In a potentiometer circuit, a cell of EMF $1.5 V$ gives the 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?

62 cm

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60 cm

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21.6 cm

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64 cm

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Solution

The correct option is B $60 cm$
From the application of potentiometer to compare two cells of emfs $E_{1}$ and $E_{2}$ by balancing lengths $l_{1}$ and $l_{2}$ .

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow l_{2} = l_{1} (\frac{E_{2}}{E_{1}})$

$= (36 cm) (\frac{2.5 V}{1.5 V}) = 60 cm$

Hence, $(B)$ is the correct answer.

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In a potentiometer circuit, a cell of EMF 1.5 V gives the 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?

The correct option is B 60 cmFrom the application of potentiometer to compare two cells of emfs E1 and E2 by balancing lengths l1 and l2. E1E2=l1l2 ⇒l2=l1(E2E1) =(36 cm)(2.5V1.5V)=60 cm Hence, (B) is the correct answer.

In a potentiometer circuit, a cell of EMF $1.5 V$ gives the 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?