List of top Physics Questions on Electricity asked in JEE Advanced

As shown in the figure, the resistance of a galvanometer $G$ can be found by the half-deflection method. Here the resistance $R_2$ is adjusted such that when the key $K$ is closed the deflection in the galvanometer becomes half of the value as compared to when $K$ is open. Half-deflection is obtained at $R_2 = 4\ \Omega$ and thus the galvanometer resistance is found to be $6\ \Omega$. In this half-deflection condition the current (in mA) through the resistor $R_1$ is:

A metal wire of cross-sectional area \(0.5\,\text{mm}^2\) and length \(100\,\text{m}\) is connected across a battery of e.m.f. \(2\,\text{V}\) and internal resistance \(1\,\Omega\). The density, atomic mass and electrical conductivity of the metal are \(6.35 \times 10^3\,\text{kg m}^{-3}\), \(63.5\,\text{gm/mole}\) and \(2 \times 10^8\,\text{mho m}^{-1}\), respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in \(\text{mm s}^{-1}\)) of the electrons in the wire is:
(Take Avogadro’s number as \(6 \times 10^{23}\) and charge of the electron as \(1.6 \times 10^{-19}\,\text{C}\).)