Question:
A potentiometer wire $P Q$ of $1 \mathrm{~m}$ length is connected to a standard cell $E_{1}$. Another cell $E_{2}$ of emf $1.02 \mathrm{~V}$ 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 $49 \mathrm{~cm}$ from $Q$. The potential gradient in the potentiometer wire is:
Correct Option: 1
Solution:
(1) Potential gradient, $x=\frac{\text { Potential drop }}{\text { length }}$
Here, Potential drop $=1.02$
Balancing length from $\mathrm{P}=100-49$
$\therefore x=\frac{1.02}{100-49}=0.02 \mathrm{volt} / \mathrm{cm}$
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