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Question

A potentiometer wire of length 10 m and resistance 10Ω per metre is connected in series with a resistance box and a 2 volt battery. If a potential difference of 100 mV is balanced across the whole length of potentiometer wire, then the resistance introduced in the resistance box will be:

A
1900Ω
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B
900Ω
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C
190Ω
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D
90Ω
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Solution

The correct option is A 1900Ω
length of potentiometer l=10m
resistance of wire R=10Ω
Emf of the cell E=2V
balancing length is l=10m (whole length or wire )
let I be the current through the potentiometer then,
I=ER+R.........................(1)
EMf of the cell balanced by whole length of the wire
V=100mV
100×103=101
by using V=IR
101=I×10I=102I=0.01A
putting the value in equation (1),we get
0.01=210+R0.1+0.01R=20.012R=1.9R=1900Ω
Hence, the option (A) is correct.

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