A potentiometer wire PQ of length 1 m is connected to a standard cell E₁. Another cell E₂ of emf 1.02V is connected with a resistance 'r' and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49cm from Q. The potential gradient in the potentiometer wire is:

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$0 .02 V / cm$

$0 .04 V / cm$

$0 .01 V / cm$

$0 .03 V / cm$

answer is A.

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Detailed Solution

Balancing length is measured from P.
So $100 - 49 = 51 cm$
$E_{2} = ϕ \times 51$
Where $ϕ =$ Potential gradient
$\begin{array}{l} 1 .02 = ϕ \times 51 \\ ϕ = 0 .02 V / cm \end{array}$

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