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Wire resistance using Ohm’s law: NEET MCQs with answers

49 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The resistance of the wire is found from the V-I graph as the:

  1. Slope of V against I
  2. Intercept on the V-axis
  3. Area under the graph
  4. Inverse of the slope of V against I
Show answer and explanation

Answer: (A) Slope of V against I

For an ohmic wire V = IR, so plotting V against I gives a straight line whose slope equals R.

NCERT Class 11

medium

2. In an Ohm's law experiment, a voltmeter of internal resistance 1000 Ω is connected in parallel with an unknown wire of resistance R. An ammeter of negligible resistance reads 0.20 A and the voltmeter reads 4.0 V. What is the true value of the wire resistance R?

  1. 20.4 Ω
  2. 20.0 Ω
  3. 19.6 Ω
  4. 21.0 Ω
Show answer and explanation

Answer: (A) 20.4 Ω

The measured current through the combination is I = 0.20 A and the voltage across both is V = 4.0 V. The current drawn by the voltmeter is Iv = V / Rv = 4.0 / 1000 = 0.004 A. Hence, the actual current flowing through the wire is IR = I - Iv = 0.20 - 0.004 = 0.196 A. Thus, the true resistance R = V / IR = 4.0 / 0.196 ≈ 20.4 Ω.

hard

3. In an experiment to determine the resistance of a uniform wire of length 100 cm, the circuit current is kept at 1.0 A. Due to continuous flow of current for 5 minutes, the temperature of the wire rises by 10 °C. If the temperature coefficient of resistance of the wire is α = 4.0 × 10⁻⁴ °C⁻¹ and its initial resistance was 20.0 Ω, the fractional change in measured resistance (ΔR / R₀) is:

  1. 4.0 × 10⁻³
  2. 2.0 × 10⁻³
  3. 8.0 × 10⁻⁴
  4. 4.0 × 10⁻²
Show answer and explanation

Answer: (A) 4.0 × 10⁻³

The variation of resistance with temperature is given by ΔR = R₀ α ΔT. Therefore, the fractional change is ΔR / R₀ = α ΔT = (4.0 × 10⁻⁴ °C⁻¹) × (10 °C) = 4.0 × 10⁻³.

medium

4. In an Ohm's law experiment, the length of the resistance wire is measured to be L = (50.0 ± 0.1) cm and its diameter is d = (1.00 ± 0.01) mm. If the measured resistance is R = (10.0 ± 0.2) Ω, the maximum relative percentage error in determining the resistivity ρ of the wire material is:

  1. 2.2%
  2. 4.2%
  3. 3.2%
  4. 1.2%
Show answer and explanation

Answer: (B) 4.2%

Resistivity is given by ρ = R × A / L = R × (πd² / 4) / L. The fractional error is Δρ/ρ = (ΔR/R) + 2(Δd/d) + (ΔL/L). Substituting the values: (0.2 / 10.0) + 2 × (0.01 / 1.00) + (0.1 / 50.0) = 0.02 + 0.02 + 0.002 = 0.042. Multiplying by 100 gives 4.2%.

easy

5. In an Ohm's law verification circuit, a standard DC moving-coil ammeter has its positive terminal incorrectly connected towards the negative terminal of the DC power supply. What will be observed when the key is closed?

  1. The pointer deflects normally in the clockwise direction.
  2. The ammeter immediately burns out without any deflection.
  3. The pointer remains stationary at zero without any torque.
  4. The pointer deflects backwards (below zero), potentially damaging the pivot.
Show answer and explanation

Answer: (D) The pointer deflects backwards (below zero), potentially damaging the pivot.

Moving-coil instruments are polarity-sensitive because the deflecting torque depends directly on the direction of current. Reversing the polarity sends current backwards through the coil, producing a counter-clockwise torque that pushes the pointer against the left mechanical stop and can damage the pivot.

hard

6. In an Ohm's law experiment, two identical uniform resistance wires, each of resistance R, are connected first in series and then in parallel across the same voltmeter-ammeter arrangement. If S_s and S_p denote the slopes of the V (y-axis) versus I (x-axis) plots for the series and parallel combinations respectively, the ratio S_s / S_p is:

  1. 4 : 1
  2. 2 : 1
  3. 1 : 4
  4. 1 : 2
Show answer and explanation

Answer: (A) 4 : 1

On a V versus I plot where V is on the vertical axis, the slope S equals the equivalent resistance R_eq. For the series combination of two identical wires, R_series = 2R, so S_s = 2R. For the parallel combination, R_parallel = R / 2, so S_p = R / 2. Therefore, S_s / S_p = (2R) / (R / 2) = 4.

medium

7. After finding resistance R of a wire of length l and cross-sectional area A, its resistivity is:

  1. ρ = l/(RA)
  2. ρ = RAl
  3. ρ = RA/l
  4. ρ = Rl/A
Show answer and explanation

Answer: (C) ρ = RA/l

R = ρl/A, so resistivity ρ = RA/l. The area A = πr² uses the radius measured with a screw gauge.

