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The length of a potentiometer wire is ℓ. A cell of emf E is balanced at a length ℓ/3 from the positive end of the wire. If the length of the wire is increased by ℓ/2. At what distance will be the same cell give a balance point.
- a)2ℓ/3
- b)ℓ/2
- c)ℓ/6
- d)4ℓ/3
Correct answer is option 'B'. Can you explain this answer?
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The length of a potentiometer wire is . A cell of emf E is balanced at...
Potential gradient in the first case 
Potential gradient in second case
From equations (i) and (ii),
Potential gradient in second case
From equations (i) and (ii),
Most Upvoted Answer
The length of a potentiometer wire is . A cell of emf E is balanced at...
Given:
- Length of the potentiometer wire = L
- Balance point for the cell of emf E = L/3 from the positive end of the wire
To find:
- The balance point when the length of the wire is increased by L/2
Solution:
1. Let's assume the balance point for the increased length of the wire is x from the positive end.
2. The potential difference across the cell can be given as:
V1 = E * (L/3)
(As the balance is achieved at L/3 from the positive end)
3. Now, according to the principle of a potentiometer, the potential difference across any length of the wire is directly proportional to the length of the wire.
So, we can write:
V1/V2 = (L/3) / (L + L/2)
V1/V2 = (L/3) / (3L/2)
V1/V2 = 2/9
(Using the fact that V1 = E * (L/3) and V2 = E * x)
4. Equating the above equation, we get:
2/9 = (L/3) / (L + L/2)
2/9 = (L/3) / (3L/2)
2/9 = 2/9
5. From the above equation, we can conclude that the value of x will be L/2.
(Therefore, the balance point for the increased length of the wire will be L/2 from the positive end)
Therefore, the correct answer is option B) /2.
- Length of the potentiometer wire = L
- Balance point for the cell of emf E = L/3 from the positive end of the wire
To find:
- The balance point when the length of the wire is increased by L/2
Solution:
1. Let's assume the balance point for the increased length of the wire is x from the positive end.
2. The potential difference across the cell can be given as:
V1 = E * (L/3)
(As the balance is achieved at L/3 from the positive end)
3. Now, according to the principle of a potentiometer, the potential difference across any length of the wire is directly proportional to the length of the wire.
So, we can write:
V1/V2 = (L/3) / (L + L/2)
V1/V2 = (L/3) / (3L/2)
V1/V2 = 2/9
(Using the fact that V1 = E * (L/3) and V2 = E * x)
4. Equating the above equation, we get:
2/9 = (L/3) / (L + L/2)
2/9 = (L/3) / (3L/2)
2/9 = 2/9
5. From the above equation, we can conclude that the value of x will be L/2.
(Therefore, the balance point for the increased length of the wire will be L/2 from the positive end)
Therefore, the correct answer is option B) /2.
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The length of a potentiometer wire is . A cell of emf E is balanced at a length /3 from the positive end of the wire. If the length of the wire is increased by /2. At what distance will be the same cell give a balance point.a)2/3b)/2c)/6d)4/3Correct answer is option 'B'. Can you explain this answer?
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Solutions for The length of a potentiometer wire is . A cell of emf E is balanced at a length /3 from the positive end of the wire. If the length of the wire is increased by /2. At what distance will be the same cell give a balance point.a)2/3b)/2c)/6d)4/3Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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