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) = (rhol_(1))/A, R_(2) = (rho l_(2))/A`
`R_(eq) = (R_(1) R_(2))/(R_(1) + R_(2))`
`rArr 8/3 =((rho l_(1))/A.(rho l_(2))/A)/((rho l_(1))/A + (rho l_(2))/A) rArr 8/3 = (rho l_(1)l_(2))/(A(l_(1) +l_(2)))`… (i)
`R_(1) + R_(2) = 12`
similarly,
`rArr 12 = rho/A (l_(1) + l_(2))`.... (ii)
from (i) and (ii)
`(8/3)/(12) = (l_(1)l_(2))/(l_(1) + l_(2))^(2)`
`rArr 2/9 =(l_(1)l_(2))/(l_(1)^(2) + l_(2)^(2) + 2l_(1) l_(2))`
`l_(1)/l_(2) = (+5 +- sqrt(25-16))/4 = (5 +- 3)/4 = 2` or `1/2`