Plot voltage V (voltmeter, parallel) vs current I (ammeter, series) for a fixed wire. A straight line through origin confirms Ohm's law; slope = R. Account for ammeter (low) and voltmeter (high) resistances and lead resistance for low-R wires.
-- NCERT Physics Lab Manual Class 12, Part 2, p. 23Experiment 11 Ohm's Law Resistance
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Experiment 11 Ohm's Law Resistance, explained for NEET
The costly habit in this experiment is treating a single pair of meter readings as the answer. One point gives V/I, not the slope of the line, and the two agree only if the graph passes exactly through the origin. Any small offset — a meter that does not rest on zero, a contact resistance at a terminal — tilts the single-point ratio while leaving the slope of the straight line intact.
Ohm's law states that the current through a conductor is proportional to the potential difference across it, provided temperature and other physical conditions stay constant. The constant of proportionality is the resistance, R = V/I, in ohms. The NCERT Physics Lab Manual Class 12, Part 2, page 24 sets out the circuit: the ammeter in series with the wire so the same current passes through both, the voltmeter across the wire so it reads the potential difference across that wire alone, and a rheostat to step the current through several values.
Take at least five well-spaced (V, I) pairs, plot V on one axis and I on the other, and draw the best straight line. Resistance is the slope of that line, read between two widely separated points on the line — not between two raw data points, and not from one reading.
The condition "temperature constant" is doing real work. Push the current high and the wire warms; a metal's resistance rises with temperature, the plot bends, and the ratio V/I climbs from point to point. Keep currents modest and the key closed only while a reading is taken.
Watch out for precision inflation: if the voltmeter resolves 0.1 V and the ammeter 0.01 A, a calculator's 4.8387 Ω is not a result. Round to the digits your meters can actually support.
Can you answer these Experiment 11 Ohm's Law Resistance MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Ohm's law, as stated for the wire used in this experiment, holds on the condition that:
Show answer and why every option is right or wrong
Answer: C. Option C. The law is stated with the proviso that temperature and other physical conditions of the conductor do not change, as given in the NCERT Physics Lab Manual Class 12, Part 2, page 24.
Why A is wrong: A is incorrect: no current limit appears in the statement of the law; low currents are used only as a practical way of keeping the wire cool.
Why B is wrong: B is incorrect: the ohmic behaviour tested here belongs to metallic conductors, while semiconductors are the standard example of non-ohmic behaviour.
Why D is wrong: D is incorrect: the duration of application is irrelevant to the law itself; short closures of the key are a precaution against heating, not a condition of the law.
MCQ 2Easy RecallPractice
In the standard circuit for finding the resistance of a wire by Ohm's law, the meters are connected so that:
Show answer and why every option is right or wrong
Answer: A. Option A. The ammeter must carry the same current as the wire and so goes in series, while the voltmeter must sample the potential difference across the wire alone and so goes in parallel, as the circuit diagram in the NCERT Physics Lab Manual Class 12, Part 2, page 24 shows.
Why B is wrong: B is incorrect: a voltmeter placed in series carries almost no current through its high resistance and would choke the circuit instead of measuring anything useful.
Why C is wrong: C is incorrect: the two connections are interchanged, and a low-resistance ammeter placed across the wire short-circuits it.
Why D is wrong: D is incorrect: an ammeter in parallel with the wire does not read the wire's current, and its low resistance diverts nearly all the current away from the wire.
MCQ 3Easy RecallPractice
In this experiment the rheostat is included in the circuit in order to:
Show answer and why every option is right or wrong
Answer: D. Option D. The rheostat steps the current through a series of values so that a set of (V, I) pairs can be plotted, as described in the NCERT Physics Lab Manual Class 12, Part 2, page 25.
Why A is wrong: A is incorrect: polarity protection comes from connecting each meter's positive terminal to the positive side of the supply, not from a rheostat.
Why B is wrong: B is incorrect: the rheostat controls current, not temperature; the wire is kept cool by using small currents and closing the key only while a reading is taken.
Why C is wrong: C is incorrect: no series component can raise a cell's emf, and the rheostat only changes how the available emf is shared.
MCQ 4Direct ApplicationPractice
A student's plot of potential difference against current for the wire is a straight line through the origin, and one point on that line is at 0.80 V and 0.40 A. The resistance of the wire is:
Show answer and why every option is right or wrong
Answer: B. Option B. The slope of the V-against-I line is the resistance, so R = 0.80 V / 0.40 A = 2.0 Ω, following the graphical method of the NCERT Physics Lab Manual Class 12, Part 2, page 26.
Why A is wrong: A is incorrect: it inverts the ratio, dividing current by potential difference, which gives the conductance in siemens rather than the resistance.
Why C is wrong: C is incorrect: it misplaces the decimal point in the current, dividing 0.80 V by 0.040 A instead of 0.40 A.
Why D is wrong: D is incorrect: the two readings have been multiplied instead of divided, and 0.80 × 0.40 = 0.32 is a power in watts, not a resistance.
MCQ 5Direct ApplicationPractice
A voltmeter resolving 0.1 V reads 1.5 V across the wire while an ammeter resolving 0.01 A reads 0.30 A through it. The resistance should be reported as:
Show answer and why every option is right or wrong
Answer: D. Option D. Both readings carry two significant figures, so the quotient is reported to two significant figures as 5.0 Ω, per the reporting rule stated in the NCERT Physics Lab Manual Class 12, Part 2, page 27.
