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Resistance, resistivity, conductivity and temperature dependence: NEET MCQs with answers

75 questions on this topic of Current Electricity (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The electrical resistivity of a metallic conductor is given by ρ = m / (n e² τ). In this expression, which of the following statements correctly explains why the resistivity of metals increases with increasing temperature?

  1. The number density n of conduction electrons decreases significantly with temperature.
  2. The average relaxation time τ decreases because the amplitude of vibration of lattice ions increases.
  3. The charge e on an electron decreases due to thermal fluctuations.
  4. The effective mass m of conduction electrons increases sharply at elevated temperatures.
Show answer and explanation

Answer: (B) The average relaxation time τ decreases because the amplitude of vibration of lattice ions increases.

For metals, the electron density n is almost independent of temperature. As temperature increases, the amplitude of thermal vibrations of lattice ions increases, causing more frequent collisions and hence reducing the relaxation time τ. Since ρ is inversely proportional to τ, resistivity increases.

medium

2. Two metal wires X and Y of the same material have lengths in the ratio 2 : 1 and masses in the ratio 3 : 2. What is the ratio of their electrical resistances (R_X : R_Y)?

  1. 3 : 1
  2. 8 : 3
  3. 4 : 3
  4. 2 : 3
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Answer: (B) 8 : 3

Mass of a wire is m = d · A · L, so A = m / (d L). Resistance is R = ρ L / A = ρ d L² / m. Since both wires are of the same material, ρ and density d are identical, giving R ∝ L² / m. Therefore, R_X / R_Y = (L_X / L_Y)² × (m_Y / m_X) = (2 / 1)² × (2 / 3) = 4 × (2 / 3) = 8 / 3.

hard

3. Two conductors 1 and 2 of resistances R₁ and R₂ have temperature coefficients of resistivity α₁ and α₂ respectively. When they are connected in parallel, the temperature coefficient of resistance of the combination at that temperature is:

  1. (α₁ R₂ + α₂ R₁) / (R₁ + R₂)
  2. (α₁ R₁ + α₂ R₂) / (R₁ + R₂)
  3. (α₁ + α₂) / 2
  4. α₁ α₂ / (α₁ + α₂)
Show answer and explanation

Answer: (A) (α₁ R₂ + α₂ R₁) / (R₁ + R₂)

For a parallel combination, 1/R_p = 1/R₁ + 1/R₂. Differentiating with respect to temperature T gives -(1/R_p²)(dR_p/dT) = -(1/R₁²)(dR₁/dT) - (1/R₂²)(dR₂/dT). Since dR/dT = αR, we get α_p / R_p = α₁ / R₁ + α₂ / R₂. Substituting 1/R_p = (R₁ + R₂) / (R₁ R₂) gives α_p = (α₁ R₂ + α₂ R₁) / (R₁ + R₂).

medium

4. A wire of resistance 13 Ω is stretched uniformly to 2 times its original length (volume unchanged). Its new resistance is:

  1. 26 Ω
  2. 52 Ω
  3. 6.5 Ω
  4. 104 Ω
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Answer: (B) 52 Ω

With volume fixed, A becomes A/2, so R ∝ L/A grows by 2² = 4. New R = 52 Ω.

easy

5. Statement I: Constantan and Manganin are widely used for making standard resistance coils. Statement II: These alloys have very high resistivity and nearly negligible temperature coefficients of resistance.

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Standard resistance coils require materials whose resistance does not change noticeably with temperature fluctuations. Alloys like manganin and constantan possess high resistivity and nearly temperature-independent resistance (very small α).

hard

6. A hollow cylindrical conductor of length L has an inner radius a and an outer radius b. If current flows radially outward from the inner cylindrical surface to the outer surface, what is the effective resistance between these surfaces in terms of resistivity ρ?

  1. [ρ / (πL)] ln(b / a)
  2. [ρ / (4πL)] (b² - a²)
  3. [ρ / (2πL)] (b - a)
  4. [ρ / (2πL)] ln(b / a)
Show answer and explanation

Answer: (D) [ρ / (2πL)] ln(b / a)

Consider a cylindrical shell of radius r, thickness dr, and length L. Its surface area normal to radial current is A = 2πrL. The resistance is dR = ρ dr / (2πrL). Integrating from r = a to r = b yields R = [ρ / (2πL)] ln(b / a).

medium

7. A wire of resistance 20 Ω is stretched uniformly to 3 times its original length (volume unchanged). Its new resistance is:

  1. 180 Ω
  2. 540 Ω
  3. 6.67 Ω
  4. 60 Ω
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Answer: (A) 180 Ω

With volume fixed, A becomes A/3, so R ∝ L/A grows by 3² = 9. New R = 180 Ω.

medium

8. A uniform wire of resistance R₀ is bent into a complete circle. Two connection terminals are placed at points on the perimeter separated by a subtended angle θ (in radians, where 0 < θ < 2π). What is the equivalent resistance between these two terminals?

  1. R₀ (2π - θ) / (2π)
  2. R₀ θ / (2π)
  3. R₀ θ (2π - θ) / (4π²)
  4. R₀ θ² / (4π²)
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Answer: (C) R₀ θ (2π - θ) / (4π²)

The circle is divided into two arcs that are in parallel. The length of one arc subtends angle θ, so its resistance is R₁ = R₀ (θ / 2π). The other arc subtends (2π - θ), so its resistance is R₂ = R₀ (2π - θ) / (2π). The parallel combination gives R_eq = (R₁ R₂) / (R₁ + R₂) = R₀² θ (2π - θ) / (4π² · R₀) = R₀ θ (2π - θ) / (4π²).

medium

9. A wire of resistance 17 Ω is stretched uniformly to 3 times its original length (volume unchanged). Its new resistance is:

  1. 459 Ω
  2. 5.67 Ω
  3. 153 Ω
  4. 51 Ω
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Answer: (C) 153 Ω

With volume fixed, A becomes A/3, so R ∝ L/A grows by 3² = 9. New R = 153 Ω.

medium

10. A wire of resistance 12 Ω is stretched uniformly to 2 times its original length (volume unchanged). Its new resistance is:

  1. 6 Ω
  2. 96 Ω
  3. 48 Ω
  4. 24 Ω
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Answer: (C) 48 Ω

With volume fixed, A becomes A/2, so R ∝ L/A grows by 2² = 4. New R = 48 Ω.

medium

11. A wire of resistance 3 Ω is stretched uniformly to 4 times its original length (volume unchanged). Its new resistance is:

  1. 12 Ω
  2. 48 Ω
  3. 0.75 Ω
  4. 192 Ω
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Answer: (B) 48 Ω

With volume fixed, A becomes A/4, so R ∝ L/A grows by 4² = 16. New R = 48 Ω.

medium

12. A wire of resistance R is elongated such that its length increases by 0.2% while keeping the volume constant. The percentage increase in its resistance is approximately:

  1. 0.1%
  2. 0.2%
  3. 0.4%
  4. 0.8%
Show answer and explanation

Answer: (C) 0.4%

For constant volume V = A L, R = ρ L² / V. Taking differentials for small changes, ΔR / R ≈ 2 (ΔL / L). Since ΔL / L = +0.2%, ΔR / R ≈ 2 × 0.2% = 0.4%.

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