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A cell of internal resistance 1.5Ω and of e.m.f. 1.5 volt balances 500 cm on a potentiometer wire. If a wire of 1.5 Ω is connected between the balance point and the cell, then the balance point will shift
- a)To zero
- b)By 500 cm
- c)By 750 cm
- d)None of the above
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A cell of internal resistance 1.5 and of e.m.f. 1.5 volt balances 500 ...
Given:
Internal resistance of the cell, r = 1.5 Ω
EMF of the cell, E = 1.5 V
Length of potentiometer wire, L = 500 cm = 5 m
To find:
Change in balance point when a wire of resistance 1.5 Ω is connected between the balance point and the cell.
Solution:
1. Before connecting the wire:
Let the balancing length be l1.
Then, by the principle of potentiometer:
E/l1 = V/L
where V is the potential difference across the wire AB.
Also, potential drop across the cell, V1 = E - Ir
2. After connecting the wire:
Let the new balancing length be l2.
Now, the potential difference across the wire AB is:
V = (l2 - l1) × V/L
The potential drop across the cell remains the same, i.e., V1 = E - Ir
3. Equating the potential differences:
V1 = V
E - Ir = (l2 - l1) × V/L
1.5 - (1.5 × 1.5)/(1.5 + 1.5) = (l2 - l1) × V/L
Simplifying, we get:
l2 - l1 = 2.25 m
Therefore, the balance point shifts by 2.25 m when the wire of 1.5 Ω resistance is connected between the balance point and the cell.
Answer:
The correct option is (d) None of the above.
Internal resistance of the cell, r = 1.5 Ω
EMF of the cell, E = 1.5 V
Length of potentiometer wire, L = 500 cm = 5 m
To find:
Change in balance point when a wire of resistance 1.5 Ω is connected between the balance point and the cell.
Solution:
1. Before connecting the wire:
Let the balancing length be l1.
Then, by the principle of potentiometer:
E/l1 = V/L
where V is the potential difference across the wire AB.
Also, potential drop across the cell, V1 = E - Ir
2. After connecting the wire:
Let the new balancing length be l2.
Now, the potential difference across the wire AB is:
V = (l2 - l1) × V/L
The potential drop across the cell remains the same, i.e., V1 = E - Ir
3. Equating the potential differences:
V1 = V
E - Ir = (l2 - l1) × V/L
1.5 - (1.5 × 1.5)/(1.5 + 1.5) = (l2 - l1) × V/L
Simplifying, we get:
l2 - l1 = 2.25 m
Therefore, the balance point shifts by 2.25 m when the wire of 1.5 Ω resistance is connected between the balance point and the cell.
Answer:
The correct option is (d) None of the above.
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Community Answer
A cell of internal resistance 1.5 and of e.m.f. 1.5 volt balances 500 ...
There will be no shift hence answer should be BY ZERO
first option is TO ZERO which is wrong
hence option d is correct.
first option is TO ZERO which is wrong
hence option d is correct.
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A cell of internal resistance 1.5 and of e.m.f. 1.5 volt balances 500 cm on a potentiometer wire. If a wire of 1.5 is connected between the balance point and the cell, then the balance point will shifta)To zerob)By 500 cmc)By 750 cmd)None of the aboveCorrect answer is option 'D'. Can you explain this answer?
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