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Question

In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5 Ω resistance and at a length of 3 m, when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is

A
1.5 Ω
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B
10 Ω
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C
15 Ω
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D
1 Ω
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Solution

The correct option is B 10 Ω
Let x be potential gradient of the wire

When the cell is shunted by 5 Ω then.
55+rE=x(2)
When the cell is shunted by 10 Ω then,
1010+rE=x(3)
By dividing both the equations, we get
10+r2(5+r)=23
or 30+3r=20+4rr=10 Ω

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