Question

The potentiometer wire is 5 m long and potential difference of 4 V is maintained between the ends. The e.m.f. of the cell which balances against a length of 200 cm of the potentiometer wire is ______.

• A. 0.4 V
• B. 0.8 V
• C. 1.2 V
• D. 1.6 V

Hint:

Ensure your units match! If you use the gradient in V/m ($0.8$), you MUST convert the balancing length to meters ($200 \text{ cm} = 2 \text{ m}$) before multiplying ($0.8 \times 2 = 1.6 \text{ V}$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the total length and total voltage across a potentiometer wire. We must find the unknown EMF of a test cell that balances at a specific length.

Step 2: Detailed Explanation:
The principle of a potentiometer states that the potential difference across any segment of the uniform wire is directly proportional to its length:
$V \propto l \implies V = k \cdot l$
where $k$ is the constant potential gradient (voltage drop per unit length).
1. Calculate the Potential Gradient ($k$):
Total potential difference ($V_{\text{total}}$) = $4 \text{ V}$
Total length of wire ($L_{\text{total}}$) = $5 \text{ m} = 500 \text{ cm}$
$k = \frac{V_{\text{total}}}{L_{\text{total}}}$
$k = \frac{4 \text{ V}}{5 \text{ m}} = 0.8 \text{ V/m}$
Convert this to Volts per cm for easier calculation with the balancing length:
$k = \frac{4 \text{ V}}{500 \text{ cm}} = 0.008 \text{ V/cm}$
2. Calculate the Unknown EMF ($E$):
The cell balances at a length $l_{\text{balance}} = 200 \text{ cm}$.
Therefore, the EMF of the cell is exactly equal to the potential drop across this 200 cm length of wire:
$E = k \times l_{\text{balance}}$
$E = 0.008 \text{ V/cm} \times 200 \text{ cm}$
$E = 1.6 \text{ V}$

Step 3: Final Answer:
The e.m.f. of the cell is 1.6 V, matching option (d).