Question: 19. In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
(1) 60 cm
(2) 21.6 cm
(3) 64 cm
(4) 62 cm
Answer: Option (1)
Explanation:
In a potentiometer, the potential difference is directly proportional to the balancing length of the wire, provided the current in the wire remains constant.
This relation can be written as E \propto l.
For the first cell, the EMF is E_1 = 1.5\,\text{V}
and the balancing length is l_1 = 36\,\text{cm}.
For the second cell, the EMF is E_2 = 2.5\,\text{V} and the balancing length is l_2.
Using the proportionality, \frac{E_1}{E_2} = \frac{l_1}{l_2}.
Substituting the given values, \frac{1.5}{2.5} = \frac{36}{l_2}.
Solving, l_2 = \frac{2.5 \times 36}{1.5} = 60\,\text{cm}.
Therefore, the balance point occurs at 60\,\text{cm},
and the correct answer is Option (1).