In a potentiometer balancing length of a $$10$$ volt cell is $$60$$ cm. Now the wire of the potentiometer is replaced by a new wire of same material but having length twice the original length and radius half the original radius. Then the balancing length of $$10$$ volt cell will be.

Question

Infinity Learn · Accepted Answer

If Eo be the emf of primary driving cell of  potentiometer, and λ be the resistance per unit length of potentiometer wire, $$thenE=Eo($$λ$$L)(l$$λ$$)=EolL$$ For the new wire, $$E=Eo$$.l'$$2L$$ ∴ Eol'$$2L=Eol2$$⇒l'$$=2l=2$$×$$60$$ $$cm=120$$ cm.

In a potentiometer balancing length of a 10 volt cell is 60 cm. Now the wire of the potentiometer is replaced by a new wire of same material but having length twice the original length and radius half the original radius. Then the balancing length of 10 volt cell will be.

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