#### A electric current of 16 A exists in a metal wire of cross section $\mathrm{10^{-6}\, m^{2}}$ and length 1m. Assuming one free electrons per atom. The drift speed of the free electrons in the wire will be ( Density of metal $\mathrm{=5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3,}$  atomic weight $\mathrm{=60}$ )Option: 1 $5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$Option: 2 $2 \times 10^{-3} \mathrm{~m} / \mathrm{s}$Option: 3 $4 \times 10^{-3} \mathrm{~m} / \mathrm{s}$Option: 4 $7.5 \times 10^{-3} \mathrm{~m} / \mathrm{s}$

$\mathrm{\frac{N}{N_A}=\frac{m}{M} \text { so } n=\frac{N}{V}=N_A \frac{m}{V M}=\frac{\rho N_A}{M}}$

$\mathrm{\text{Hence total number of atoms }\mathrm{n}=\frac{6 \times 10^{23} \times 5 \times 10^3}{60 \times 10^{-3}}}$

$\mathrm{=5 \times 10^{28} / \mathrm{m}^3}$
As $\mathrm{\mathrm{I}=\mathrm{n}_{\mathrm{e}}\, \mathrm{eA} \, \mathrm{v}_{\mathrm{d}}}$

$\mathrm{\text{Hence drift velocity }v_d=\frac{1}{n_e e A}}$
$\mathrm{v_d=\frac{16}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}}$
$\mathrm{=2 \times 10^{-3} \mathrm{~m} / \mathrm{s}}$