A potentiometer wire PQ of 1,m length is conn | vedclass

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(A) In a potentiometer,the balancing length $l$ is always measured from the high potential end $P$.Given the total length of the wire $PQ = 100\,cm$.T

Vedclass · Accepted Answer

$0.02$

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(A) In a potentiometer,the balancing length $l$ is always measured from the high potential end $P$.Given the total length of the wire $PQ = 100\,cm$.The null point is obtained at $49\,cm$ from $Q$.Therefore,the balancing length from $P$ is $l = 100\,cm - 49\,cm = 51\,cm$.The emf of the cell $E_{2}$ is balanced by the potential drop across the length $l$ of the potentiometer wire.The formula for the emf is $E_{2} = \phi 	imes l$,where $\phi$ is the potential gradient.Substituting the given values: $1.02\,V = \phi 	imes 51\,cm$.Solving for $\phi$: $\phi = \frac{1.02}{51}\,V/cm = 0.02\,V/cm$.Thus,the potential gradient in the potentiometer wire is $0.02\,V/cm$.

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