Potentiometer wire of length 1 m is connected in series with $450 \Omega$ resistance and 2 V battery. If 2 mV/cm is the potential gradient, then resistance of the potentiometer wire is:

$50 \Omega$

The correct answer is Option (1) → $50 \Omega$

Potential gradient, $K = \frac{Voltage\,drop\,across\,the\,wire}{Length\,of\,the\,wire}$

V (Voltage drop across the wire) = $K×l$

$=0.002×100$

$=0.2V$

∴ Current (I) = $\frac{0.2}{R_P}=\frac{2}{450+R_P}$

$⇒\frac{0.2}{R_P}=\frac{2}{450+R_P}$

$⇒0.2(450+R_P)=2R_P$

$⇒R_P=50Ω$