Question

A potentiometer wire of \(400\ \text{cm}\) length has a resistance of \(20\ \Omega\). It is connected in series with a resistance of \(15\ \Omega\) and a accumulator of EMF \(6\ \text{V}\) (negligible internal resistance). The potential gradient \(k=\frac{x}{7}\ \text{V/m}\). The value of \(x\) is

• A. \(6\)
• B. \(7\)
• C. \(4\)
• D. \(8\)

Hint:

Find the current in the series circuit, then the potential drop across the wire, and divide by the wire length in metres.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The potential gradient is the potential drop across the wire divided by its length, \(k=\dfrac{V}{L}\).

Step 2: Current
Total resistance \(=20+15=35\ \Omega\), so \(I=\dfrac{6}{35}\ \text{A}\).

Step 3: Potential drop on the wire
\(V=I\times20=\dfrac{120}{35}=\dfrac{24}{7}\ \text{V}\).

Step 4: Gradient
\(k=\dfrac{24/7}{4}=\dfrac{6}{7}\ \text{V/m}\), the length being \(4\ \text{m}\). Comparing with \(\dfrac{x}{7}\) gives \(x=6\).

Final Answer:
The value of x is 6. \[ \boxed{\text{(A) }6} \]