AB is a potentiometer wire. If the value of R is increased, then in w…

The correct answer is Option (2) → Towards B

The rheostat R is in the primary circuit, in series with the driver cell E and the potentiometer wire AB. At balance, the emf of the cell in the galvanometer branch equals the drop across the balancing length \( l \) measured from A:

    \( \displaystyle \varepsilon = k\,l, \qquad k = \frac{V_{AB}}{L} = \frac{E}{R + R_{AB}}\cdot\frac{R_{AB}}{L} \)

Here \( R_{AB} \) and \( L \) are the resistance and length of the wire.

Increasing R reduces the current \( I \) in the wire, so the potential gradient \( k \) decreases. To balance the same emf \( \varepsilon \), a longer length is needed:

    \( \displaystyle l = \frac{\varepsilon}{k} \;\text{increases} \)

So the balance point moves away from A, towards B.

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 – “… A current I flows through the wire which can be varied by a variable resistance (rheostat, R) in the circuit. 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 φ is the potential drop per unit length.”