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Question

Two batteries of emf's ε1 and ε2 are connected in a circuit as shown in figure. The galvanometer is kept undeflected always. Here AB=l. When the switch S is connected to point S1, BP=x1 and when the switch S is connected to point S2, BP=x2. Here, the ratio of emf's of the two batteries, i.e., ε1/ε2 equal/s :

290253_7dffe2145e6b4fb39ce01ff59fa6c7e6.png

A
x1/x2
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B
x2/x1
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C
(lx1)(lx2)
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D
(lx2)(lx1)
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Solution

The correct option is C (lx1)(lx2)
AP=lx
We know that EMFlength of wire across battery(i.e., AP)
EAP
For first battery
E1=(lx1)
For second battery
E2=(lx2)
E1E2=lx1lx2
Answer C
791303_290253_ans_45998ee5b0df413d98d18cd9945e4ab4.png

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