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NEET PHYSICSMedium

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 \Omega 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 k\Omega 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:

A

1.33 V

B

1.50 V

C

1.24 V

D

1.07 V

Step-by-Step Solution

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: εEstd=llstd\frac{\varepsilon}{E_{std}} = \frac{l}{l_{std}}.

Given: Estd=1.02 VE_{std} = 1.02 \text{ V} lstd=67.3 cml_{std} = 67.3 \text{ cm} l=82.3 cml = 82.3 \text{ cm}

Substituting the values: ε=1.02×82.367.3≈1.02×1.2229≈1.247 V\varepsilon = 1.02 \times \frac{82.3}{67.3} \approx 1.02 \times 1.2229 \approx 1.247 \text{ 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 k\Omega ) 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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