Question

Figure shows a potentiometer wire AB having resistance of \(5 \Omega\) and length 10 m. An e.m.f. is \(0.4\) V of battery, the balancing length AP is (internal resistance is negligible)

• A. \(8\) m
• B. \(10\) m
• C. \(6\) m
• D. \(4\) m

Hint:

Find the potential gradient of the wire from the driver circuit, then divide the cell emf by it.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The figure shows a driver battery of 5 V in series with a 45 ohm resistor, a key and the potentiometer wire AB of 5 ohm and length 10 m. A cell of emf 0.4 V is connected through a galvanometer to the sliding point P.

Step 2: Current in the wire
\[ I=\frac{5}{45+5}=0.1\ \text{A} \]

Step 3: Potential across the wire
\[ V_{AB}=I\times5=0.1\times5=0.5\ \text{V} \]

Step 4: Potential gradient
\[ k=\frac{0.5}{10}=0.05\ \text{V/m} \]

Step 5: Balancing length
At balance, the cell emf equals the potential drop along AP:
\[ 0.4=k\,l\Rightarrow l=\frac{0.4}{0.05}=8\ \text{m} \]
This is option (A).

Final Answer:
The wire carries 0.1 A, so it drops 0.5 V over 10 m, and 0.4 V is balanced at 8 m, option (A). \[ \boxed{8\ \text{m}} \]