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Class 9 Science formula sheet 2026-27: motion, force, energy, sound, atoms, with units
This sheet puts the formulas, units and key numbers of the new NCERT Class 9 Science book, Exploration, in one place. Physics comes first (motion, force, work and energy, sound), then chemistry (solutions, the atom, formulae and masses), then the few numbers that the cell and Earth chapters give. Every line names the textbook page, so you can go back to the chapter. At the end are 8 worked questions of our own and three common mistakes.
To print it, use your browser's Print option; the page prints without colours.
The layout, wording and questions on this page are our own, drafted with AI help. Every formula and number was checked against the NCERT page given next to it, and every answer was re-computed by a script, but no practising teacher has reviewed the page yet. If you find a mistake, please tell us.
Page numbers are the printed page numbers of NCERT Exploration, Textbook of Science for Grade 9 (First Edition, April 2026). Chapters 1, 3, 11 and 12 have no formulas, so they do not appear below.
Which formulas carry the most marks
CBSE's Class IX Science curriculum for 2026-27 splits the 80-mark annual paper into four units (PDF page 4). Motion, Force, Work and Sound (Chapters 4, 6, 7 and 10) is worth 23 marks, and Matter: Its Nature and Behaviour (Chapters 5, 8 and 9) is worth 25. Almost every formula on this sheet sits in those two units. World of Living (27 marks) and Earth as a System (5 marks) use only a few numbers. Class 9 has no board exam, so your school sets the papers.
Physics, Chapter 4: Describing motion around us
| Formula | Symbols and units | Page |
|---|---|---|
| Average speed = total distance travelled ÷ time interval | distance in m, time in s, speed in m s⁻¹. Speed has no direction (Eq. 4.1) | 52 |
| Average velocity: v_av = s/t | s is the displacement, so velocity has a direction, shown by + or − in a straight line (Eq. 4.2) | 53 |
| Unit of speed and velocity: m s⁻¹ (m/s); also km h⁻¹ | to change km h⁻¹ to m s⁻¹, multiply by 1000/3600, which is 5/18: 36 km h⁻¹ = 10 m s⁻¹ | 53, 55 |
| Average acceleration: a = (v − u)/(t₂ − t₁) | u velocity at t₁, v velocity at t₂; unit m s⁻² (Eq. 4.3) | 54 |
| Acceleration due to gravity: g = 9.8 m s⁻² | for a freely falling object, pointing down | 56 |
| Slope of a position-time graph = velocity | a steeper line means a faster object | 60, 62 |
| Slope of a velocity-time graph = acceleration; area under it = displacement | a flat line on the v-t graph means zero acceleration | 62-63 |
| v = u + at | first kinematic equation, for constant acceleration (Eq. 4.4a) | 63 |
| s = ut + ½at² | second, from the area under the v-t graph (Eq. 4.4b) | 64 |
| v² = u² + 2as | third, with t eliminated (Eq. 4.4c) | 64 |
| Circular motion, one round: average speed = 2πR/T | R radius, T time for one revolution (Eq. 4.5). Average velocity over a full round is 0 | 66 |
The three kinematic equations hold only when the acceleration stays constant. In uniform circular motion the speed stays the same, but the direction of the velocity keeps changing, so the object is accelerating (pages 66-67).
More practice: Describing Motion Around Us questions with answers.
Physics, Chapter 6: How forces affect motion
| Formula or rule | Symbols and units | Page |
|---|---|---|
| Unit of force: the newton (N) | small forces are often in millinewtons, 10⁻³ N | 95 |
| Net force of two forces in a line | same direction: add them; opposite directions: subtract, and the net force points along the larger one | 96 |
| Newton's second law: a = F/m, so F = ma | F net force in N, m mass in kg, a in m s⁻²; a points the same way as the net force (Eq. 6.1, 6.2) | 104 |
| 1 N = 1 kg m s⁻² | the force that gives 1 kg an acceleration of 1 m s⁻² | 104 |
| Weight: F = mg | g = 9.8 m s⁻², or 10 m s⁻² for quick estimates; g does not depend on the mass (Eq. 6.3) | 104 |
| Newton's third law | forces come in equal and opposite pairs, and the two forces of a pair act on different objects | 108 |
| Two objects pulled together: a = F/(m₁ + m₂) | treat them as one system; the string's tension is an internal force (Eq. 6.4) | 111 |
More practice: How Forces Affect Motion questions with answers.
