Physics Multiple Choice Questions on “Current Electricity – Temperature Dependence of Resistivity”.
1. Identify the material which is suitable for making standard resistors.
a) Silver
b) Copper
c) Constantan
d) Germanium
Answer: c
Clarification: Alloys like constantan or manganin are used for making standard resistance coils due to their high resistivity values and very small temperature coefficient.
2. What is the unit of the temperature coefficient of resistance (α)?
a) oC
b) oC-1
c) (frac {Omega}{ ^{circ} C})
d) (frac { ^{circ} C}{Omega})
Answer: b
Clarification: The temperature coefficient of resistivity is defined as the increase in resistivity per unit resistivity per degree rise in temperature.
The unit of the temperature coefficient of resistance is oC-1.
3. Which of the following relation is significant for metals when the temperature increases?
a) Resistivity increases and conductivity decreases
b) Resistivity decreases and conductivity decreases
c) Resistivity and conductivity do not change with temperature
d) Temperature dependence is non-linear
Answer: a
Clarification: The resistivity of a metal increases and the conductivity decreases with the increase in temperature. With an increase in temperature, the free electrons collide more frequently with the metal ions. The mean collision time also decreases.
4. Identify the type of material based on the T-ρ graph given below.
a) Silicon
b) Polymer
c) Nichrome
d) Copper
Answer: d
Clarification: For metals, the temperature coefficient of resistivity is positive. At lower temperatures, the resistivity of a pure metal increases as a higher power of temperature. So, the answer is copper, which is a metal.
5. Which among the following has weak temperature dependence values with resistivity?
a) Silver
b) Copper
c) Nichrome
d) Germanium
Answer: c
Clarification: Alloys have high resistivity. The resistivity of nichrome has weak temperature dependence. At absolute zero, a pure metal has negligibly small resistivity while an alloy like nichrome has some residual resistivity.
6. The resistivity of semiconductors and insulators decreases linearly with the increase of temperature.
a) True
b) False
Answer: b
Clarification: The resistivity of semiconductors and insulators decreases exponentially with the increase in temperature. This is because the number density of free electrons increases exponentially with the increase in temperature.
7. The resistivity of ‘X’ decreases with temperature and its coefficient of resistivity is negative. Identify X.
a) Silver
b) Silicon
c) Copper
d) Nichrome
Answer: b
Clarification: The coefficient of resistivity is negative for semiconductors and their resistivity decreases with temperature. The relaxation time does not change with temperature but the number density of free electrons increases exponentially with the increase in temperature. Consequently, the resistivity decreases exponentially with the increase in temperature.
8. A wire has a resistance of 5.5 Ω at 19oC and 21.5 Ω at 200oC. Find the temperature coefficient of resistivity(α) of the material.
a) 0.016 oC-1
b) 0.160 oC-1
c) 1.600 oC-1
d) 16.00 oC-1
Answer: a
Clarification: Temperature coefficient α=(frac {(R_2-R_1)}{R_2(T_2-T_1)}).
α = (frac {(21.5-5.5)}{(5.5(200-19))})
= 0.01607 oC-1.
9. Which of the following is not a valid reason for using alloys to make standard resistors?
a) Alloys have a high value of resistivity
b) They are least affected by air and moisture
c) Alloys have a large temperature coefficient
d) Their contact potential with copper is small
Answer: c
Clarification: Alloys have a high value of resistivity. They have a very small temperature coefficient. So their resistance does not change appreciably even for several degrees rise of temperature. That leaves the answer – alloys have a large temperature coefficient.
10. Identify the temperature at which the resistance of copper would be double of its resistance at oC. Given α (temperature coefficient of resistivity) for copper=3.9 x 10-3oC-1.
a) 125oC
b) 256oC
c) 1080oC
d) 273oC
Answer: b
Clarification: α=(frac {(R_2-R_1)}{R_2(T_2-T_1)}).
α=(frac {(2R_0-R_0)}{R_0(T-0)} = frac {1}{T}).
T=(frac {1}{alpha})
= (frac {1}{3.9}) x 10-3oC-1
= 256oC.
Therefore, the required temperature is 256oC.