# A potentiometer wire has length 4 m and resistance 8 ohm. 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: Option 1) 44 ohm Option 2) 48 ohm Option 3) 32  ohm Option 4) 40  ohm

As we learnt in

$V=iR= \left (\frac{e}{R+R_{n}+r} \right )R$

$x=\frac{V}{L}=\frac{e}{(R+R_{n}+r)}\frac{R}{L}$

- wherein

$r-$ internal resistance

$Potential \:gradient=\frac{dv}{dr}=1mV/cm$

So for 400 cm $\Delta V=\frac{V}{R+Rs}\times R$

$\\0.4=\frac{2}{8+R}\times 8$

$=8+R=\frac{16}{0.4}=40$

$R=32\ ohm$

Option 1)

44 ohm

Incorrect

Option 2)

48 ohm

Incorrect

Option 3)

32  ohm

Correct

Option 4)

40  ohm

Incorrect

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