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NEET Physics Numericals: A Step-by-Step Method with 10 Worked Examples
NEET physics numericals are solved most reliably with one fixed routine: read and list the data, sketch, convert to SI units, choose the formula and check its conditions, solve in symbols, substitute with sensible approximations, and check the answer against the options. The physics in NEET numericals is usually one or two steps deep. Most lost marks come from units, signs, conditions and arithmetic done without a calculator, and a routine catches those.
This guide is part of our NEET Physics tutor in Gurgaon series. It sets out the method, a mental-maths toolkit, 10 worked examples from across the syllabus, the traps our Home Tutors Team sees most often, and a five-question practice set with full solutions. Every number in the examples and the practice set has been checked by hand and again with a short computer script.
What NEET physics numericals look like
Every NEET physics numerical is a multiple-choice question with four options and one correct answer. According to the NEET (UG) 2026 Information Bulletin, the paper has 180 compulsory questions in 180 minutes, including 45 physics questions worth 180 marks. Each correct answer earns +4, each incorrect answer −1, and an unanswered question 0. The exam is in pen-and-paper mode, and calculators, slide rules and log tables are among the items not allowed in the hall.
Three things follow from those rules:
- Time is tight. 180 questions in 180 minutes is one minute per question on average across the whole paper. A numerical that takes five minutes costs time you need elsewhere.
- Arithmetic must be done by hand. Questions are usually set so that the numbers work out neatly, but only if you simplify in symbols first and use standard approximations.
- Wrong answers cost marks. A careless slip does not just lose 4 marks; it subtracts 1. Checking the answer before marking it matters more than solving one extra question.
The syllabus for these questions is the NMC's Syllabus for NEET (UG) 2026, with 20 physics units. No 2027 syllabus had been published when we checked on 29 September 2026, so the 2026 version is the reference for now.
The step-by-step method
We teach the same seven steps for every numerical. On easy questions they take seconds.
| Step | What to do | What it catches |
|---|---|---|
| 1. Read and list | Write the given quantities with symbols and units, and underline what is asked. Note words such as "not", "maximum", "ratio" and "just". | Answering a different question from the one asked |
| 2. Sketch | Draw a free-body diagram, circuit, ray diagram or energy-level diagram. Mark a sign convention. | Missing forces, wrong directions, sign errors |
| 3. Convert to SI | cm to m, g to kg, μF to F, °C to K, eV to J only if needed. Write the conversion beside the data. | Unit slips, the commonest single error |
| 4. Choose the formula and check its conditions | Name the principle (conservation of energy, Kirchhoff, lens formula) and check the condition: constant acceleration, ideal gas, series or parallel, battery connected or not. | Formulas used where they do not hold |
| 5. Solve in symbols | Rearrange for the unknown before putting in numbers. Look for ratios in which constants cancel. | Long arithmetic and rounding errors |
| 6. Substitute and approximate | Put numbers in once, using standard approximations (g = 10 m/s2 if allowed, π2 ≈ 10, hc ≈ 1240 eV nm). | Time lost to heavy calculation |
| 7. Check | Units of the answer, order of magnitude, a limiting case, and the options. If two options differ only by a factor of 2 or a sign, recheck that step. | Factor-of-2 and sign errors before they cost −1 |
Steps 1 to 4 are thinking; steps 5 to 7 are calculating; skipping straight to step 6 is the usual cause of avoidable errors. For the formulas and their conditions, keep our NEET physics formula sheet beside you while you practise.
