The resistance of a potentiometer wire of length 4 m is 10 Ω. A resistance box and 2 V accumulator are connected in series with it. What resistance should be introduced in the box to have a potential drop of 1μV/mm of the potentiometer wire?

4990 Ω

The correct answer is Option (4) → 4990 Ω

Length of Potentiometer wire = $L = 4000 mm$

Resistance of Potentiometer wire = $R_{pot}=10Ω$

Accumulator Voltage, $V_{acc}=2V$

Desire potential drop, $ΔV_{mm}=1×10^{-6}V/mm$

$ΔV_{mm}=I×\frac{R_{pot}}{L}$

$⇒I=\frac{ΔV_{mm}×L}{R_{pot}}$  ...(1)

and,

$I=\frac{V_{acc}}{R_{pot}+R_{box}}$  ...(2)

from (1) and (2)

$\frac{1×10^{-6}×4000}{10}=\frac{2}{10+R_{box}}$

$R_{box}=5000-10=4990Ω$