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In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, then the emf of the second cell is:

A

1.27 V

B

2.25 V

C

3.27 V

D

3.25 V

Step-by-Step Solution

The potentiometer works on the principle that the potential drop across a length of a uniform wire is directly proportional to its length (VlV \propto l). When used to compare the EMF of two cells, the ratio of their EMFs is equal to the ratio of their balancing lengths: ε2ε1=l2l1\frac{\varepsilon_2}{\varepsilon_1} = \frac{l_2}{l_1}. Substituting the given values: ε2=ε1×l2l1=1.25 V×63.0 cm35.0 cm=1.25×1.8=2.25 V\varepsilon_2 = \varepsilon_1 \times \frac{l_2}{l_1} = 1.25 \text{ V} \times \frac{63.0 \text{ cm}}{35.0 \text{ cm}} = 1.25 \times 1.8 = 2.25 \text{ V}.

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