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Q.

A resistance of 4  Ω and a wire of length 5 metres and resistance 5  Ω are joined in series and connected to a cell of e.m.f. 10 V and internal resistance 1  Ω.  A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is

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a

1.5 V

b

3.0 V

c

0.67 V

d

1.33 V

answer is B.

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Detailed Solution

E=xl =  Vl =iRL × l  ⇒  E= eR+Rh +r ×  RL  × l⇒  E =  105+4+1 ×  55×  3=3 V
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