Question

Two cells A and B are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively 360 cm and 420 cm. If emf of A is 2.4 V, the emf of the second cell B is

• A. 2.8 V
• B. 3.2 V
• C. 3.0 V
• D. 2.6 V

Hint:

Principle of potentiometer: $E \propto L$ (emf is proportional to balancing length).
For two cells, $\frac{E_1}{E_2} = \frac{L_1}{L_2}$.
Ensure the primary circuit (driving cell and rheostat settings) remains unchanged when comparing emfs.
Units for length must be consistent for the ratio $L_B/L_A$, but conversion to meters is not necessary if both are in cm.

Answer 1

The Correct Option is A

In a potentiometer, the emf ($E$) of a cell connected in the secondary circuit is directly proportional to the balancing length ($L$) obtained on the potentiometer wire, assuming the potential gradient along the wire is constant. So, $E \propto L$, or $E = kL$, where $k$ is the potential gradient. Let $E_A$ and $E_B$ be the emfs of cells A and B, respectively. Let $L_A$ and $L_B$ be their respective balancing lengths. Then, we have the relations: $E_A = k L_A$ $E_B = k L_B$ Dividing the second equation by the first (assuming $k$ is the same for both measurements, which is true if the primary circuit is unchanged): $\frac{E_B}{E_A} = \frac{k L_B}{k L_A} = \frac{L_B}{L_A}$. So, $E_B = E_A \times \frac{L_B}{L_A}$. Given values: Balancing length for cell A, $L_A = 360 \text{ cm}$. Balancing length for cell B, $L_B = 420 \text{ cm}$. Emf of cell A, $E_A = 2.4 \text{ V}$. Substitute these values to find $E_B$: $E_B = 2.4 \text{ V} \times \frac{420 \text{ cm}}{360 \text{ cm}}$. The units 'cm' cancel out. $\frac{420}{360} = \frac{42}{36}$. Both are divisible by 6: $\frac{42 \div 6}{36 \div 6} = \frac{7}{6}$. So, $E_B = 2.4 \text{ V} \times \frac{7}{6}$. $E_B = \frac{2.4 \times 7}{6} \text{ V}$. Since $2.4 / 6 = 0.4$: $E_B = 0.4 \times 7 \text{ V} = 2.8 \text{ V}$. \[ \boxed{2.8 \text{ V}} \]