# Manufacturing Engineering Questions & Answers – Electrical Properties of Materials

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This set of Manufacturing Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Electrical Properties Of Materials”.

1. Which of the following is the de Broglie’s equation (λ = wavelength, v = velocity, m = mass, h = Planck’s constant)?
a) λ = h/mv
b) λ = h/m
c) λ = h/v
d) λ = mv/h

Explanation: The equation, λ = h/mv is known as de Broglie’s equation. This equation represents the wave-particle duality of an electron.

2. With an increase in the mean free path of electrons, materials conductivity ____________
a) increases
b) decreases
c) remains constant
d) first increases and then decreases

Explanation: Conductivity of a material increases with an increase in the mean free path of electrons. Mean free path is nothing but the average distance travelled by an electron between two successive collisions.

3. What is the resistivity of silver (in Ohm-m) at room temperature?
a) 1.59 × 10-2
b) 1.59 × 10-3
c) 1.59 × 10-5
d) 1.59 × 10-9

Explanation: Silver has lowest conductivity. The resistivity of silver is 1.59 × 10-9 Ω-m.

4. What is the resistivity of iron (in Ohm-m) at room temperature?
a) 9.71 × 10-2
b) 9.71 × 10-8
c) 9.71 × 10-5
d) 9.71 × 10-9

Explanation: Iron has medium conductivity. The resistivity of iron is 9.71 × 10-8 Ω-m.

5. What is the resistivity of bismuth (in Ohm-m) at room temperature?
a) 1.29 × 10-6
b) 1.29 × 10-3
c) 1.29 × 10-5
d) 1.29 × 10-9

Explanation: Bismuth has low conductivity. The resistivity of iron is 1.29 × 10-6 Ω-m.

6. What is the energy gap (Eg) of an insulator?
a) Eg < 3 ev
b) Eg < 1 ev
c) Eg < 0 ev
d) Eg > 3 ev

Explanation: The energy gap (Eg) of an insulator is approximately equal to 15 ev.

7. What is the energy gap (Eg) of a metal?
a) Eg < 3 ev
b) Eg < 1 ev
c) Eg > 1 ev
d) Eg > 3 ev

Explanation: Metals are nothing but electrical conductors. The energy gap of a conductor is very small and equal to 1 ev.

8. If a copper wire of 200 m length having 0.44mm diameter has resistance of 20 Ohm, then find its specific resistance (in Ohm-m).
a) 1.520 × 10-8
b) 1.520 × 10-7
c) 1.520 × 10-6
d) 1.520 × 10-5

Explanation: A = 3.14×d2/4 and resistivity = R×A/l, where ‘A’ is the area of cross section of wire, ρ is specific resistance, R = resistance, d = diameter, l = length.
Thus, A = 3.14×d2/4 = 3.14×(0.44)2/4 = 0.152 mm2, resistivity = R×A/l = (20×0.152/200) × 10-6 = 1.520 × 10-8.

9. If a copper wire of 10 m length having 0.44mm diameter has resistance of 2 Ohm, then what will be the specific resistance (in Ohm-m)?
a) 1.520 × 10-8
b) 3.041 × 10-8
c) 1.520 × 10-6
d) 1.520 × 10-5

Explanation: A = 3.14×d2/4 and resistivity = R×A/l, where ‘A’ is the area of cross section of wire, ρ is specific resistance, R = resistance, d = diameter, l = length.
Thus, A = 3.14×d2/4 = 3.14×(0.44)2/4 = 0.152 mm2, resistivity = R×A/l = (2×0.152/10) × 10-6 = 3.041 × 10-8.

10. With an increase in temperature, the resistance of a semiconductor ____________
a) increases
b) decreases
c) remains Constant
d) first increases and then decreases 