In a potentiometer experiment, if the balance point is found at 4 m with a 2 V b

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Question: In a potentiometer experiment, if the balance point is found at 4 m with a 2 V battery, what is the EMF of the cell being measured?

Options:

Correct Answer: 4 V

Solution:

The EMF of the cell is equal to the potential gradient multiplied by the length to the balance point. Here, it is 2 V.

In a potentiometer experiment, if the balance point is found at 4 m with a 2 V battery, what is the EMF of the cell being measured?

In a potentiometer experiment, if the balance point is found at 4 m with a 2 V battery, what is the EMF of the cell being measured?

Correct Answer: 2 V

Step 1: Understand that a potentiometer is used to measure the EMF (Electromotive Force) of a cell.
Step 2: Identify that the balance point is where the potential difference across the potentiometer wire equals the EMF of the cell being measured.
Step 3: Note that the balance point is found at 4 meters with a 2 V battery.
Step 4: Recognize that the potential gradient (voltage per meter) can be calculated using the 2 V battery and the length of the wire (4 m).
Step 5: Calculate the potential gradient: 2 V / 4 m = 0.5 V/m.
Step 6: Use the potential gradient to find the EMF of the cell: EMF = potential gradient × length to balance point.
Step 7: Since the length to the balance point is 4 m, the EMF = 0.5 V/m × 4 m = 2 V.

Potentiometer Principle – The principle that the EMF of a cell can be determined by comparing it to a known voltage using a potentiometer, where the balance point indicates the potential difference.
Potential Gradient – The rate of change of potential difference per unit length along the wire of the potentiometer.
EMF Calculation – The calculation of EMF using the formula: EMF = Potential Gradient × Length to Balance Point.