Calculating without a calculator
Because calculators are not allowed, NEET physics numericals reward students who know a small set of approximations and use ratios instead of full values.
| Tool | Value or rule | Where it helps |
|---|---|---|
| g | Use the value the question gives; 10 m/s2 when stated, otherwise 9.8 | Mechanics, fluids, pendulums |
| π2 | ≈ 10 (true value about 9.87) | Pendulum and spring periods, g = 4π2l/T2 |
| Roots | √2 ≈ 1.414; √3 ≈ 1.732; √5 ≈ 2.236 | Projectiles, vectors, rms values |
| Photon energy | E (eV) ≈ 1240/λ (nm) | Photoelectric effect, spectra, LEDs |
| Hydrogen atom | En = −13.6/n2 eV | Transition energies |
| Binomial approximation | (1 + x)n ≈ 1 + nx for small x | Small changes, g at small heights, percentage change |
| Percentage change | If y ∝ xn, a small p% change in x gives about np% change in y | "If the length is increased by 2%..." questions |
| Powers of ten | Write every number as a × 10b before multiplying | Electrostatics, modern physics |
| Ratios | Write the formula for both cases and divide; constants cancel | "How many times..." and comparison questions |
The ratio method is the biggest time-saver. If a question asks how the period of a pendulum changes when its length is made four times larger, T ∝ √l gives T2/T1 = √4 = 2 without any value of g or π.
From our tutors: many NEET aspirants who are confident in biology tell us they are "bad at physics numericals". When we watch them solve, the physics is often fine; the trouble is arithmetic with powers of ten and fractions, done in the head and in a hurry. We give them ten minutes a day of calculation drills with no calculator, alongside the physics, and their accuracy usually improves within a few weeks.
Worked examples 1 to 6: mechanics and heat
These examples are written by our team in NEET style. They are not previous-year questions. Try each one before reading the solution.
Example 1: errors in measurement (Unit 1)
The density of a cube is found by measuring its mass and the length of its side. The maximum errors in the measurement of mass and length are 2% and 1% respectively. The maximum error in the density is: (a) 3% (b) 4% (c) 5% (d) 7%
Answer: (c) 5%. Step 4: density ρ = m/L3. Step 5: relative errors add, and a power multiplies its error, so Δρ/ρ = Δm/m + 3ΔL/L. Step 6: 2% + 3 × 1% = 5%. Trap: option (a) comes from forgetting the power 3; errors never subtract, even though L is in the denominator.
Example 2: vertical motion from a tower (Unit 2)
A ball is thrown vertically upwards at 20 m/s from the top of a tower 25 m high. How long does it take to reach the ground? (g = 10 m/s2) (a) 2 s (b) 4 s (c) 5 s (d) 6 s
Answer: (c) 5 s. Step 2: take upwards as positive, with the origin at the top of the tower. Then u = +20 m/s, a = −10 m/s2, and the ground is at s = −25 m. Step 5: −25 = 20t − 5t2, so t2 − 4t − 5 = 0, giving (t − 5)(t + 1) = 0. Step 7: reject the negative root, so t = 5 s. Check: the ball rises 20 m in 2 s, then falls 45 m from rest in 3 s (½ × 10 × 32 = 45 m); 2 + 3 = 5 s. Trap: taking s = +25 m, which gives no sensible root, or answering 4 s, the time to return to the top.
Example 3: two blocks over a pulley (Unit 3)
Blocks of 3 kg and 2 kg hang from the two ends of a light string passing over a smooth, light pulley. Find the acceleration of the blocks and the tension in the string. (g = 10 m/s2) (a) 2 m/s2, 24 N (b) 2 m/s2, 20 N (c) 10 m/s2, 30 N (d) 5 m/s2, 25 N
Answer: (a) 2 m/s2, 24 N. Step 2: draw a free-body diagram for each block; the 3 kg block moves down. Step 5: 3g − T = 3a and T − 2g = 2a. Adding gives a = (3 − 2)g/5 = 2 m/s2. Then T = 2(g + a) = 2 × 12 = 24 N. Step 7: check with the other block: 3(g − a) = 3 × 8 = 24 N. The tension lies between the two weights (20 N and 30 N), as it must. Trap: option (b) takes the tension equal to the lighter weight, which is true only if the system is at rest.
