Ohm's Law, Resistance and Resistivity – Ohm's law: NEET MCQs with Answers

14 free questions on Ohm's Law, Resistance and Resistivity – Ohm's law for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Ohm's law. Electrical resistance, V-l characteristics of Ohmic and non-ohmic conductors, Electrical resistivity and conductivity, Temperature dependence of resistance

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Question 1

Previous-year question
Solids possessing a negative temperature coefficient of resistance are [NEET 2020]
  1. insulators only
  2. semiconductors only
  3. both insulators and semiconductors
  4. metals
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Answer: (C) both insulators and semiconductors
Explanation: In insulators and semiconductors conduction requires charge carriers to be thermally excited across an energy gap. Raising the temperature creates many more carriers, so the resistance falls and α is negative. Metals already have a fixed, large carrier density, so for them increased lattice scattering dominates and α is positive.

Question 2

Previous-year question
A carbon resistor of (47 ± 4.7) kΩ is to be identified by coloured rings. The correct colour sequence will be [NEET 2018]
  1. Violet – Yellow – Orange – Silver
  2. Yellow – Violet – Orange – Silver
  3. Green – Orange – Violet – Gold
  4. Yellow – Green – Violet – Gold
Show answer
Answer: (B) Yellow – Violet – Orange – Silver
Explanation: 47 kΩ = 47 × 10³ Ω and the tolerance 4.7 kΩ is 10 % of 47 kΩ. The first significant figure 4 is yellow, the second figure 7 is violet, the multiplier 10³ is orange and the ± 10 % tolerance band is silver.

Question 3

Previous-year question
A wire has resistance R ohm. If it is melted and drawn out to n times its original length, its new resistance will be [NEET 2017]
  1. R/n
  2. n²R
  3. R/n²
  4. nR
Show answer
Answer: (B) n²R
Explanation: Melting and redrawing conserves the volume, so when the length becomes nℓ the area becomes A/n. Then R′ = ρ(nℓ)/(A/n) = n²ρℓ/A = n²R.

Question 4

Two coils each have a resistance of 100 Ω at 20 °C. One is copper (α = 3.9 × 10⁻³ °C⁻¹) and the other is manganin (α = 2 × 10⁻⁵ °C⁻¹). When the room warms to 30 °C, how do the resistances change, and which coil is the better standard resistor?
  1. Copper rises by about 3.9 Ω and manganin falls by about 0.02 Ω; both are equally good standards
  2. Copper falls by about 3.9 Ω and manganin rises by about 0.02 Ω; copper is the better standard
  3. Copper rises by about 3.9 Ω and manganin by about 0.02 Ω; manganin is the better standard
  4. Copper rises by about 0.39 Ω and manganin by about 0.2 Ω; copper is the better standard since it conducts better
Show answer
Answer: (C) Copper rises by about 3.9 Ω and manganin by about 0.02 Ω; manganin is the better standard
Explanation: For a small temperature change, ΔR ≈ R₀αΔT. For copper this is 100 × 3.9 × 10⁻³ × 10 = 3.9 Ω, and for manganin it is 100 × 2 × 10⁻⁵ × 10 = 0.02 Ω. Alloys such as manganin, constantan and nichrome have a very weak dependence of resistivity on temperature. A wire-bound standard resistor must hold its value, so such an alloy is preferred.

Question 5

On heating, the resistivity of silicon falls while that of copper rises. Which statement correctly explains the difference?
  1. In silicon the carrier density rises steeply and outweighs the fall in relaxation time, while in copper the carrier density hardly changes.
  2. In copper the carrier density falls with temperature, while in both metals and semiconductors the relaxation time stays constant.
  3. Heating widens the band gap of silicon and closes the band gap of copper.
  4. In silicon the relaxation time increases with temperature, while in copper it decreases.
Show answer
Answer: (A) In silicon the carrier density rises steeply and outweighs the fall in relaxation time, while in copper the carrier density hardly changes.
Explanation: Resistivity is ρ = m/(ne²τ), and in both materials collisions become more frequent on heating, so τ falls. In a metal the free electron density n is nearly independent of temperature, so ρ rises. In a semiconductor n increases very rapidly as more electrons cross the band gap, and this outweighs the fall in τ, so ρ falls.

Question 6

Which one of the following observations is NOT a departure from Ohm's law?
  1. The resistance of a uniform metal wire at a fixed temperature doubles when its length is doubled
  2. In a device, the current stops being proportional to the voltage as the voltage is raised
  3. For a device, the current for a given size of voltage changes greatly when the polarity of the voltage is reversed
  4. For a material such as GaAs, the same current is obtained at more than one value of voltage
Show answer
Answer: (A) The resistance of a uniform metal wire at a fixed temperature doubles when its length is doubled
Explanation: Ohm's law fails when V is not proportional to I, when the V-I relation depends on the sign of V, or when the relation is not unique. A diode and GaAs are examples of the last two. A wire whose resistance grows with length still has V proportional to I at each length, so this is ordinary ohmic behaviour with R = ρl/A.

