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Question

If a cell of internal resistance 1.5 Ω and emf of 1.5 V balances 500 cm on a potentiometer wire, now by what length would the balance point shift if a cell of internal resistance 2 Ω and emf of 2 V is connected in parallel (along the same polarity) with the prior cell?

A
71.4 cm towards left
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B
71.4 cm towards right
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C
125 cm towards right
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D
Does not shift
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Solution

The correct option is B 71.4 cm towards right
Case-1:
Case-2:

using kirchhoff's current law:
i=Erequi=E1r1+E2r2=1+1=2
total resistance= 2×1.52+1.5=67

So, the EMF of the circuit,
7E6=2E=127
Here, extra emf is adding for balancing, hence meter point will move towards right.

for the case-1
1.5500 cm
12712/7×5001.5=571.4 cm

Now, the shift of the meter Δx=571.4500=71.4 cm

so, option B is correct.

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