Example 4: stopping distance with friction (Unit 4)
A 2 kg block slides on a rough horizontal floor with an initial speed of 6 m/s. The coefficient of kinetic friction is 0.3. How far does it slide before stopping? (g = 10 m/s2) (a) 3 m (b) 6 m (c) 12 m (d) 18 m
Answer: (b) 6 m. Step 4: use the work–energy theorem, since friction does negative work: Wfriction = ΔK. Step 5: −μmg s = 0 − ½mv2, so s = v2/(2μg); the mass cancels. Step 6: s = 36/(2 × 0.3 × 10) = 36/6 = 6 m. Step 7: friction 6 N × 6 m = 36 J, the initial kinetic energy. Trap: option (c) forgets the 2 in the denominator.
Example 5: escape velocity on another planet (Unit 6)
A planet has the same mean density as the Earth but twice its radius. If the escape velocity from the Earth is 11.2 km/s, the escape velocity from the planet is: (a) 5.6 km/s (b) 11.2 km/s (c) 15.8 km/s (d) 22.4 km/s
Answer: (d) 22.4 km/s. Step 5: ve = √(2GM/R), and M = (4/3)πR3ρ, so ve = R√(8πGρ/3), which is proportional to R when the density is fixed. Step 6: doubling R doubles ve: 2 × 11.2 = 22.4 km/s. Trap: option (c) (about 11.2 × √2) comes from using ve = √(2gR) with g unchanged, but g = (4/3)πGρR also doubles. If the mass were held fixed instead, ve would fall to about 7.9 km/s. Read whether the mass, the density or g is held constant.
Example 6: rms speed and temperature (Unit 9)
The temperature of an ideal gas is 27 °C. To what temperature must it be heated so that the rms speed of its molecules doubles? (a) 54 °C (b) 108 °C (c) 927 °C (d) 1200 °C
Answer: (c) 927 °C. Step 3: convert to kelvin: 27 °C = 300 K. Step 5: vrms = √(3RT/M), so vrms ∝ √T; doubling the speed needs four times the absolute temperature. Step 6: T = 4 × 300 = 1200 K = 927 °C. Trap: option (b) multiplies the Celsius temperature by 4, and option (d) is the right number in the wrong unit. Both are placed there on purpose.
Does your child understand the physics but still lose marks in numericals? Book a free NEET Physics demo class in Gurgaon. The tutor will watch a few questions being solved and show exactly which step is going wrong.
Book a Free NEET Physics Demo +91 92204 75088Worked examples 7 to 10: electricity, optics and modern physics
Example 7: capacitors in series (Unit 11)
Capacitors of 2 μF and 3 μF are connected in series across a 10 V battery. The potential difference across the 2 μF capacitor is: (a) 4 V (b) 5 V (c) 6 V (d) 10 V
Answer: (c) 6 V. Step 4: in series, both capacitors carry the same charge. Step 5: Ceq = (2 × 3)/(2 + 3) = 1.2 μF, so Q = CeqV = 1.2 × 10 = 12 μC. Step 6: V2 = Q/C = 12/2 = 6 V and V3 = 12/3 = 4 V. Step 7: 6 + 4 = 10 V, the battery voltage. Trap: option (a) assumes the larger voltage goes across the larger capacitor. In series, the smaller capacitor takes the larger share. Here there is no need to convert μF to F, because the μ cancels in Q/C.
Example 8: charged particles in a magnetic field (Unit 13)
A proton and an alpha particle enter the same uniform magnetic field perpendicular to it, with equal kinetic energies. The ratio of the radii of their circular paths, rp : rα, is: (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : √2
Answer: (a) 1 : 1. Step 5: r = mv/(qB), and mv = √(2mK), so r = √(2mK)/(qB) ∝ √m/q for the same K and B. Step 6: for the proton √m/q = √1/1 = 1; for the alpha particle √4/2 = 1 (mass 4 units, charge 2 units). So the ratio is 1 : 1. Trap: option (b) is the answer for equal speeds, not equal kinetic energies. Read which quantity is equal: speed, momentum, kinetic energy or accelerating voltage each gives a different ratio.
