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 resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is then
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a
1.5 Ω
b
10 Ω
c
15 Ω
d
1Ω
answer is B.
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Detailed Solution
In case of internal resistance measurement by potentiometer,V1V2=l1l2=ER1/R1+rER2/R2+r=R1R2+rR2R1+rHere l1=2m,l2=3m,R1=5Ω and R2=10Ω∴ 23=5(10+r)10(5+r)or 20+4r=30+3ror r=10Ω