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A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at length
- a)L/2 from the positive end
- b)L/5 from the positive end
- c)L/3 from the positive end
- d)L/4 from the positive end
Correct answer is option 'A'. Can you explain this answer?
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A potentiometer wire has length L For a given cell of emf E, the bala...
For the cell of emf E balancing length
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Increased length of wire 
Since the cell is the same, therefore the new balancing length = 1/3 (new increased length)
= L/2
Therefore, for the same cell the balance point will be as length L/2 from positive end.
Most Upvoted Answer
A potentiometer wire has length L For a given cell of emf E, the bala...
Given:
- Length of the potentiometer wire: L
- Balancing length: L/3 from the positive end
- Cell emf: E
To find:
- Balance point when the length of the potentiometer wire is increased by 50%
Solution:
Step 1: Relationship between balancing length and emf:
- The balancing length is the point on the wire where the potential difference across the wire is zero.
- At this point, the potential drop across the wire is equal to the emf of the cell.
- According to the question, the balancing length is L/3 from the positive end of the wire.
- Therefore, the potential drop across this length is E.
Step 2: Relationship between potential difference and length:
- The potential difference across a wire is directly proportional to its length.
- Therefore, if the length of the wire is increased by 50%, the potential difference across the wire will also increase by the same percentage.
Step 3: Finding the new balance point:
- Let the new length of the wire be L'.
- Since the potential difference across the wire is directly proportional to its length, the potential difference across the new length L' will also be increased by 50%.
- Therefore, the potential drop across the new balancing length L'/2 from the positive end will be 1.5E (50% more than the original potential drop).
- We need to find the balance point where the potential difference across the length L'/2 is 1.5E.
Step 4: Calculation:
- According to the question, the potential drop across the length L/3 is E.
- Since the potential difference across a wire is directly proportional to its length, the potential drop across the length L'/2 will be 1.5E.
- Equating the potential drops across L/3 and L'/2, we get:
E = 1.5E
L/3 = L'/2
L' = (3/2)L
- Therefore, the new length of the potentiometer wire is (3/2)L.
- The balance point is L/2 from the positive end, which is the correct answer (option A).
Hence, the correct answer is option A, L/2 from the positive end.
- Length of the potentiometer wire: L
- Balancing length: L/3 from the positive end
- Cell emf: E
To find:
- Balance point when the length of the potentiometer wire is increased by 50%
Solution:
Step 1: Relationship between balancing length and emf:
- The balancing length is the point on the wire where the potential difference across the wire is zero.
- At this point, the potential drop across the wire is equal to the emf of the cell.
- According to the question, the balancing length is L/3 from the positive end of the wire.
- Therefore, the potential drop across this length is E.
Step 2: Relationship between potential difference and length:
- The potential difference across a wire is directly proportional to its length.
- Therefore, if the length of the wire is increased by 50%, the potential difference across the wire will also increase by the same percentage.
Step 3: Finding the new balance point:
- Let the new length of the wire be L'.
- Since the potential difference across the wire is directly proportional to its length, the potential difference across the new length L' will also be increased by 50%.
- Therefore, the potential drop across the new balancing length L'/2 from the positive end will be 1.5E (50% more than the original potential drop).
- We need to find the balance point where the potential difference across the length L'/2 is 1.5E.
Step 4: Calculation:
- According to the question, the potential drop across the length L/3 is E.
- Since the potential difference across a wire is directly proportional to its length, the potential drop across the length L'/2 will be 1.5E.
- Equating the potential drops across L/3 and L'/2, we get:
E = 1.5E
L/3 = L'/2
L' = (3/2)L
- Therefore, the new length of the potentiometer wire is (3/2)L.
- The balance point is L/2 from the positive end, which is the correct answer (option A).
Hence, the correct answer is option A, L/2 from the positive end.
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A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. Can you explain this answer?
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A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. 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 A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. Can you explain this answer?.
A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. 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 A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at lengtha)L/2 from the positive endb)L/5 from the positive endc)L/3 from the positive endd)L/4 from the positive endCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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