A ring is made of a wire having a resistance `R_(0) = 12 Omega`. Find the point `A` and `B`, as shown in the figure, at which a current carrying conductor should be connected so that the resistance `R` of the subcircuit between these points is equal to `(8)/(3) Omega`

Options

A. `(l_1)/(l_2) = (5)/(8)`
B. `(l_1)/(l_2) = (1)/(3)`
C. `(l_1)/(l_2) = (3)/(8)`
D. `(l_1)/(l_2) = (1)/(2)`

Answer

D

Explanation

`R_1` : resistance of length `l_1`
`R_1 + R_2 = R_0 = 12` …(i)
`(R_1 R_2)/(R_1 + R_2) = (8)/(3) rArr (R_1 R_2)/(12) = (8)/(3) rArr R_1 R_2 = 32` …(ii)
`(R_2 - R_1)^2 =(R_2 + R_1)^2 - 4R_1 R_2 = (12)^2 - 4 xx 32`
`= 144 - 128 = 16`
`R_2 - R_1 = 4` ...(iii)
`R_2 = 8 Omega, R_1 = 4 Omega`
`(R_1)/(R_2) = (l_1)/(l_2) = (4)/(8) = (1)/(2)`.