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Question

# Two cells of emf ε1 and ε2 (ε1>ε2) are connected as shown in figure. When a potentiometer is connected between A and B, the 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 ε1:ε2 is

A
3 : 1
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B
1 : 3
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C
2 : 3
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D
3 : 2
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Solution

## The correct option is D 3 : 2When potentiometer is connected between A and B, then it measures only ε1 and when connected between A and C, then it measures ε1−ε2. ∴ε1ε1−ε2=l1l2,ε1−ε2ε1=l2l1or 1−ε2ε1=100300 or ε2ε1=1−13or ε2ε1=23 or ε1ε2=32

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