Molecular mass of copper $M=63 g m=0.063 k g$
Number of copper atoms per unit volume $n=\frac{\rho N_{A}}{M}$
$\therefore n=\frac{9 \times 10^{3} \times 6.023 \times 10^{23}}{0.063}=8.60 \times 10^{28} \mathrm{m}^{-3}$
As one copper atom gives one free electron, thus number of electrons per unit volume is $8.6 \times 10^{28} \mathrm{m}^{-3}$
Cross-section area of wire $A=\frac{\pi d^{4}}{4}=\frac{3.14 \times(0.001)^{2}}{4}=7.85 \times 10^{-7} \mathrm{m}^{2}$
Drift velocity $v_{d}=\frac{I}{A n e}$
$\therefore v_{d}=\frac{1.1}{\left(7.85 \times 10^{-7}\right)\left(8.6 \times 10^{28}\right)\left(1.6 \times 10^{-19}\right)}=0.01 \times 10^{-2} \mathrm{m} / \mathrm{s}=0.1 \mathrm{mm} / \mathrm{s}$
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(Assume, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^{+9}\; Nm ^2 C ^{-2}$)