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)
Show Hint
Find the potential gradient of the wire from the driver circuit, then divide the cell emf by it.
Step 1: Approach
Use a proportion between the cell emf and the voltage across the whole wire.
Step 2: Voltage across the wire
The total resistance of the driver loop is 50 ohm, so the wire (5 ohm) takes $\dfrac5{50}=\dfrac1{10}$ of the 5 V, which is 0.5 V.
Step 3: Proportion
At the null point $\dfrac{l}{10}=\dfrac{0.4}{0.5}$, so $l=10\times0.8=8$ m.
Step 4: Answer
Option (A). The 10 m option would need an emf equal to the full 0.5 V.
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).
\[ \boxed{8\ \text{m}} \]
