Question

A (2.5 V) battery is connected to a potentiometer wire. A cell of e.m.f. (1.08 V) is balanced by the voltage drop across (2.16 m) of wire. The length of the potentiometer wire is

• A. (2.5 m)
• B. (3 m)
• C. (5 m)
• D. (6 m)

Hint:

Potential gradient is constant for a uniform wire with a steady current.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A potentiometer is used to balance a secondary cell. We need to find the total length of the wire given the driving e.m.f. and the balancing length.
Step 2: Key Formula or Approach:
The potential gradient \(k\) of the potentiometer wire is \(k = \frac{V_{wire}}{L_{total}}\).
At balance, the e.m.f. of the cell is \(E = k \times l_{\text{balance}}\).
Therefore, \(E = \frac{V \times l}{L}\).
Step 3: Detailed Explanation:
Given:
Driving e.m.f. \(V = 2.5 \text{ V}\)
Cell e.m.f. \(E = 1.08 \text{ V}\)
Balancing length \(l = 2.16 \text{ m}\)
We need to find total length \(L\).
Using the relation:
\[ 1.08 = \frac{2.5 \times 2.16}{L} \]
Rearrange for \(L\):
\[ L = \frac{2.5 \times 2.16}{1.08} \]
Notice that \(2.16 = 2 \times 1.08\).
\[ L = 2.5 \times 2 \]
\[ L = 5 \text{ m} \]
Step 4: Final Answer:
The length of the potentiometer wire is \(5 \text{ m}\).