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In an experiment to measure the internal resistance of a cell, by a potentiometer, 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 3m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is
  • a)
    1 Ω
  • b)
    1.5 Ω
  • c)
    10 Ω
  • d)
    15 Ω
Correct answer is option 'C'. Can you explain this answer?
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In an experiment to measure the internal resistance of a cell, by a po...
Ohm resistor. If the balancing length becomes 1.8m, when a resistance of 10 ohm is used in shunt, calculate the internal resistance of the cell.

Let R be the internal resistance of the cell.
When a 5 ohm resistor is used in shunt, the balancing length is 2m.
According to the principle of potentiometer, the emf of the cell is given by E = R * balancing length / total length.
Therefore, E = R * 2 / (2 + 5).
Simplifying, we get E = 2R / 7.

When a 10 ohm resistor is used in shunt, the balancing length is 1.8m.
Using the same principle, we have E = R * 1.8 / (1.8 + 10).
Simplifying, we get E = 1.8R / 11.8.

Since the emf of the cell remains constant, we can equate the two expressions for E:
2R / 7 = 1.8R / 11.8.

Cross multiplying, we get 11.8 * 2R = 7 * 1.8R.
Simplifying, we get 23.6R = 12.6R.
Subtracting 12.6R from both sides, we get 11R = 0.
Dividing by 11, we get R = 0.

Therefore, the internal resistance of the cell is 0 ohms.
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In an experiment to measure the internal resistance of a cell, by a potentiometer, 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 3m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell isa)1 Ωb)1.5 Ωc)10 Ωd)15 ΩCorrect answer is option 'C'. Can you explain this answer?
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In an experiment to measure the internal resistance of a cell, by a potentiometer, 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 3m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell isa)1 Ωb)1.5 Ωc)10 Ωd)15 ΩCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In an experiment to measure the internal resistance of a cell, by a potentiometer, 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 3m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell isa)1 Ωb)1.5 Ωc)10 Ωd)15 ΩCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment to measure the internal resistance of a cell, by a potentiometer, 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 3m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell isa)1 Ωb)1.5 Ωc)10 Ωd)15 ΩCorrect answer is option 'C'. Can you explain this answer?.
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