Q 12-03-160JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
A potentiometer wire of length $10\ \text{m}$ and resistance $20\ \Omega$ is connected in series with a $25\ \text{V}$ battery and an external resistance $30\ \Omega$. A cell of emf $E$ in the secondary circuit is balanced by $250\ \text{cm}$ long potentiometer wire. The value of $E$ (in volt) is $\dfrac{x}{10}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 25
Current in the wire $= \dfrac{25}{20+30} = 0.5\ \text{A}$; potential drop across the wire $= 0.5\times20 = 10\ \text{V}$, so the gradient is $1\ \text{V m}^{-1}$.
$$E = 1\times2.5 = 2.5\ \text{V} = \frac{25}{10}\ \Rightarrow\ x = 25$$
Solution by Sreeraj P, M.Sc Physics