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In a potentiometer a cell of e.m.f 2V gives a balance point at 30cm. If the cell is replaced by another cell and the balance point shifts to 60cm.What is the e.m.f of the other cell?
- a)0.04 V
- b)40.0 V
- c)0.4 V
- d)4.0 V
Correct answer is 'D'. Can you explain this answer?
Most Upvoted Answer
In a potentiometer a cell of e.m.f 2V gives a balance point at 30cm. I...
Given:
- e.m.f of first cell = 2V
- Balance point with first cell = 30cm
- Balance point with second cell = 60cm
To find:
- e.m.f of the second cell
Explanation:
To understand this problem, we need to understand the working principle of a potentiometer. A potentiometer is a device used to measure potential difference (or voltage) accurately. It consists of a long wire of uniform cross-section, a jockey, and a galvanometer.
When the jockey is moved along the wire, the galvanometer shows a deflection. At a certain point, the deflection becomes zero, indicating that the potential difference across that point is equal to the potential difference across the terminals of the cell being tested.
In this problem, the balance point with the first cell is at 30cm. This means that the potential difference across 30cm of the wire is equal to the e.m.f of the first cell, which is 2V.
Now, when the cell is replaced by another cell, the balance point shifts to 60cm. This means that the potential difference across 60cm of the wire is equal to the e.m.f of the second cell.
To find the e.m.f of the second cell, we can use the concept of proportionality. The potential difference across the wire is directly proportional to the length of the wire. So, we can set up the following proportion:
Potential difference across 30cm / Length of the wire = Potential difference across 60cm / Length of the wire
Since the length of the wire is the same on both sides of the equation, we can simplify the proportion to:
Potential difference across 30cm = Potential difference across 60cm
Substituting the given values:
2V = Potential difference across 60cm
Therefore, the e.m.f of the second cell is 2V.
Answer:
The e.m.f of the other cell is 4.0V (Option D).
- e.m.f of first cell = 2V
- Balance point with first cell = 30cm
- Balance point with second cell = 60cm
To find:
- e.m.f of the second cell
Explanation:
To understand this problem, we need to understand the working principle of a potentiometer. A potentiometer is a device used to measure potential difference (or voltage) accurately. It consists of a long wire of uniform cross-section, a jockey, and a galvanometer.
When the jockey is moved along the wire, the galvanometer shows a deflection. At a certain point, the deflection becomes zero, indicating that the potential difference across that point is equal to the potential difference across the terminals of the cell being tested.
In this problem, the balance point with the first cell is at 30cm. This means that the potential difference across 30cm of the wire is equal to the e.m.f of the first cell, which is 2V.
Now, when the cell is replaced by another cell, the balance point shifts to 60cm. This means that the potential difference across 60cm of the wire is equal to the e.m.f of the second cell.
To find the e.m.f of the second cell, we can use the concept of proportionality. The potential difference across the wire is directly proportional to the length of the wire. So, we can set up the following proportion:
Potential difference across 30cm / Length of the wire = Potential difference across 60cm / Length of the wire
Since the length of the wire is the same on both sides of the equation, we can simplify the proportion to:
Potential difference across 30cm = Potential difference across 60cm
Substituting the given values:
2V = Potential difference across 60cm
Therefore, the e.m.f of the second cell is 2V.
Answer:
The e.m.f of the other cell is 4.0V (Option D).
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Community Answer
In a potentiometer a cell of e.m.f 2V gives a balance point at 30cm. I...
We can use the formula E1/E2=L1/L2
E1=2v,E2=?,l1=30,l2=60
2/E2=30/60
E2=120/30=4v
E1=2v,E2=?,l1=30,l2=60
2/E2=30/60
E2=120/30=4v
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In a potentiometer a cell of e.m.f 2V gives a balance point at 30cm. If the cell is replaced by another cell and the balance point shifts to 60cm.What is the e.m.f of the other cell?a)0.04 Vb)40.0 Vc)0.4 Vd)4.0 VCorrect answer is 'D'. Can you explain this answer?
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