A wire AB (of length 1m, area of cross section `pi m^(2)`) is used in potentiometer experiment to calculate emf and internal resistance (r) of battery.
The emf and internal resistance of driving battery are 15 V and `3 Omega` respectively.The resistivity of wire AB varies as `rho = rho_(0)x` (where x is distance form A in meters and `rho_(0) = 24 pi Omega`)

The distance of null point from A obtained at `sqrt((2)/(3))`m when switch 'S' is open and at `(1)/(sqrt(2m))` m when switch is closed.
The resistance of whole wire AB is-

Options

A. `6 Omega`
B. `12 Omega`
C. `18 Omega`
D. `24 Omega`

Answer

B

Explanation

Resitance of wire AB is -
`R_(AB) = ((rho_(0))/(2)) (l)/(A) = (24pi)/(2pi) = 12Omega " " {R = underset(0)int (rhodx)/(A)}`
Current in wire AB is `I = (15)/(12 + 3) = 1A`
when switch is open, null point at `C (AC = x)`
`R_(AC ) = ((rho_(0)x)/(2)) ((x)/(A)) = (rho_(0)x^(2))/(2A) = ((24pi (2)/(3))/(2pi)) 8Omega`
EMF `E = 1 xx 8 = 8V`
when switch closed null point at `D (AD = x)`
`R_(AD) = ((rho_(0))/(2)) ((x)/(A)) = (rho_(0)x^(2))/(2A) = ((24 - (1)/(2))/(2 pi)) = thetaOmega`
`DeltaV_("battery") = 6 xx 1`
`8 - (8)/(r + 3) r = 6`
`r = 1 Omega`