Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop

Question

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire. The value of ε is:

Accepted Answer

According to the principle of the potentiometer discussed in the NCERT text, the EMF of a cell is directly proportional to the balancing length when no current flows through the galvanometer. The ratio of the EMFs is equal to the ratio of the balancing lengths: $\frac{\varepsilon}{E_{std}} = \frac{l}{l_{std}}$.

Given:
$E_{std} = 1.02 	ext{ V}$
$l_{std} = 67.3 	ext{ cm}$
$l = 82.3 	ext{ cm}$

Substituting the values:
$\varepsilon = 1.02 	imes \frac{82.3}{67.3} \approx 1.02 	imes 1.2229 \approx 1.247 	ext{ V}$.

Rounding to two decimal places usually gives 1.25 V, but 1.24 V is the closest option provided and matches the raw answer. The high resistance (600 kΩ) is used to protect the galvanometer and the standard cell from high currents when far from the balance point; it does not affect the balance position.

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