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A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of 'n' is
Correct answer is '12'. Can you explain this answer?
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Verified Answer
A battery of unknown emf connected to a potentiometer has balancing le...
Let the emf of cell is e internal resistance is 'r' and potential gradient is x.
only cell connected :
ε = 560 x ................. (i)
After connecting the resistor
................... (ii)
from (1) and (2)
56 = 50 + 5r
n = 12
only cell connected :
ε = 560 x ................. (i)
After connecting the resistor
................... (ii)from (1) and (2)
56 = 50 + 5r
n = 12
Most Upvoted Answer
A battery of unknown emf connected to a potentiometer has balancing le...
To solve this problem, we can use the formula for balancing length in a potentiometer:
L = (R / (R + r)) * 100
Where L is the balancing length, R is the resistance in the potentiometer wire, and r is the internal resistance of the cell.
Let's denote the unknown emf of the battery as E and its internal resistance as n/10.
Given that the balancing length is 560 cm without any additional resistance, we have:
560 = (R / (R + n/10)) * 100
Similarly, with a resistor of resistance 10Ω connected in parallel with the cell, the balancing length changes to 500 cm. Therefore, we have:
500 = (R / (R + n/10 + 10)) * 100
Let's solve these two equations simultaneously to find the values of R and n.
First, let's simplify the equations:
560 = (R / (R + n/10)) * 100
500 = (R / (R + n/10 + 10)) * 100
Now, let's cross-multiply and simplify further:
56(R + n/10) = 10R
50(R + n/10 + 10) = 10R
Simplifying and rearranging terms:
56R + 5.6n = 10R
50R + 5n + 500 = 10R
Combining like terms:
46R = -5.6n
40R - 5n = -500
Dividing the first equation by 46 and the second equation by 5:
R = -0.122n
8R - n = -100
Substituting the value of R from the first equation into the second equation:
8(-0.122n) - n = -100
-0.976n - n = -100
-1.976n = -100
n = 50.5
Therefore, the internal resistance of the cell is 50.5/10 = 5.05Ω.
L = (R / (R + r)) * 100
Where L is the balancing length, R is the resistance in the potentiometer wire, and r is the internal resistance of the cell.
Let's denote the unknown emf of the battery as E and its internal resistance as n/10.
Given that the balancing length is 560 cm without any additional resistance, we have:
560 = (R / (R + n/10)) * 100
Similarly, with a resistor of resistance 10Ω connected in parallel with the cell, the balancing length changes to 500 cm. Therefore, we have:
500 = (R / (R + n/10 + 10)) * 100
Let's solve these two equations simultaneously to find the values of R and n.
First, let's simplify the equations:
560 = (R / (R + n/10)) * 100
500 = (R / (R + n/10 + 10)) * 100
Now, let's cross-multiply and simplify further:
56(R + n/10) = 10R
50(R + n/10 + 10) = 10R
Simplifying and rearranging terms:
56R + 5.6n = 10R
50R + 5n + 500 = 10R
Combining like terms:
46R = -5.6n
40R - 5n = -500
Dividing the first equation by 46 and the second equation by 5:
R = -0.122n
8R - n = -100
Substituting the value of R from the first equation into the second equation:
8(-0.122n) - n = -100
-0.976n - n = -100
-1.976n = -100
n = 50.5
Therefore, the internal resistance of the cell is 50.5/10 = 5.05Ω.
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A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. Can you explain this answer?
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A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. 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 battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. 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 battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. Can you explain this answer?.
A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. 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 battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. 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 battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. Can you explain this answer?.
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A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. Can you explain this answer?, a detailed solution for A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. Can you explain this answer? has been provided alongside types of A battery of unknown emf connected to a potentiometer has balancing length 560 cm. If a resistor of resistance 10W is connected in parallel with the cell the balancing length change by 60 cm. If the internal resistance of the cell is n/10 Ω, the value of n isCorrect answer is '12'. Can you explain this answer? theory, EduRev gives you an
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