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: Understanding the Concept:
In a potentiometer the wire carries a steady current from the driving cell. The potential drop is uniform along the wire, so the balancing length gives the unknown emf.

Step 2: Key Formula or Approach:
Total resistance in the circuit \(= 5 + 992 + 3 = 1000\ \Omega\) (wire, series resistor and internal resistance). Current \(I = \frac{4}{1000} = 4\) mA.

Step 3: Detailed Explanation:
Potential drop across the whole wire: \(V = I \times 5 = 4 \times 10^{-3} \times 5 = 20\) mV.
Potential gradient \(= \frac{20\ \text{mV}}{4\ \text{m}} = 5\) mV/m.
\[ E = 5 \times 0.75 = 3.75\ \text{mV} \]

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