Example 9: image by a convex lens (Unit 16)
An object is placed 30 cm in front of a convex lens of focal length 20 cm. The image is: (a) virtual, 60 cm from the lens, magnification +2 (b) real, 60 cm from the lens, magnification −2 (c) 12 cm from the lens, magnification +0.4 (d) real, 60 cm from the lens, magnification +2
Answer: (b). Step 2: Cartesian convention, light travelling left to right: u = −30 cm, f = +20 cm. Step 5: 1/v − 1/u = 1/f gives 1/v = 1/20 − 1/30 = 1/60, so v = +60 cm. Step 6: m = v/u = 60/(−30) = −2. Step 7: a positive v means a real image on the far side; a negative m means inverted and magnified, which matches an object between f and 2f. Trap: option (c) comes from putting u = +30 cm, dropping the sign (1/v = 1/20 + 1/30 gives v = 12 cm). Enter every sign before solving.
Example 10: photoelectric effect (Unit 17)
Light of wavelength 310 nm falls on a metal with a work function of 2.5 eV. The stopping potential is: (a) 1.5 V (b) 2.5 V (c) 4.0 V (d) 6.5 V
Answer: (a) 1.5 V. Step 6: photon energy E ≈ 1240/310 = 4.0 eV. Step 5: Kmax = E − φ = 4.0 − 2.5 = 1.5 eV, and eV0 = Kmax, so V0 = 1.5 V. Step 7: the threshold wavelength is 1240/2.5 = 496 nm; 310 nm is shorter, so emission does occur. Trap: option (c) gives the photon energy and (d) adds the work function instead of subtracting it. Working in eV throughout avoids converting to joules at all.
For more practice with light and atoms, our guide to NEET physics tips and tricks collects short methods that work across units.
Common traps in NEET physics numericals
In our experience, these traps account for most lost marks in NEET physics numericals. Each has a simple fix.
| Trap | Example | Fix |
|---|---|---|
| Mixed units | Using cm with SI constants; molar mass 32 instead of 0.032 kg/mol | Convert everything in step 3, before any formula |
| Celsius in gas laws | Doubling "27 °C" to 54 °C | Kelvin in every thermodynamics and kinetic theory formula |
| Sign conventions | Ball thrown up from a tower; mirror and lens distances | Mark the positive direction on the sketch and enter every sign |
| Formula outside its conditions | Equations of motion with changing acceleration; g(1 − 2h/R) for large h | Say the condition aloud in step 4 |
| Peak versus rms | Using 220 V as the peak of household AC | AC values are rms unless the question says peak or amplitude |
| Series versus parallel | Capacitors combined like resistors | Capacitors in series add as reciprocals, resistors in series add directly |
| Which quantity is equal | Equal speed, momentum or kinetic energy in ratio questions | Underline the equal quantity in step 1 |
| Static friction taken as μsN | Block at rest under a small push | Find the friction needed first; compare with μsN |
| "Not" and "incorrect" | Choosing the first true statement in a "which is incorrect" question | Circle negative words before reading options |
| Early rounding | Rounding √3 to 2 in the first line | Keep symbols until the last step; round once |
Keep an error log with one line per mistake: the question, the wrong step, and the trap from this table. Within a few weeks, two or three traps usually stand out; drill those.
Using the options, and when to skip
NEET options are designed to catch predictable mistakes, as the worked examples show. Our NEET physics tips and tricks guide covers each check in detail; here is the short version. Use them in three ways:
- Units and dimensions. If an option has the wrong dimensions, cross it out straight away.
- Limiting cases. Put an extreme value into each option: θ = 0, a very large mass, zero friction. An option that gives a physically impossible result in the limit is wrong.
- Order of magnitude. A rough estimate often rules out two options without full calculation.
