Potentiometer

Question

In a potentiometer experiment, the balancing length for a cell of EMF 1.5V is 60cm. If the cell is replaced by another cell of EMF E, the balancing length becomes 80cm. The value of E is:

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1. **Potentiometer Principle**: For a cell of EMF $E$, the balancing length $L$ follows $E \propto L \implies \frac{E_1}{E_2} = \frac{L_1}{L_2}$.
2. **Identify Given Values**: $E_1 = 1.5	ext{ V}$, $L_1 = 60	ext{ cm}$, and $L_2 = 80	ext{ cm}$.
3. **Calculation**: 
   $$E_2 = E_1 	imes \frac{L_2}{L_1} = 1.5 	imes \frac{80}{60} = 1.5 	imes \frac{4}{3} = 2.0	ext{ V}$$
4. **Conclusion**: The EMF of the second cell is 2.0 V.

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