Q 12-03-267JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
In a conductor, if the number of conduction electrons per unit volume is $8.5\times10^{28}\ \text{m}^{-3}$ and mean free time is $25\ \text{fs}$ (femto second), its approximate resistivity is ($m_e = 9.1\times10^{-31}\ \text{kg}$)
Answer: (D) $10^{-8}\ \Omega\,\text{m}$
$$\rho = \frac{m_e}{ne^2\tau} = \frac{9.1\times10^{-31}}{8.5\times10^{28}\times(1.6\times10^{-19})^2\times25\times10^{-15}}$$
$$\rho = \frac{9.1\times10^{-31}}{5.44\times10^{-23}} \approx 1.7\times10^{-8}\ \Omega\,\text{m}$$
which is of the order of $10^{-8}\ \Omega\,\text{m}$.
Solution by Sreeraj P, M.Sc Physics