Experimental Skills — NEET Physics MCQs with answers

Experimental Skills is a Class 11 Physics chapter in the NEET (UG) syllabus. NEET720 has 462 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Vernier callipers and screw gauge · Mechanics experiments · Thermal and fluid experiments · Vernier caliper and screw gauge practicals · Electrical experiments · Simple pendulum experiment (measurement of g) · Optics and electronics experiments · Coefficient of viscosity by Stokes' method · Specific heat capacity by method of mixtures · Ohm's law verification and metre bridge · Resonance tube (speed of sound in air) · Focal length of concave mirror / convex lens · Characteristic curves of a p-n junction diode · Refractive index using travelling microscope (real and apparent depth) · General experimental skills and error analysis · Young's modulus by Searle's method · Surface tension by capillary rise · Viscosity by Stokes' method · Specific heat by method of mixtures · Resistivity of a wire using a meter bridge · Resistance by voltmeter-ammeter method · Potentiometer to compare emfs · Potentiometer for internal resistance · Focal length by u-v method (concave mirror) · Focal length by u-v method (convex lens) · Refractive index using a travelling microscope · Refractive index using minimum deviation (prism) · Characteristics of a diode · Characteristics of a transistor · Speed of sound using resonance tube · Error analysis and percentage error · Use of a multimeter · Meter bridge principle (Wheatstone bridge) · Meter bridge · Calorimetry · Young's modulus · Refractive index · Vernier Callipers · Screw Gauge · Vernier/Screw Gauge

8 free Experimental Skills practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Electrical experiments

   In a circuit set up to verify Ohm's law, the ammeter is always connected in series with the resistor under test and the voltmeter is always connected in parallel across it. If these connections were reversed (ammeter in parallel, voltmeter in series), the most likely outcome is

   • A.no difference in circuit behaviour, since both instruments still measure the same circuit
   • B.the ammeter (very low resistance) placed in parallel would draw an extremely large current and could damage the ammeter or short the circuit, while the voltmeter (very high resistance) in series would block nearly all current, showing almost zero deflection on both meters
   • C.only the voltmeter reading would be affected; the ammeter would work normally either way
   • D.both instruments would still give correct readings, just with reversed sign conventions
   Show answer and explanation

   Answer: B. the ammeter (very low resistance) placed in parallel would draw an extremely large current and could damage the ammeter or short the circuit, while the voltmeter (very high resistance) in series would block nearly all current, showing almost zero deflection on both meters

   An ammeter has very low internal resistance and is designed to carry the full circuit current in series; placing it in parallel creates a near short-circuit. A voltmeter has very high resistance and is designed to draw negligible current in parallel; placing it in series would choke off the circuit current almost entirely.

   An ideal ammeter has (near) zero resistance so that, in series, it does not alter the current it measures. If instead connected in parallel across a resistor, its very low resistance would short out most of the circuit current through itself, causing an unsafe current surge and possible damage. A voltmeter, by contrast, has very high resistance so that, in parallel, it draws negligible current and does not disturb the circuit. If placed in series, its huge resistance would drastically reduce the circuit current, making both the ammeter and voltmeter read close to zero. This is exactly why correct wiring (ammeter in series, voltmeter in parallel) is essential, not just a convention.

   Common mistake: Treating the series/parallel placement of meters as an arbitrary convention rather than a consequence of their internal resistance

2. Question 2 · medium · Electrical experiments

   A student performing the metre bridge experiment ignores the small, roughly constant end resistance of the copper strips and connectors, treating the scale reading of the balance length as exact. This assumption introduces the largest percentage error in the deduced value of X when the balance point happens to fall

   • A.near either end of the wire (close to 0 cm or 100 cm), because the fixed end resistance then forms a proportionally larger fraction of the resistance in that short arm
   • B.exactly at the 50 cm mark, because the bridge wire is thinnest there
   • C.nowhere in particular, since the end resistance shifts every balance length reading by the same absolute amount, so percentage error is identical everywhere
   • D.only when a high-value resistance box (R > 100 Ω) is used, regardless of where the balance point falls
   Show answer and explanation

   Answer: A. near either end of the wire (close to 0 cm or 100 cm), because the fixed end resistance then forms a proportionally larger fraction of the resistance in that short arm

   A fixed end resistance contributes a fixed absolute error, but this becomes a much larger percentage of a short arm's resistance than of a long arm's — so error is worst when the balance point is near either end.

