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Question

Two cells of e.m.f. E1 and E2(E1>E2) are connected as shown in the 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 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
The balancing length of the potentiometer connected between A and B is given by
lAB=300 cm
The balancing length of the potentiometer connected between A and C is given by
lAC=100 cm
The principal of potentiometer states that emf is directly proportional to the balancing length
El

E1=K×300.........(i)
(where K is constant of proportionality)
and E1E2=K×100......(ii)
On solving (i) and (ii) we get
E1E2=32

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