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Question

AB ia a potentiate wire of length 100 cm and is resistance is 10Ω. It is connected in series with a resistance R=40Ω and a battery of emf 2 V and negligible internal resistance. If a source of unknown potentiate wire, the value of E is:
1063466_514130298f7b4e918021b3f553e5a20d.png

A
0.8 V
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B
1.6 V
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C
0.06 V
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D
0.16 V
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Solution

The correct option is D 0.16 V

Given,

Cell voltage, V=2V

I=VR+r=240+10=0.04A

Potential in wire= Ir=0.04×10=0.4V

Potential gradient, 0.4100=0.004Vcm1

Emf of cell is equal to length 40 cm =0.004×40=0.16V

Hence, value of E is 0.16V


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