JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 3)
In a conductor, if the number of conduction
electrons per unit volume is 8.5 × 1028 m–3 and
mean free time is 25ƒs (femto second), it's
approximate resistivity is :-
(me = 9.1 × 10–31 kg)
(me = 9.1 × 10–31 kg)
10–8 $$\Omega $$m
10–7 $$\Omega $$m
10–5 $$\Omega $$m
10–6 $$\Omega $$m
Explanation
$$\rho = {{2m} \over {n{e^2}\tau }}$$
= 3.34 × 10–8 $$\Omega $$ m
= 3.34 × 10–8 $$\Omega $$ m
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