Question

A potentiometer is used to measure the potential difference between A and B, the null point is obtained at 0.9 m. Now potential difference between A and C is measured, the null point is obtained at 0.3 m. The ratio \( \frac{E_2}{E_1} \) is \( (E_1>E_2) \)

• A. 3:1
• B. 3:2
• C. 2:3
• D. 1:3

Hint:

In a potentiometer, the null point length is directly proportional to the potential difference being measured.

Answer 1

The Correct Option is C

Step 1: Using the potentiometer principle.
The potentiometer works on the principle of comparing potential differences. The ratio of the lengths at the null points is proportional to the ratio of the potential differences.
Step 2: Applying the given data.
For the first case, the null point is at 0.9 m, and for the second case, the null point is at 0.3 m. Hence, the ratio of the potential differences is: \[ \frac{E_2}{E_1} = \frac{0.3}{0.9} = \frac{2}{3} \] Step 3: Conclusion.
The correct answer is (C), 2:3.