Drift Velocity and Mobility Numerical Class-12 Nootan ISC Physics Ch-5 Electric-Resistance-and-Ohms-Law Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Drift Velocity and Mobility Numerical Class-12 Nootan ISC Physics Ch-5
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-5 | Electric Resistance and Ohm’s Law |
| Topics | Numericals on Drift Velocity and Mobility |
| Academic Session | 2025-2026 |
Numericals on Drift Velocity and Mobility
Que-2. The cross-sectional area of a wire is 1.0 x 10^-7 m² and the density of free electrons is 2.0 × 10^28 m^-3. What will be the drift velocity of the free electrons for a current of 3.2 A in the wire?
Ans-2 i = nAVde
=> Vd = i / nAe
=> 3.2 / 2 x 10^28 x 1 x 10^-7 x 1.6 x 10^-19
=> 1/10^2 = 1 x 10^-2 m/s
Que-3. 90 C charge flows for 75 minutes in a silver wire of diameter 1.0 mm. The free-electron density in the wire is 5.8 x 10^28 m^-3. Find the current and the drift velocity of free electrons in the wire.
Ans-3 i = Q/A = 90 / 75 x 60 = 0.02 A
Vd = i / nAe = 0.02 / 5.8 x 10^28 x π x (0.05 x 10^-3)^2 x 1.6 x 10^-19
=> 2.7 x 10^-6 m/s
Que-4. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10^-7 m² carrying a current of 1.5 A. Assume that each copper atom contributes one conduction electron. The density of copper is 9.0 x 10^3 kg m^-3 and its atomic mass is 63.5. Take Avogadro’s number N = 6.0 x 10^23 mol^-1
Ans-4 No. of atoms in 1 m^3 = (9 x 10^3 x 10^3 / 63.5) x N
=> 9 x 10^6 x 6.022 x10^23 / 63.5 = 8.5 x 10^28
Since each atom contributes 1 electron therefore electron density = 8.5 x 10^28 m^-3
Therefore Vd = i / nAe
=> 1.5 / 8.5 x 10^28 x 1 x 10^-7 x 1.6 x 10^-19
=> 1.1 x 10^-3 m/s
Que-5. A 3.0 m long copper wire is carrying a current of 3.0 A. How long does it take for an electron to drift from one end of the wire to the other? The cross-sectional area of the wire is 2.0 x 10^-6 m^-2 and the number of conduction electrons in copper is 8.5 x 10^28 m^-3.
Ans-5 Vd = i / nAe = 3 / 8.5 x 10^28 x 2 x 10^-6 x 1.6 x 10^-19
=> (3 / 0.5 x 2 x 1.6) x 10^-3
=> 1.1 x 10^-4 m/s
t = l / Vd = 3 / 1.1 x 10^-4 = 27272 sec
=> 27272 / 60 x 60 = 7.5 hrs
— : End of Drift Velocity and Mobility Numerical Class-12 Nootan ISC Physics Ch-5. :–
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