On skipping: with +4 for a correct answer and −1 for a wrong one, the expected score of a guess depends on how many options you have eliminated. This is simple arithmetic, not a strategy we guarantee:
| Options left after elimination | Chance of a correct guess | Expected marks from guessing |
|---|---|---|
| 4 (no elimination) | 1 in 4 | +4 × ¼ − 1 × ¾ = +0.25 |
| 3 | 1 in 3 | +4 × ⅓ − 1 × ⅔ ≈ +0.67 |
| 2 | 1 in 2 | +4 × ½ − 1 × ½ = +1.5 |
A blind guess gains very little on average and adds risk; most of the gain comes from elimination, which comes from knowing the physics. If a numerical has taken more than about two minutes with no clear path, mark it, move on, and return if time allows.
Practice set: 5 questions with solutions
These five questions are written by our team for practice. They are not NEET previous-year questions. Solve all five before reading the solutions, and time yourself: aim for about two minutes each.
Questions
- Current electricity (Unit 12). A cell of emf 12 V and internal resistance 1 Ω is connected to a 5 Ω resistor. Find the current, the terminal voltage and the power delivered to the resistor.
- Alternating current (Unit 14). A series LCR circuit has R = 30 Ω, XL = 80 Ω and XC = 40 Ω, connected to a 200 V (rms) supply. Find the impedance, the rms current, the power factor and the average power.
- Atoms (Unit 18). An electron in a hydrogen atom falls from n = 3 to n = 2. Find the energy of the emitted photon and its approximate wavelength.
- Oscillations (Unit 10). A 0.2 kg mass on a spring of constant 20 N/m oscillates with an amplitude of 5 cm. Find the angular frequency, the maximum speed, the total energy, and the speed when the displacement is 3 cm.
- Thermodynamics (Unit 8). Two moles of a monatomic ideal gas are heated at constant pressure so that the temperature rises by 50 K. Find the work done by the gas, the change in internal energy and the heat supplied. (R = 8.3 J mol−1 K−1)
Solutions
Solution 1: current, terminal voltage and power
I = 2 A; V = 10 V; P = 20 W. I = E/(R + r) = 12/(5 + 1) = 2 A. Terminal voltage V = E − Ir = 12 − 2 × 1 = 10 V (check: IR = 2 × 5 = 10 V). Power in the resistor = I2R = 4 × 5 = 20 W. Check: the cell supplies EI = 24 W, of which 4 W is lost in the internal resistance. Trap: using V = 12 V across the resistor, which ignores the internal resistance and gives 28.8 W.
Solution 2: series LCR circuit
Z = 50 Ω; I = 4 A; cosφ = 0.6; P = 480 W. Z = √[302 + (80 − 40)2] = √(900 + 1600) = √2500 = 50 Ω. Irms = 200/50 = 4 A. Power factor = R/Z = 30/50 = 0.6. Average power = VrmsIrmscosφ = 200 × 4 × 0.6 = 480 W. Check: I2R = 16 × 30 = 480 W, because only the resistor dissipates power. Note that VL = 320 V is larger than the supply voltage; that is normal in LCR circuits, because VL and VC partly cancel.
Solution 3: hydrogen transition from n = 3 to n = 2
E ≈ 1.89 eV; λ ≈ 656 nm. E = 13.6(1/22 − 1/32) = 13.6 × (1/4 − 1/9) = 13.6 × 5/36 ≈ 1.89 eV. λ ≈ 1240/1.89 ≈ 656 nm, in the visible region (red), as expected for a Balmer line. Trap: using 13.6(1/3 − 1/2), without squaring n, or reversing the order and getting a negative energy.
Solution 4: spring–mass oscillation
ω = 10 rad/s; vmax = 0.5 m/s; E = 0.025 J; v = 0.4 m/s at x = 3 cm. ω = √(k/m) = √(20/0.2) = √100 = 10 rad/s. vmax = ωA = 10 × 0.05 = 0.5 m/s. E = ½kA2 = ½ × 20 × 0.0025 = 0.025 J. At x = 3 cm, v = ω√(A2 − x2) = 10 × √(25 − 9) cm/s = 10 × 4 = 40 cm/s = 0.4 m/s. Check: ½mvmax2 = ½ × 0.2 × 0.25 = 0.025 J. Trap: leaving A in centimetres in ½kA2, which gives an energy 10,000 times too large.
