JEE MAIN - Physics (2015 (Offline) - No. 13)
When $$5V$$ potential difference is applied across a wire of length $$0.1$$ $$m,$$ the drift speed of electrons is $$2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}.$$ If the electron density in the wire is $$8 \times {10^{28}}\,\,{m^{ - 3}},$$ the resistivity of the material is close to :
$$1.6 \times {10^{ - 6}}\Omega m$$
$$1.6 \times {10^{ - 5}}\Omega m$$
$$1.6 \times {10^{ - 8}}\Omega m$$
$$1.6 \times {10^{ - 7}}\Omega m$$
Explanation
$$V = IR = \left( {neA{v_d}} \right)\rho {\ell \over A}$$
$$\therefore$$ $$\rho = {V \over {{V_d}\ln e}}$$
Here $$V=$$ potential difference
$$l = $$ length of wire
$$n=$$ no. of electrons per unit volume of conductor.
$$e=$$ no. of electrons
Placing the value of above parameters we get resistivity
$$\rho = {5 \over {8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}}$$
$$ = 1.6 \times {10^{ - 5}}\Omega m$$
$$\therefore$$ $$\rho = {V \over {{V_d}\ln e}}$$
Here $$V=$$ potential difference
$$l = $$ length of wire
$$n=$$ no. of electrons per unit volume of conductor.
$$e=$$ no. of electrons
Placing the value of above parameters we get resistivity
$$\rho = {5 \over {8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}}$$
$$ = 1.6 \times {10^{ - 5}}\Omega m$$
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