   The end resistance adds a small, roughly constant extra resistance to whichever arm's effective length is being measured. If the balance length l (or 100−l) is small — i.e., the null point is near one end — this same fixed extra resistance represents a much larger fraction of that short arm's true resistance, causing a large percentage error in the deduced X. When the balance point is near the middle, both arms are comparably long, so the fixed end resistance is a small percentage of either arm, and the error is minimized. This is exactly why the resistance box value is chosen (see the companion practice on choosing R) to keep the balance point away from the extreme ends of the wire.

   Common mistake: Assuming a fixed absolute error translates into a uniform percentage error regardless of the arm length it is compared against

3. Question 3 · easy · Electrical experiments

   In the Ohm's law verification experiment, before finally locating a stable, safe operating point, a high protective resistance is kept in series with the galvanometer-free ammeter-voltmeter circuit (or a high resistance is kept in the rheostat), and current is increased gradually rather than switching on full circuit current immediately. The main purpose of this practice is to

   • A.permanently increase the sensitivity of the ammeter and voltmeter
   • B.compensate for the resistance of the connecting wires used in the circuit
   • C.ensure the balance point of the circuit always occurs near the midpoint of the rheostat
   • D.limit the initial current through the circuit and protect the ammeter and voltmeter from damage due to a sudden large current or voltage
   Show answer and explanation

   Answer: D. limit the initial current through the circuit and protect the ammeter and voltmeter from damage due to a sudden large current or voltage

   Starting with high resistance in the circuit and increasing current gradually prevents a current surge that could damage the delicate ammeter and voltmeter coils.

   Ammeters and voltmeters have delicate coils that can be damaged by a sudden large current. By keeping the rheostat at its highest resistance setting when the circuit is first switched on, the initial current is kept small; the experimenter then gradually reduces the rheostat resistance (or increases current) while watching the meters, ensuring readings stay within safe limits at every step. This is a standard safety practice in all such circuit experiments, not a way to boost sensitivity, compensate wire resistance, or control any 'balance point' (which is not even a relevant concept in a simple Ohm's law circuit).

   Common mistake: Attributing this safety practice to an unrelated experimental goal such as sensitivity, wire-resistance compensation, or balance-point control

4. Question 4 · medium · Mechanics experiments

   In the simple pendulum experiment, students are instructed to time a large number of oscillations (e.g. 20) with a stopwatch and divide by that number, rather than timing a single oscillation directly. The main reason for this practice is

   • A.it makes the total measured time longer, which is easier to write down
   • B.it changes the pendulum's period to a more measurable value
   • C.the stopwatch's own least count becomes negligible when spread over many oscillations
   • D.the fixed reaction-time error in starting/stopping the stopwatch becomes a much smaller fraction of the total time, reducing the percentage error in the period
   Show answer and explanation

   Answer: D. the fixed reaction-time error in starting/stopping the stopwatch becomes a much smaller fraction of the total time, reducing the percentage error in the period

   A roughly constant absolute error from human reaction time contributes a much smaller percentage error when spread over a long total time for many oscillations.

   A stopwatch reading always carries a small, roughly fixed absolute error (from human reaction time in starting/stopping it), say ~0.2–0.3 s. If only a single oscillation (T ~ 2 s) is timed, this fixed error is a large fraction of T (~10-15%). By timing, say, 20 oscillations (total time ~40 s) and dividing by 20, the same fixed absolute error is now a much smaller fraction of the total time (~0.5-0.75%), so the percentage error in the derived single-oscillation period T is greatly reduced. This is a general error-reduction strategy: measure a large multiple of a small quantity to minimize the relative effect of a fixed instrumental/reaction-time error.

   Common mistake: Conflating stopwatch least count error with human reaction-time error as the main source being minimized

5. Question 5 · medium · Mechanics experiments

   In the metre-scale experiment to find its mass using the principle of moments, the scale used must be uniform (of constant cross-section and material) because

   • A.a non-uniform scale would have a different total mass regardless of the method used
   • B.uniformity ensures the scale balances only at the 50 cm mark, and nowhere else, even when unloaded incorrectly
   • C.uniformity is required so that the vernier callipers used to measure its length gives a correct reading
   • D.only for a uniform scale does its weight act exactly at the geometric centre (the 50 cm mark), simplifying the torque equation
   Show answer and explanation

   Answer: D. only for a uniform scale does its weight act exactly at the geometric centre (the 50 cm mark), simplifying the torque equation

   A uniform scale's weight acts exactly at its geometric centre, letting the experimenter use the known 50 cm mark directly in the torque balance equation.

   The moments-based method locates the scale's weight at a single known point to set up the torque balance equation: (mass hung) × (its distance from pivot) = (scale's weight) × (distance of scale's centre of gravity from pivot). For a uniform scale, symmetry guarantees the centre of gravity is exactly at the midpoint (50 cm mark), a known quantity requiring no separate measurement. For a non-uniform scale, the centre of gravity could lie anywhere and would first need to be located separately (e.g., by balancing it unloaded), adding an extra step — but the scale would still balance somewhere, just not necessarily at 50 cm, so option C is too strong.

