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Question

A potentiometer wire has length 4 m and resistance $$8\Omega$$. The resistance that must be connected in series with the wire and an accumulator of e.m.f 2V, so as to get a potential gradient 1 mV per cm on the wire is


A
44Ω
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B
48Ω
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C
32Ω
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D
40Ω
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Solution

The correct option is C $$32\Omega$$
the total resistance after adding resistance R in series with potentiometer wire 
=$$8\Omega+R\Omega$$
the current in the circuit will be 
$$i=V/R_{eq}=\frac{2V}{R+8}$$
the potential drop across potentiometer wire will be $$iR_{potentiometer}=\frac{2V}{R+8}\times 8$$
the potential gradient along length 4 m is given as 
$$=V_potetiometer=\frac{\frac{2V}{R+8}\times 8}{4}$$ 
the gradient is given as $$1mV/cm=.1 V/m$$
$$\frac{\frac{2V}{R+8}\times 8}{4}=.1$$
$$R=32\Omega$$
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Physics

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