AB is `1` meter long uniform wire of `10 Omega` resistance. Other data are shown in the diagram. Calculate (i) potential gradient along `AB` (ii) length `AO` when galvanometer shown deflection

Answer

(i)`0.016 V cm^(-1)`(ii)`31.2 cm`

Explanation

(i) potential gradent along `AB`
`= ((4)/(15v + 10)) (10)/(100) = 0.016 V cm^(-1)`
(ii) current through `0.6 Omega = (1.5)/(1.2 xx 0.6) = (5)/(6) A`
pot diff across `0.3 Omega =(3)/(6) xx 0.6 = 0.5 V`
Across length `AO, V = 0.3 V `is to be balance Let `l` be the length `AO` then
`0.5 = 0.016 xx l or I = (0.5)/(0.016) = 31.2 cm`