Potentiometer wire of length 1 m is connected in series with 450 Ω 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) → \( 50\,\Omega \)
The potential gradient of a potentiometer is the potential drop per unit length of the wire. The wire and the 450 Ω resistance form a potential divider across the 2 V battery, so the same current flows through both.
Potential drop across the whole wire
\( V_w = 2\,\mathrm{mV\,cm^{-1}}\times100\,\mathrm{cm} = 200\,\mathrm{mV} = 0.2\,\mathrm{V} \)
Resistance of the wire (let it be \( R_w \))
\( \displaystyle I = \frac{2\,\mathrm{V}}{450\,\Omega + R_w},\qquad V_w = IR_w \)
\( \displaystyle 0.2 = \frac{2R_w}{450 + R_w} \)
\( 0.2\times450 + 0.2R_w = 2R_w \)
\( 90 = 1.8R_w \)
\( R_w = 50\,\Omega \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Potentiometer – “… Since the wire is uniform, the potential difference between A and any point at a distance l from A is
\( \displaystyle \varepsilon(l) = \phi\, l \qquad (3.89) \)
where \( \phi \) is the potential drop per unit length.”