A potentiometer wire has length $4$ m and resistance $8 Ω$ .What should be the resistance that must be connected in series with the wire and an accumulator of emf $2$ V, so as to get a potential gradient $1$ mV per cm on the wire?

32 Ω

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36 Ω

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40 Ω

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30 Ω

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Solution

The correct option is A $32 Ω$
Given, $l = 4$ m
$R =$ Potentiometer wire resistance $= 8 Ω$
Potential gradient $= \frac{d V}{d r} = 1$ mVper cm
So, for $400$ cm, $Δ V = 400 \times 1 \times 10^{- 3} = 0.4$ V
Let a resistor $R_{s} =$ connected in series, so as
$Δ V = \frac{V}{R + R_{s}} \times R$
$0.4 = \frac{2}{8 + R} \times 8$
$\Rightarrow 8 + R = \frac{16}{0.4} = 40$
$\Rightarrow R = 32 Ω$

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A potentiometer wire has length 4m and resistance 8Ω.What should be the resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1mV per cm on the wire?

The correct option is A 32ΩGiven,l=4mR=Potentiometer wire resistance =8ΩPotential gradient=dVdr=1mVper cmSo, for 400cm, ΔV=400×1×10−3=0.4VLet a resistor Rs=connected in series, so asΔV=VR+Rs×R0.4=28+R×8⇒8+R=160.4=40⇒R=32Ω

A potentiometer wire has length $4$ m and resistance $8 Ω$ .What should be the resistance that must be connected in series with the wire and an accumulator of emf $2$ V, so as to get a potential gradient $1$ mV per cm on the wire?