When a cell of emf 5 volt and internal resistance 1 Ω is short circuited by an external resistance of 5 Ω, the balancing length of a potentiometer wire is 80 cm. What will be the balancing length of the same potentiometer wire for a cell of emf 10 volt having internal resistance 2 Ω is short circuited by an external resistance of 10 Ω ?

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140 cm

120 cm

40 cm

160 cm

answer is D.

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Detailed Solution

Potential drop across the external resistance E.R(R+r)∝l l is the balancing length ∴l2l1=E2E1R2R1R1+r1R2+r2⇒l280=1051055+110+2⇒l2=2 × 80 cm=160 cm

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