A cell can be balanced against 110 cm and 100 cm of potentiometer wire respectively when in open circuit and when short circuited through a resistance of 10Ω . The internal resistance of the cell is
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a
0.9Ω
b
1.0Ω
c
1.5Ω
d
2.0Ω
answer is B.
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Detailed Solution
Balancing length l1=110 cm Balancing length l2=100 cm Resistance R=10 Ω Internal resistance r=? r=l1−l2l2R = 110−100100×10 ∴ r=1 Ω