NCERT Class 11

easy

8. In the circuit used for finding the resistance of a wire via Ohm's law, a rheostat is connected in series with the battery and the test wire. When the sliding contact of the rheostat is moved to increase its introduced resistance, how do the readings of the ammeter (I) and voltmeter (V across the wire) change?

  1. Both I and V decrease.
  2. I decreases while V increases.
  3. I increases while V decreases.
  4. Both I and V remain unchanged.
Show answer and explanation

Answer: (A) Both I and V decrease.

Increasing the rheostat's resistance increases the total equivalent resistance of the series circuit. Consequently, the total circuit current I decreases, and the potential difference across the test wire V = I × R_wire also decreases proportionally.

easy

9. A student performs the Ohm's law experiment using a standard DC voltmeter that has 20 divisions between the 0 V and 1 V marks. What is the least count of this voltmeter?

  1. 0.1 V
  2. 0.05 V
  3. 0.02 V
  4. 0.01 V
Show answer and explanation

Answer: (B) 0.05 V

The least count of an instrument is the value of one smallest scale division. Here, 20 divisions correspond to 1 V, so 1 division = 1 V / 20 = 0.05 V.

easy

10. While plotting the V-I characteristics of a metallic wire of resistance R, two students plot graphs. Student A plots V along the y-axis and I along the x-axis, obtaining an angle of inclination θ₁ with the x-axis. Student B plots I along the y-axis and V along the x-axis, obtaining an angle θ₂ with the x-axis. Which of the following is correct?

  1. tan θ₁ = 1/R and tan θ₂ = R
  2. tan θ₁ = R and tan θ₂ = 1/R
  3. tan θ₁ · tan θ₂ = R²
  4. tan θ₁ + tan θ₂ = R
Show answer and explanation

Answer: (B) tan θ₁ = R and tan θ₂ = 1/R

For Student A, the slope of the V versus I graph is ΔV/ΔI = tan θ₁ = R. For Student B, the slope of the I versus V graph is ΔI/ΔV = tan θ₂ = 1/R.

medium

11. In an Ohm's law experiment, the resistance of a wire is calculated as R = V / I. The measured values are V = (12.0 ± 0.3) V and I = (2.00 ± 0.02) A. The value of resistance R with absolute error limit is:

  1. (6.0 ± 0.3) Ω
  2. (6.0 ± 0.2) Ω
  3. (6.0 ± 0.1) Ω
  4. (6.0 ± 0.4) Ω
Show answer and explanation

Answer: (B) (6.0 ± 0.2) Ω

The nominal resistance is R = 12.0 / 2.00 = 6.0 Ω. The relative error is ΔR/R = ΔV/V + ΔI/I = 0.3/12.0 + 0.02/2.00 = 0.025 + 0.010 = 0.035. The absolute error is ΔR = 6.0 × 0.035 = 0.21 Ω ≈ 0.2 Ω. Thus, R = (6.0 ± 0.2) Ω.

hard

12. A student determines the resistance of an unknown wire using the voltmeter-ammeter method with a battery of emf 6.0 V and internal resistance 1.0 Ω. The ammeter has resistance R_A = 0.5 Ω and the voltmeter has resistance R_V = 1000 Ω. The voltmeter is placed directly across the unknown wire of true resistance R = 100 Ω. What is the percentage error in the determined resistance due to the loading effect of the voltmeter?

  1. +9.1%
  2. -9.1%
  3. -0.5%
  4. -1.0%
Show answer and explanation

Answer: (B) -9.1%

The voltmeter measures the potential drop V across the parallel combination of R and R_V. The equivalent resistance measured is R_eq = (R × R_V) / (R + R_V) = (100 × 1000) / (1100) = 10000 / 110 = 90.91 Ω. The percentage error in the determined resistance is [(R_measured - R_true) / R_true] × 100 = [(90.91 - 100) / 100] × 100 = -9.09% ≈ -9.1%.

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