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Q.

Two cells of emfs E1 and E2 are connected as shown in figure (E1>E2), when a potentiometer is connected between A and B, balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio E1/E2 is

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a

4/3

b

2/3

c

3/2

d

3/4

answer is C.

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Detailed Solution

E1E1−E2=300100⇒2E1=3E2⇒E1E2=32
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