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Question

In fig., AB is a potentiometer wire of length 10 m and resistance 2 Ω. With key (K) open the balancing length is 5.5 m. However, on closing key (K) the balancing length reduces to 5 m. The internal resistance of the cell (E1) is –

A
0.01 Ω
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B
0.1 Ω
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C
0.2 Ω
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D
1 Ω
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Solution

The correct option is B 0.1 Ω
When K is open we are balancing the emf E1, hence E1=VL×5.5=210×5.5.......(1)
When K is closed we are balancing the potential across the discharging battery which is rr+r1E1 where r1 is the internal resistance of the battery.
Hence,
E1r+r1=210×5.....(2)
On dividing (1) and (2), we get
r1+1=1.1r1=0.1 Ω

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