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Question
A potential drop per unit length along a wire is 5 × 10−3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.
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Solution
Data: K = 5 × 10-3 `"V"/"m"`, L = 216 cm = 216 × 10-2 m
E = KL
∴ E = 5 × 10-3 × 216 ×10-2
= 1080 × 10-5
= 0.01080 V
The emf of the cell is 0.01080 volL
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