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Question

Two cells of emfs E1 and E2 (E1>E2) are connected as shown in fig.


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 E1E2 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 3:2
For balanced potentiometer,
EABEAC=lABlAC
E1E1E2=300100=3

or E1=3E13E2
or 3E2=2E1
or E1E2=32

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