In this guide
- A five-step routine for every numerical
- Technique 1: the ratio method
- Technique 2: dimensional checks
- Technique 3: limiting cases
- Technique 4: approximate when the options allow it
- Technique 5: pick the shortest principle
- Technique 6: eliminate, then commit
- Standard results to know without deriving
- Practice set: use the fastest method
- What to do next
NEET gives you 180 minutes for 180 questions. Biology questions are often answered in well under a minute each, which leaves some extra time for physics. Even so, a physics numerical that takes three minutes costs you somewhere else in the paper. Most candidates who run short of time in physics are not slow at arithmetic. They are slow at deciding what to do, and they do full calculations where a ratio or a quick check would settle the answer.
This guide is about method. Each technique below is shown on a problem of the kind NEET asks, so you can see where it saves time.
A five-step routine for every numerical
- Read the last line first. Know exactly what is asked: speed or velocity, the ratio or the value, the change or the final value.
- List the givens with units. Write "m = 2 kg, v = 36 km/h" on the rough sheet. This is where you catch km/h, μF, mA and cm.
- Name the principle before the formula. Conservation of energy? Momentum? Newton's second law on one body? The principle decides the formula, not the other way round.
- Convert units, then substitute. Use SI unless the whole question sits comfortably in another system (eV and nm in modern physics, for example).
- Check the answer is sensible. Right unit, right order of magnitude, right sign. A car moving at 3,000 m/s means something went wrong.
Steps 1 and 5 take a few seconds each, and they prevent most careless mistakes.
Technique 1: the ratio method
Many NEET questions ask how a quantity changes when another is doubled, halved or multiplied. You don't need the full formula, only the proportionality.
Example. A satellite's orbital radius is made four times larger. What happens to its period?
- Kepler's third law: T² ∝ r³, so T ∝ r³ᐟ², i.e. T ∝ r√r.
- New T = (√4)³ × old T = 2³ × old T = 8 × old T.
This takes ten seconds. Computing both periods from GM would take two minutes and invite errors.
Watch what stays constant. If a wire is stretched to twice its length, its volume is unchanged, so the area halves. R = ρl/A becomes ρ(2l)/(A/2) = 4R, not 2R. The question of what is held constant is where ratio questions are won or lost.
Technique 2: dimensional checks
If the options differ in form, check dimensions before you calculate anything.
Example. The speed of a transverse wave on a string of tension T and mass per unit length μ is: (a) √(T/μ) (b) √(Tμ) (c) T/μ (d) √(μ/T).
- [T] = MLT⁻², [μ] = ML⁻¹.
- T/μ has dimensions L²T⁻², so √(T/μ) has LT⁻¹, a speed.
- Only (a) is dimensionally correct.
A dimensional check can't tell you about pure numbers such as 2π or ½, but it removes wrong options in seconds.
Technique 3: limiting cases
Push a variable to an extreme where you know the answer, and see which option survives.
Example. Two masses m₁ > m₂ hang over a light, frictionless pulley. Which expression can be the acceleration?
(a) (m₁ − m₂)g/(m₁ + m₂) (b) (m₁ + m₂)g/(m₁ − m₂) (c) m₁g/(m₁ + m₂) (d) (m₁ − m₂)g/m₂
- If m₁ = m₂, the system balances, so a = 0. This rules out (b), which blows up, and (c), which does not vanish.
- If m₂ = 0, m₁ falls freely, so a = g. (d) blows up; (a) gives g.
- The answer is (a).
Technique 4: approximate when the options allow it
If the options are far apart, don't grind decimals. Standard shortcuts: g = 10 m s⁻² (unless 9.8 is specified), π² ≈ 10, √2 ≈ 1.41, √3 ≈ 1.73.
Example. Find the period of a simple pendulum 1 m long.
- T = 2π√(l/g) = 2π√(1/10) = 2π/√10.
- Since √10 ≈ π, T ≈ 2π/π = 2 s.
