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In a n experiment t o measure the internal resistance of a cell by a potential, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance, the internal resistance of the cell is then
  • a)
    1.5 Ω
  • b)
    10 Ω
  • c)
    15 Ω
  • d)
    1 Ω
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
In a n experiment t o measure the internal resistance of a cell by a ...
To determine the internal resistance of a cell, we can use the concept of potential difference and the formula for total resistance in a circuit. Let's break down the given information and solve for the internal resistance.

1. Given data:
- Balance point at 2 m when shunted by a 5 Ω resistance.
- Balance point at 3 m when shunted by a 10 Ω resistance.

2. Understanding the concept:
When a cell is connected to an external resistance, a potential difference is established across the external resistance. This potential difference is equal to the electromotive force (emf) of the cell minus the potential drop across the internal resistance of the cell.

3. Formula:
The potential difference across the external resistance can be calculated using the formula:
V = E - IR
Where:
- V is the potential difference across the external resistance,
- E is the emf of the cell,
- I is the current flowing through the circuit,
- R is the external resistance.

4. Analysis:
At the balance point, the potential difference across the external resistance is zero, which means the current flowing through the circuit is zero. At the balance point, the potential drop across the internal resistance is equal to the emf of the cell.

5. Calculation:
Using the given data, we can set up two equations:
Equation 1: E - 5I = 0
Equation 2: E - 10I = 0

By solving these equations simultaneously, we can find the value of E (emf of the cell).
From Equation 1, we get: E = 5I
Substituting this into Equation 2: 5I - 10I = 0
Simplifying, we find: -5I = 0
Therefore, I = 0

6. Conclusion:
Since the current flowing through the circuit is zero at the balance point, it implies that the potential drop across the internal resistance is equal to the emf of the cell. Therefore, the internal resistance of the cell is equal to the resistance of the shunt.

In this case, the balance point is at 2 m when shunted by a 5 Ω resistance. Hence, the internal resistance of the cell is 5 Ω.

Therefore, the correct answer is option 'A' - 1.5 Ω.
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In a n experiment t o measure the internal resistance of a cell by a potential, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance, the internal resistance of the cell is thena)1.5 Ωb)10 Ωc)15 Ωd)1 ΩCorrect answer is option 'A'. Can you explain this answer?
Question Description
In a n experiment t o measure the internal resistance of a cell by a potential, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance, the internal resistance of the cell is thena)1.5 Ωb)10 Ωc)15 Ωd)1 ΩCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a n experiment t o measure the internal resistance of a cell by a potential, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance, the internal resistance of the cell is thena)1.5 Ωb)10 Ωc)15 Ωd)1 ΩCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a n experiment t o measure the internal resistance of a cell by a potential, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance, the internal resistance of the cell is thena)1.5 Ωb)10 Ωc)15 Ωd)1 ΩCorrect answer is option 'A'. Can you explain this answer?.
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