Q37 / 54 · JEE Advanced 2026 (Paper 2) · Physics
A metal wire of cross-sectional area $0.5$ mm$^2$ and length $100$ m is connected across a battery of e.m.f. $2$ V and internal resistance $1\ \Omega$. The density, atomic mass and electrical conductivity of the metal are $6.35 \times 10^3$ kg m$^{-3}$, $63.5$ gm/mole and $2 \times 10^8$ mho m$^{-1}$, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s$^{-1}$) of the electrons in the wire is:
[Take Avogadro's number as $6 \times 10^{23}$ and charge of the electron as $1.6 \times 10^{-19}$ C.]
[Take Avogadro's number as $6 \times 10^{23}$ and charge of the electron as $1.6 \times 10^{-19}$ C.]
- A$0.052$
- B$0.104$
- Correct answerC$0.208$
- D$0.156$
Solution
The number density of conduction electrons $n$ is given by the number of atoms per unit volume, since there is one conduction electron per atom:
$n = \dfrac{d \times N_A}{M}$
Substituting the given values:
$n = \dfrac{6.35 \times 10^3 \times 6 \times 10^{23}}{63.5 \times 10^{-3}} = 6 \times 10^{28} \text{ m}^{-3}$
The resistance of the wire $R$ is:
$R = \dfrac{L}{\sigma A}$
$R = \dfrac{100}{2 \times 10^8 \times 0.5 \times 10^{-6}} = \dfrac{100}{10^2} = 1\ \Omega$
The total resistance of the circuit is $R_{total} = R + r = 1 + 1 = 2\ \Omega$.
The current in the circuit is:
$I = \dfrac{E}{R_{total}} = \dfrac{2}{2} = 1 \text{ A}$
The drift velocity $v_d$ is given by the relation $I = n e A v_d$:
$v_d = \dfrac{I}{n e A}$
$v_d = \dfrac{1}{6 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.5 \times 10^{-6}}$
$v_d = \dfrac{1}{4.8 \times 10^3} \text{ m s}^{-1}$
Converting to mm s$^{-1}$:
$v_d = \dfrac{1000}{4800} \text{ mm s}^{-1} = \dfrac{5}{24} \text{ mm s}^{-1} \approx 0.208 \text{ mm s}^{-1}$
Answer: $0.208$
$n = \dfrac{d \times N_A}{M}$
Substituting the given values:
$n = \dfrac{6.35 \times 10^3 \times 6 \times 10^{23}}{63.5 \times 10^{-3}} = 6 \times 10^{28} \text{ m}^{-3}$
The resistance of the wire $R$ is:
$R = \dfrac{L}{\sigma A}$
$R = \dfrac{100}{2 \times 10^8 \times 0.5 \times 10^{-6}} = \dfrac{100}{10^2} = 1\ \Omega$
The total resistance of the circuit is $R_{total} = R + r = 1 + 1 = 2\ \Omega$.
The current in the circuit is:
$I = \dfrac{E}{R_{total}} = \dfrac{2}{2} = 1 \text{ A}$
The drift velocity $v_d$ is given by the relation $I = n e A v_d$:
$v_d = \dfrac{I}{n e A}$
$v_d = \dfrac{1}{6 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.5 \times 10^{-6}}$
$v_d = \dfrac{1}{4.8 \times 10^3} \text{ m s}^{-1}$
Converting to mm s$^{-1}$:
$v_d = \dfrac{1000}{4800} \text{ mm s}^{-1} = \dfrac{5}{24} \text{ mm s}^{-1} \approx 0.208 \text{ mm s}^{-1}$
Answer: $0.208$
Hence, the correct answer is option: 'C'.