Question

In a potentiometer circuit, a cell of EMF 1.5 V gives a balance point at 36 cm length of wire. If another cell of EMF 2.5V replaces the first cell, then at what length of the wire, the balance point occurs

• A. 62 cm
• B. 60 cm
• C. 21.6 cm
• D. 64 cm

Answer 1

The Correct Option is B

To solve the problem of determining the new balance point when a cell with a different electromotive force (EMF) replaces the initial cell in a potentiometer circuit, we need to understand the principle behind a potentiometer. 

The key principle is that the potentiometer measures the EMF of the cell in terms of voltage \((V)\) across a specific length of the wire \((L)\). This is guided by the formula:

\(V = k \cdot L\)

where:

• \(V\) is the EMF of the cell.
• \(k\) is the potential gradient (voltage per unit length).
• \(L\) is the length of wire at balance point.

Initially, a cell of EMF \(1.5 \, \text{V}\) gives a balance point at \(36 \, \text{cm}\). The equation becomes:

\(1.5 = k \cdot 36\)

Solving for \(k\), we have:

\(k = \frac{1.5}{36}\)

Now, when the cell with EMF \(2.5 \, \text{V}\) is used, the equation becomes:

\(2.5 = k \cdot L_{\text{new}}\)

Substitute \(k\) from above:

\(2.5 = \left(\frac{1.5}{36}\right) \cdot L_{\text{new}}\)

Solve for \(L_{\text{new}}\):

\(L_{\text{new}} = \frac{2.5 \times 36}{1.5}\)

\(L_{\text{new}} = 60 \, \text{cm}\)

Thus, when the EMF is 2.5V, the balance point occurs at 60 cm.

Therefore, the correct answer is 60 cm.