The resistance of the meter bridge AB in given figure is 4 `Omega` .With a cell of emf `epsilon=0.5 V` and rheostat resistance `R_(h) = 2 Omega ` the null point is obatined at some point J is found for `R_(h) = 6 Omega " The emf "epsilon _(2)` is

Options

A. 0.5 V
B. 0.3 V
C. 0.4 V
D. 0.6 V

Answer

B

Explanation

In circuit, when `R_(h) = 2 Omega`
`i_(1) =6/(4 + 2) = 1 A, E= (1 xx 4)/(AB) xx AJ`
`AJ =(AB)/4 xx E`
Whem `R_(h) = 6 Omega`
`i_(2) = 6/(4 + 6) =0.6 A`
`E_(2)=(0.6 xx 4)/(AB) xx AK`
`AJ =(E_(2) xx AB)/(0.6 xx 4)`…… (ii)
From (i) and (ii)
`(E_(2) xx AB)/(0.6 xx 4) = (E xx AB)/4`
`E_(2) = 0.6 xx E = 0.6 xx 0.5 = 0.3 V`