Question

A potentiometer wire of length \(4 \text{m}\) and resistance \(5 \Omega\) is connected in series with a resistance of \(992 \Omega\) and a cell of e.m.f. \(4 \text{V}\) with internal resistance \(3 \Omega\). The length of \(0.75 \text{m}\) on potentiometer wire balances the e.m.f. of

• A. \(2.50 \text{mV}\)
• B. \(3 \text{mV}\)
• C. \(3.75 \text{mV}\)
• D. \(4 \text{mV}\)

Hint:

Find the current in the wire, then the potential gradient.

Answer 1

The Correct Option is C

Step 1: Use a fraction of the wire:
The balancing length is $\frac{0.75}{4}$ of the wire.

Step 2: Take the same fraction of the wire drop:
The wire drop is $20$ mV, so $E = 20 \times \frac{0.75}{4} = 3.75$ mV.

Final Answer:
The emf is $3.75$ mV, option (C). \[ \boxed{3.75\ \text{mV}} \]