Question

A potentiometer wire is 4m long and potential difference of 2V is maintained between the ends. The e.m.f. of the cell which balances against a length of 80 cm of the potentiometer wire is

• A. \(0.25\) V
• B. \(0.30\) V
• C. \(0.40\) V
• D. \(0.80\) V

Hint:

EMF = potential gradient x balancing length.

Answer 1

The Correct Option is C

Step 1: Fraction of wire
$80$ cm is $0.8/4 = 1/5$ of the wire.

Step 2: Share of voltage
The cell balances a fraction $1/5$ of the 2 V drop.

Step 3: Result
$2/5 = 0.4$ V. Option (C).

Final Answer:
Option (C). \[ \boxed{0.4 \text{ V}} \]