Solution 5: isobaric heating of a monatomic gas
W = 830 J; ΔU = 1245 J; Q = 2075 J. At constant pressure, W = PΔV = nRΔT = 2 × 8.3 × 50 = 830 J. For a monatomic gas Cv = (3/2)R, so ΔU = nCvΔT = 2 × 1.5 × 8.3 × 50 = 1245 J. By the first law, Q = ΔU + W = 1245 + 830 = 2075 J, which matches nCpΔT with Cp = (5/2)R. Trap: using Cp for ΔU. The change in internal energy of an ideal gas is always nCvΔT, whatever the process.
From our tutors: when a student gets a practice question wrong, we do not let them simply read the solution. We ask them to find the exact step of the seven where they went off track and write it in the error log. It is slower in the first week, but students who do this begin to catch their own mistakes during the exam, which is where the marks are.
How to practise numericals week by week
A simple routine for students preparing alongside Class 11 or 12 school work:
- After each chapter, solve 15 to 20 NCERT-level numericals using all seven steps written out in full, even the easy ones.
- Then move to previous-year questions for that chapter, timed at about two minutes each. Our guide to NEET physics PYQs explains how to use them without simply memorising answers.
- Every week, do one mixed set of 20 to 25 physics questions from chapters already finished, in one sitting and without a calculator.
- Review the error log at the end of the week and drill the two traps that appeared most often.
- In the final months, practise full 45-question physics sections inside full-length mock papers, so that physics timing is tested alongside chemistry and biology.
Students in Gurgaon and across Gurugram often have heavy Class 12 school schedules; two focused numerical sessions a week with an error log are worth more than daily rushed ones.
Frequently asked questions
How many physics questions are there in NEET?
According to the NEET (UG) 2026 Information Bulletin, there are 45 physics questions worth 180 marks, within a paper of 180 compulsory questions in 180 minutes. Each correct answer earns +4 and each incorrect answer −1. Check the bulletin for the year you are appearing, as NTA can change the pattern.
Is a calculator allowed for NEET physics numericals?
No. The NEET (UG) 2026 Information Bulletin lists calculators, slide rules and log tables among items not allowed in the examination hall. Numericals are solved by hand, which is why simplifying in symbols and using approximations such as π2 ≈ 10 and hc ≈ 1240 eV nm matter so much.
How can I solve NEET physics numericals faster?
Solve in symbols before substituting, use ratios so that constants cancel, and learn a small set of standard approximations. Speed comes from reducing arithmetic, not from skipping the thinking steps.
Should I attempt a physics numerical if I am not sure?
On the arithmetic of +4 and −1, a blind guess gains only +0.25 marks on average, while eliminating two options raises that to +1.5. So attempt when you can rule out wrong options with physics, and skip genuine blind guesses unless you have decided on a different policy in your mock tests.
Which units have the most numericals in NEET physics?
Mechanics, electrostatics, current electricity, optics and modern physics are all calculation-heavy in the syllabus, while units such as electromagnetic waves and parts of electronic devices are more factual. We do not give question counts here; our method is to practise numericals in every unit and let your own mock-test errors decide the priority.
Are the worked examples on this page previous-year NEET questions?
No. All ten worked examples and the five practice questions were written by our team in NEET style, and every answer has been checked by hand and with a computer script. Use previous-year papers from official sources for real exam questions.
Can a home tutor help with NEET physics numericals?
Yes. A one-to-one tutor can watch you solve and see exactly which step goes wrong, which is hard to spot in a large class. Ajay Vatsyayan Classes offers NEET Physics tuition at home across Gurgaon (Gurugram) and online, with male and female tutors available.
Want a tutor to build this method into your child's NEET physics practice? Book a free NEET Physics demo with Ajay Vatsyayan Classes, Saraswati kunj II, Wazirabad, Sector 52, Gurugram, Haryana 122003. Male and female tutors are available, at home or online.
Book a Free NEET Physics Demo +91 92204 75088