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(C) Let $x$ be the potential gradient of the potentiometer wire. The emf $E$ of a cell is proportional to the balancing length $l$,i.e.,$E = xl$.For c

Vedclass · Accepted Answer

$1:1.2:2$

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(C) Let $x$ be the potential gradient of the potentiometer wire. The emf $E$ of a cell is proportional to the balancing length $l$,i.e.,$E = xl$.For cells $A, B, C$ in series: $E_A + E_B + E_C = x(420) \quad (1)$For cells $A, B$ in series: $E_A + E_B = x(220) \quad (2)$For cells $B, C$ in series: $E_B + E_C = x(320) \quad (3)$Subtracting equation $(2)$ from $(1)$: $E_C = x(420 - 220) = x(200)$.Substituting $E_C$ into equation $(3)$: $E_B + x(200) = x(320) \implies E_B = x(120)$.Substituting $E_B$ into equation $(2)$: $E_A + x(120) = x(220) \implies E_A = x(100)$.Thus,the ratio $E_A : E_B : E_C = 100 : 120 : 200 = 1 : 1.2 : 2$.

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