` is stretched uniform in a times `R` is then cut to from five as The effective resistance between `A` and `C` is

Options

A. `(n^(2)R)/(3)`
B. `(n^(2)R)/(5)`
C. `(nR)/(5)`
D. `(nR)/(3)`

Answer

b

Explanation

Here, `R = (rhol_(1))/(A_(1))` when `l_(2) = n l_(1)`
Then `A_(1) l_(1) =A_(1) l_(1) or A_(2) =(A_(1) l_(1))/(l_(2)) = (A_(1) l_(1))/(nl_(1)) = (A_(1))/(n)`
New resistance of the stretched wire
`R' = (rhol_(2))/(A_(2)) = (rhonl_(2))/((A_(1)//n)) = n^(2)R`
Resistance of each arrange in from a balance Wheastone bridge so
`R_(AB) = (2R_(1)xx2R_(1))/(2R_(1) + 2R_(1)) = R_(1) = (n^(2)R)/(5)`