Exams › JEE Advanced › Physics
If n is the electron density, e the electronic charge, tau the average relaxation time, and m the electron mass, write the expression for the resistance of a conducting wire of length l and cross-sectional area A.
- m*l / (n*e²*tau*A)
- m*tau*A / (n*e²*l)
- n*e²*tau*A / (2*m*l)
- n*e²*A / (2*m*tau*l)
Correct answer: m*l / (n*e²*tau*A)
Solution
The Drude resistivity is rho = m/(n e² tau); substituting into R = rho*l/A gives R = m*l/(n e² tau A).
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