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Q.

A potentiometer wire of length 100 cm  has a resistance 5 ohm . It is connected in series with a resistance and a cell of emf 2 V  and of negligible internal resistance. A source of emf 5 mV  balanced by 10 cm  length of potentiometer wire. The value of external resistance is_____

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a

540 Ω

b

195 Ω

c

190 Ω

d

990 Ω

answer is B.

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Detailed Solution

Potentiometer wire of length L=100 cm Resistance R=5 Ω Cell of emf E=2 V Source of emf E1=5 mV Balancing length l=10 cm E1=(ER+RS) RLl 5×10−3=(25+RS) 5100×10 5×10−3=2×5×10(5+RS)100 ∴RS= 195 Ω
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A potentiometer wire of length 100 cm  has a resistance 5 ohm . It is connected in series with a resistance and a cell of emf 2 V  and of negligible internal resistance. A source of emf 5 mV  balanced by 10 cm  length of potentiometer wire. The value of external resistance is_____