Question 7

The V–I graph of a metal wire, with V on the vertical axis and I on the horizontal axis, is a straight line through the origin at each temperature. At 0 °C the line makes an angle of 30° with the I-axis, and at temperature T (in °C) it makes an angle of 60° with the I-axis. If α is the temperature coefficient of resistance referred to 0 °C, and the change in the wire's dimensions is ignored, what is T?
  1. 1/α
  2. (√3 − 1)/α
  3. 2/α
  4. 3/α
Show answer
Answer: (C) 2/α
Explanation: The slope of a V–I line gives the resistance, so R_T/R₀ = tan 60°/tan 30° = 3. Using R_T = R₀(1 + αT) gives 1 + αT = 3. Hence T = 2/α.
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Question 8

At 0°C, wire A has resistance 20 Ω and temperature coefficient of resistance 4.0 × 10⁻³ /°C, and wire B has resistance 30 Ω and temperature coefficient 1.0 × 10⁻³ /°C. The wires are joined in series. Assuming linear variation with temperature, what is the temperature coefficient of the combination, referred to 0°C?
  1. 2.8 × 10⁻³ /°C
  2. 2.5 × 10⁻³ /°C
  3. 2.2 × 10⁻³ /°C
  4. 5.0 × 10⁻³ /°C
Show answer
Answer: (C) 2.2 × 10⁻³ /°C
Explanation: At temperature t the series resistance is 20(1 + 4.0 × 10⁻³ t) + 30(1 + 1.0 × 10⁻³ t) = 50 + 0.11t. Writing this as 50(1 + αt) gives α = 0.11/50 = 2.2 × 10⁻³ /°C. The result is the average of the two coefficients weighted by the resistances, not their simple mean.

Question 9

A uniform wire of resistance 20 Ω is cut into 5 identical parts. Each part is then stretched uniformly to three times its length, with no change in its volume, and all five stretched parts are joined in parallel. What is the resistance of the combination?
  1. 7.2 Ω
  2. 2.4 Ω
  3. 36 Ω
  4. 180 Ω
Show answer
Answer: (A) 7.2 Ω
Explanation: Each part has a resistance of 20/5 = 4 Ω. At constant volume, stretching to n times the length raises resistance n² times, because the length rises and the area falls by the same factor, so each part becomes 4 × 9 = 36 Ω. Five equal resistors in parallel give 36/5 = 7.2 Ω.

Question 10

A rectangular copper block measures 1 cm × 2 cm × 5 cm. A fixed potential difference is applied between a pair of opposite faces, and the pair of faces is then changed. What is the ratio of the largest possible current to the smallest possible current?
  1. 50 : 1
  2. 10 : 1
  3. 5 : 1
  4. 25 : 1
Show answer
Answer: (D) 25 : 1
Explanation: Using R = ρl/A, the resistance is least when current crosses the shortest length through the largest face: l = 1 cm, A = 10 cm², so l/A = 0.1 per cm. It is greatest for l = 5 cm, A = 2 cm², so l/A = 2.5 per cm. At fixed voltage the current varies inversely as R, so the current ratio is 2.5/0.1 = 25.

Question 11

Current density j is plotted against electric field E for wires of materials P and Q at the same temperature. Both graphs are straight lines through the origin, and the slope for P is greater than the slope for Q. Which conclusion is correct?
  1. The slope gives the resistance of the sample, so it changes with the length of the wire
  2. P has the lower resistivity, so of the two it suits connecting wires better
  3. P has the higher resistivity, so of the two it suits connecting wires better
  4. For the same electric field, Q carries the larger current density
Show answer
Answer: (B) P has the lower resistivity, so of the two it suits connecting wires better
Explanation: From j = σE, the slope of the graph is the conductivity, a property of the material and not of the wire's dimensions. A larger slope therefore means higher conductivity and lower resistivity. Connecting wires should waste as little energy as possible, so the material with the lower resistivity suits them better.

Question 12

Two wires P and Q have equal length and equal resistance. P has resistivity 3 × 10⁻⁸ Ω m and density 3000 kg m⁻³, while Q has resistivity 2 × 10⁻⁸ Ω m and density 9000 kg m⁻³. What is the ratio of the mass of P to the mass of Q?
  1. 3 : 2
  2. 1 : 3
  3. 9 : 2
  4. 1 : 2
Show answer
Answer: (D) 1 : 2
Explanation: For equal R and L, the area needed is A = ρL/R, so A is proportional to resistivity. Mass = density × AL, so mass is proportional to (resistivity × density). The ratio is (3 × 3000)/(2 × 9000) = 1/2, which is why a low-density metal can be the lighter choice even with a higher resistivity.

Question 13

A uniform conducting material fills the space between two coaxial metal cylinders of radii 1 cm and 4 cm and equal length. A steady current flows radially from the inner cylinder to the outer one. What is the ratio of the electric field in the material at radius 1.5 cm to that at radius 3 cm?
  1. 4 : 1
  2. 1 : 2
  3. 1 : 1
  4. 2 : 1
Show answer
Answer: (D) 2 : 1
Explanation: In steady flow the same current crosses every coaxial cylindrical surface of area 2πrL, so the current density J = I/(2πrL) varies as 1/r. In a uniform material E = ρJ, so the field also varies as 1/r. The ratio is 3 : 1.5 = 2 : 1.

Question 14

A carbon resistor (temperature coefficient −5.0 × 10⁻⁴ /°C) is joined in series with a copper coil (temperature coefficient +4.0 × 10⁻³ /°C). The total resistance at the reference temperature is 90 Ω and it must not change for small changes of temperature. What should the two resistances be?
  1. Carbon 80 Ω, copper 10 Ω
  2. Carbon 10 Ω, copper 80 Ω
  3. Carbon 45 Ω, copper 45 Ω
  4. Carbon 72 Ω, copper 18 Ω
Show answer
Answer: (A) Carbon 80 Ω, copper 10 Ω
Explanation: For a small temperature rise ΔT the changes are R_c α_c ΔT and R_Cu α_Cu ΔT, and in series they must cancel. So R_c × 5.0 × 10⁻⁴ = R_Cu × 4.0 × 10⁻³, which gives R_c/R_Cu = 8. With a total of 90 Ω, the carbon resistor is 80 Ω and the copper coil is 10 Ω.

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