If the options are close together (say 1.98 s and 2.01 s), then do the exact arithmetic. Look at the spacing of the options before you choose.
Technique 5: pick the shortest principle
The same problem can often be solved several ways. Energy methods skip time and direction; momentum skips forces during collisions.
Example. A ball is dropped from 20 m. Find its speed on reaching the ground.
- Kinematics works, but energy is one line: ½mv² = mgh, so v = √(2gh) = √(2 × 10 × 20) = 20 m/s.
- The same energy line answers "speed at the bottom of a frictionless curved track" directly, where the kinematics equations do not apply at all because the acceleration is not constant.
Technique 6: eliminate, then commit
With +4 for a correct answer and −1 for a wrong one, removing two options and choosing between the remaining two is often a reasonable bet. The decision to skip is personal, and the negative marking guide sets out the maths. What matters here is that elimination techniques (dimensions, limiting cases, sign, order of magnitude) often leave only one option standing, and then it is not a guess at all.
Standard results to know without deriving
| Situation | Result |
|---|---|
| Max height and range of a projectile | H = u² sin²θ/2g, R = u² sin 2θ/g; R is greatest at 45° |
| Resistors in parallel, two only | R₁R₂/(R₁ + R₂) |
| Capacitors in series, two only | C₁C₂/(C₁ + C₂) |
| Springs in series | 1/k = 1/k₁ + 1/k₂ |
| Kinetic energy and momentum | K = p²/2m |
| Escape speed | v_e = √(2gR) ≈ 11.2 km/s for Earth |
| Photon energy | E (eV) ≈ 1240/λ (nm) |
| Bohr energy levels | Eₙ = −13.6 Z²/n² eV |
| Elastic collision, equal masses, one at rest | Velocities exchange |
Write these on your formula sheet and revise them weekly. The NEET physics formula sheet has the full list with conditions.
Practice set: use the fastest method
- The momentum of a body doubles. Its kinetic energy becomes: (a) 2 times (b) 4 times (c) √2 times (d) unchanged
- The mass on a spring is made four times larger. The period of oscillation becomes: (a) 4 times (b) 2 times (c) ½ (d) unchanged
- A 2 μF capacitor is charged to 100 V. The energy stored is: (a) 0.01 J (b) 0.02 J (c) 100 J (d) 10⁴ J
- A projectile is launched at 20 m/s (g = 10 m s⁻²). Its greatest possible range is: (a) 20 m (b) 40 m (c) 80 m (d) 400 m
- Resistors of 6 Ω and 3 Ω are in parallel. The combination is: (a) 9 Ω (b) 4.5 Ω (c) 2 Ω (d) 0.5 Ω
- A planet has the same density as Earth but twice its radius. Its escape speed is: (a) the same (b) √2 times (c) 2 times (d) 4 times
- Which has the dimensions of energy? (a) hν (b) h/ν (c) hλ (d) h/λ
- A wire of resistance R is stretched to three times its length. The new resistance is: (a) 3R (b) 6R (c) 9R (d) R/3
Answers
- (b) K = p²/2m, so K ∝ p².
- (b) T ∝ √m, and √4 = 2.
- (a) ½ × 2 × 10⁻⁶ × (100)² = 10⁻⁶ × 10⁴ = 0.01 J.
- (b) Maximum R = u²/g = 400/10 = 40 m, at 45°.
- (c) 6 × 3/(6 + 3) = 18/9 = 2 Ω.
- (c) v_e = √(2GM/R) and M ∝ R³ at fixed density, so v_e ∝ R.
- (a) E = hν, the photon energy.
- (c) Volume fixed: l → 3l, A → A/3, so R → 9R.
What to do next
- For one week, write the principle (energy, momentum, Newton, conservation of charge) beside every numerical before you solve it.
- Do 20 mixed numericals a day, timed, and note which chapters take the longest.
- Keep a "wrong because" column in your error log, and read the common slips in physics mistakes in NEET.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the National Testing Agency website .
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