JEE PYQ: Current Electricity Question 46
Question 46 - 2019 (09 Apr 2019 Shift 2)
In a conductor, if the number of conduction electrons per unit volume is $8.5 \times 10^{28}$ m$^{-3}$ and mean free time is 25 fs (femto second), its approximate resistivity is: ($m_e = 9.1 \times 10^{-31}$ kg)
(1) $10^{-6},\Omega$m
(2) $10^{-7},\Omega$m
(3) $10^{-8},\Omega$m
(4) $10^{-5},\Omega$m
Show Answer
Answer: (3)
Solution
$\rho = \frac{m}{ne^2\tau} = \frac{9.1 \times 10^{-31}}{8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 25 \times 10^{-15}} \approx 10^{-8},\Omega$m.
BETA