Why A is wrong: A is incorrect: writing a bare 5 drops a digit the meters genuinely supplied and understates the precision achieved.
Why B is wrong: B is incorrect: three significant figures claim a precision neither meter can deliver.
Why C is wrong: C is incorrect: four significant figures simply copy the calculator display, and the extra digits have no measurement behind them.
MCQ 6Direct ApplicationPractice
With the circuit open, the ammeter pointer rests at 0.02 A instead of at zero. During a trial it reads 0.42 A. The current to be used in the calculation is:
Show answer and why every option is right or wrong
Answer: A. Option A. A positive offset of 0.02 A is subtracted from every reading, giving 0.42 A − 0.02 A = 0.40 A before R = V/I is applied, as the precautions in the NCERT Physics Lab Manual Class 12, Part 2, page 25 require.
Why B is wrong: B is incorrect: the offset has been added, and adding applies only when the resting pointer sits below zero, not above it.
Why C is wrong: C is incorrect: it uses the raw reading and leaves the instrument's offset inside every calculated resistance.
Why D is wrong: D is incorrect: halving the reading has no basis here, because the correction is an offset rather than a scale factor.
MCQ 7CalculationPractice
A student's best straight line on the V-against-I graph passes through the points (0.20 A, 1.20 V) and (0.50 A, 2.40 V), and its extension meets the voltage axis a little above the origin. The best estimate of the wire's resistance from this graph is:
Show answer and why every option is right or wrong
Answer: C. Option C. Resistance is the slope, so R = (2.40 − 1.20) V / (0.50 − 0.20) A = 1.20 / 0.30 = 4.0 Ω, which is the value the graphical method of the NCERT Physics Lab Manual Class 12, Part 2, page 26 is designed to recover when the line misses the origin.
Why A is wrong: A is incorrect: it takes V/I at the first point alone, so the non-zero intercept is carried straight into the answer.
Why B is wrong: B is incorrect: it takes V/I at the second point alone, and the single-point ratio drifts with current whenever the line misses the origin.
Why D is wrong: D is incorrect: averaging the two single-point ratios still averages in the offset instead of cancelling it, which is what taking a difference does.
MCQ 8CalculationPractice
A student records 2.00 V at 0.50 A early in the run, then keeps the key closed for some minutes and records 4.40 V at 1.00 A with the wire noticeably warm. The pair of results is best explained by:
Show answer and why every option is right or wrong
Answer: B. Option B. The first pair gives 4.00 Ω and the second 4.40 Ω; the wire warmed under the sustained current and a metal's resistance rises with temperature, which is exactly the condition the NCERT Physics Lab Manual Class 12, Part 2, page 24 attaches to the law.
Why A is wrong: A is incorrect: a meter offset shifts readings by a fixed amount and would not raise the computed ratio in step with a visible rise in temperature.
Why C is wrong: C is incorrect: the law is simply being applied outside its stated condition here, and once the temperature is held steady metallic wires plot as straight lines through the origin.
Why D is wrong: D is incorrect: a voltmeter in series would have collapsed the current to a negligible value rather than allowing a reading of 1.00 A.
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How do you solve a Experiment 11 Ohm's Law Resistance question? A worked example
- 1
Given.
A wire is connected in series with an ammeter, a key, a cell and a rheostat, with a voltmeter across the wire. Voltmeter least count 0.1 V, ammeter least count 0.01 A. Four trials give (I, V) = (0.10 A, 0.5 V), (0.20 A, 1.0 V), (0.30 A, 1.4 V), (0.40 A, 2.0 V). With the circuit open the ammeter rests at 0.01 A.
- 2
Required.
The resistance of the wire, reported to a precision the meters support.
- 3
Concept.
For a conductor at constant temperature, V is proportional to I and the constant is R. Plotting V against I and taking the slope uses every trial and cancels a constant offset, which a single V/I ratio cannot.
- 4
Formula.
R = V/I, and graphically R = ΔV / ΔI between two widely separated points on the best straight line.
- 5
Substitution.
Correct each current for the 0.01 A offset: 0.09, 0.19, 0.29, 0.39 A. The best straight line through the four points passes close to (0.09 A, 0.5 V) and (0.39 A, 2.0 V). So ΔV = 2.0 V − 0.5 V = 1.5 V and ΔI = 0.39 A − 0.09 A = 0.30 A.
- 6
Calculation.
R = 1.5 / 0.30 = 5.0 Ω. The offset 0.01 A is read directly off the same scale as the currents, and the count of two points used for the slope is a counting integer (exact); neither contributes to the significant-figure count.
- 7
Final answer.
R = 5.0 Ω — two significant figures, matching the two-figure voltmeter readings. Writing 5.000 Ω would claim precision the 0.1 V scale cannot supply.
- 8
Common trap.
Taking the third trial alone gives 1.4 / 0.29 = 4.8 Ω and the first gives 0.5 / 0.09 = 5.6 Ω. A student who quotes either believes the wire's resistance changed between readings. It did not; the scatter is measurement noise, and the slope is what averages it out.
- 9
Similar NEET-style question.
A V-against-I plot for a wire is a straight line cutting the V axis at 0.20 V and passing through (0.50 A, 2.70 V). What resistance does the graph give, and why does the single-point ratio at that reading overstate it?
What to remember before solving Experiment 11 Ohm's Law Resistance questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 11 Ohm's Law Resistance questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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