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:
Answer & explanation
Correct answer: option 1
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) = $\frac{0.2}{R_P}=\frac{2}{450+R_P}$
$⇒\frac{0.2}{R_P}=\frac{2}{450+R_P}$
$⇒0.2(450+R_P)=2R_P$
$⇒R_P=50Ω$