Q.

In an experiment with potentiometer to measure the internal resistance of a cell, when the cell is shunted by 5Ω , the null point is obtained at2 m . when cell is shunted by 20Ω  the null point is obtained at3 m .The internal resistance of cell is

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a

2 Ω

b

4 Ω

c

6 Ω

d

8 Ω

answer is B.

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Detailed Solution

Resistance R1=5 Ω Resistance R2=20  Ω Balancing length l1=2 m Balancing length l2=3 m Internal resistance r=? l1l2=R1R2[R2+rR1+r] 23=520(20+r5+r) ∴  r=4 Ω
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