   Common mistake: Confusing uniformity with total mass, or assuming a non-uniform scale would not balance at all

6. Question 6 · medium · Mechanics experiments

   In Searle's apparatus for determining Young's modulus, two identical wires (an experimental wire and a reference wire) are suspended side by side from the same rigid support. The main purpose of using a second, reference wire is to

   • A.cancel out errors from temperature changes and any yielding/sagging of the support, since both wires experience the same conditions and only their differential extension is measured
   • B.double the total load the apparatus can safely carry
   • C.allow two different Young's moduli to be measured simultaneously from the two wires
   • D.provide a spare wire in case the experimental wire breaks
   Show answer and explanation

   Answer: A. cancel out errors from temperature changes and any yielding/sagging of the support, since both wires experience the same conditions and only their differential extension is measured

   Both wires, hung from the same support, experience identical temperature changes and support sag; measuring the extension of the experimental wire relative to the (unchanging-load) reference wire cancels these common errors.

   If only a single wire were used, any change in room temperature would cause it to expand or contract, and any sagging of the rigid support under load would add a spurious extension — both would be mistaken for extension due to the applied load. By hanging an identical, unloaded (fixed-load) reference wire from the same support, both wires undergo the same temperature-driven length change and the same support movement. Measuring the extension of the experimental wire relative to the reference wire (using the Vernier arrangement between them) cancels out these common systematic effects, leaving only the extension caused by the added load.

   Common mistake: Believing the reference wire shares the experimental load or measures a separate modulus

7. Question 7 · medium · Mechanics experiments

   In the metre-scale moments experiment, the scale is balanced on a knife edge with a mass hung on one side. If a second, larger mass is now hung at some point on the same side as the first (without moving the knife edge or the first mass), the scale will

   • A.remain balanced automatically, since both masses are on the same side
   • B.tip down on the opposite side, since adding mass always shifts balance away from where it is added
   • C.tip down on the side where the masses were added, since the total anticlockwise (or clockwise) torque on that side now exceeds the opposing torque from the scale's own weight and any other loads
   • D.remain balanced as long as the total mass on the scale is less than the mass of the scale itself
   Show answer and explanation

   Answer: C. tip down on the side where the masses were added, since the total anticlockwise (or clockwise) torque on that side now exceeds the opposing torque from the scale's own weight and any other loads

   Adding torque on one side without a compensating change elsewhere unbalances the system, tipping it toward the side with the greater net torque.

   Balance about the knife edge requires the total clockwise torque to equal the total anticlockwise torque. Hanging an additional mass on the same side as an existing one increases the total torque on that side (its own weight times its distance from the pivot) without any compensating change on the other side, so the sum of torques is no longer balanced — the scale rotates and tips down on the side with the now-larger net torque, exactly as expected from the torque balance condition.

   Common mistake: Assuming balance is unaffected as long as masses are added, without considering the resulting torque imbalance

8. Question 8 · medium · Mechanics experiments

   In Searle's apparatus, the reference wire carries only a small, fixed load throughout the experiment (just enough to keep it taut), while all the additional weights are hung on the experimental wire. The small elongation of the experimental wire relative to the reference wire is measured using

   • A.an ordinary metre scale held alongside the two wires
   • B.a vernier callipers applied directly across the diameter of the experimental wire
   • C.a stopwatch, by timing how long the wire takes to stretch under the added load
   • D.a spirit level and micrometer screw arrangement mounted between the two wires, which detects and measures the small relative displacement precisely
   Show answer and explanation

   Answer: D. a spirit level and micrometer screw arrangement mounted between the two wires, which detects and measures the small relative displacement precisely

   Searle's apparatus uses a sensitive spirit-level-and-micrometer arrangement bridging the two wires to detect the tiny differential elongation with high precision.

   The elongation produced by typical loads in this experiment is only a fraction of a millimetre, far too small to read reliably with an ordinary scale. Searle's apparatus solves this using a frame connecting the two wires, fitted with a spirit level whose tilt (caused by the relative displacement of the two wires) is restored to horizontal by turning a calibrated micrometer screw; the screw's rotation, read against a circular scale, gives a precise measure of the elongation. The wire's diameter, by contrast, is measured separately beforehand using a screw gauge, not during the loading process.

   Common mistake: Assuming an ordinary scale or the diameter-measuring instrument is used to measure elongation

Questions about Experimental Skills for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (462 active practice questions in this chapter today).