Physics, Chapter 7: Work, energy and simple machines
| Formula | Symbols and units | Page |
|---|---|---|
| Work: W = F × s | F constant force in N, s displacement in m along the force (Eq. 7.2) | 117 |
| 1 J = 1 N × 1 m = 1 kg m² s⁻² | the joule is the unit of work. A force at right angles to the motion does no work | 118 |
| Work-energy theorem: the work done on an object equals the change in its energy | energy is also measured in joules (Eq. 7.3) | 120 |
| W = ½m(v² − u²) | work done by a constant net force that changes the speed from u to v (Eq. 7.5) | 122 |
| Kinetic energy: K = ½mv² | m in kg, v in m s⁻¹, K in J. Double the speed and K becomes four times as large (Eq. 7.6) | 122 |
| Potential energy: U = mgh | h height above the ground in m; holds only near the Earth's surface (Eq. 7.8) | 125-126 |
| Mechanical energy = K + U | stays constant when only gravity does work: a falling object's U turns into K | 126-127 |
| Power: P = W/t | unit watt (W): 1 W = 1 J s⁻¹. One horsepower is 746 W (Eq. 7.11) | 129 |
| Mechanical advantage = load/effort | a fixed pulley has MA = 1; it only changes the direction of the force (Eq. 7.12) | 130 |
| Inclined plane: MA = mg/F′ = L/h | L length of the slope, h height. Longer slope, smaller effort (Eq. 7.13) | 132 |
| Lever: F₁ × d₁ = F₂ × d₂ | effort × effort arm = load × load arm (Eq. 7.14, 7.15) | 133-134 |
| Lever: MA = load/effort = effort arm/load arm | a machine reduces the force, not the total work (Eq. 7.16) | 133-134 |
More practice: Work, Energy and Simple Machines questions with answers.
Physics, Chapter 10: Sound waves
| Formula or value | Symbols and units | Page |
|---|---|---|
| Wavelength λ, unit metre | distance between two crests (or two troughs) in a row | 194 |
| Frequency ν, unit hertz (Hz) = s⁻¹ | oscillations per second | 194 |
| ν = 1/T | T the time period in s (Eq. 10.1) | 194 |
| Speed of sound: v = λν | speed = wavelength × frequency (Eq. 10.2). Also v = λ/T | 196 |
| Speed of sound in air | dry air: about 331 m s⁻¹ at 0 °C, rising to nearly 344 m s⁻¹ at 22 °C; it rises with temperature and humidity | 196 |
| Speeds at 15 °C (Table 10.1) | air 340 m s⁻¹, water 1500 m s⁻¹, steel 5000 m s⁻¹: fastest in solids, slowest in gases | 197 |
| Human hearing: 20 Hz to 20 kHz | below 20 Hz is infrasound, above 20 kHz is ultrasound | 199 |
| Loudness in decibels (dB) | ordinary talk is about 60 dB; firecrackers can go above 100 dB | 199 |
| Octave | the higher note has twice the frequency of the lower one | 200 |
| Echo: at least 0.1 s gap, so the wall must be at least 17 m away | at 340 m s⁻¹, sound covers 34 m in 0.1 s, there and back | 201 |
| Echo or sonar distance = v × t/2 | t is the time for the sound to go and come back | 201, 203 |
| Reverberation | reflections arriving less than 0.05 s apart, which blur the sound | 201 |
More practice: Sound Waves questions with answers.
Chemistry, Chapter 5: Concentration of a solution
| Formula | What it tells you | Page |
|---|---|---|
| Solution mass = solute mass + solvent mass | add the water to the solute; never divide by the water alone | 75 |
| % m/m = (mass of solute ÷ mass of solution) × 100 | grams of solute in 100 g of solution (Eq. 5.1) | 75 |
| % m/v = (mass of solute ÷ volume of solution) × 100 | grams of solute in 100 mL of solution; mass in g, volume in mL (Eq. 5.2) | 75 |
| % v/v = (volume of solute ÷ volume of solution) × 100 | mL of solute in 100 mL, for one liquid in another (Eq. 5.3) | 76 |
| Hospital saline: 0.9% m/v | 0.9 g of salt in every 100 mL of solution | 76 |
| % w/w is the same number as % m/m | industry writes weight percentage | 76 |
More practice: Exploring Mixtures and their Separation questions with answers.
Chemistry, Chapter 8: Journey inside the atom
| Formula or rule | What it means | Page |
|---|---|---|
| Electron charge: −1.602 × 10⁻¹⁹ C, written as −1 | relative charges: electron −1, proton +1, neutron 0 (Table 8.1) | 142, 147 |
| Atomic number Z = number of protons | in a neutral atom it also equals the number of electrons | 149-150 |
| Mass number A = protons + neutrons | so neutrons = A − Z. Electrons are too light to count | 150 |
| Notation: A on top, Z below, then the symbol | carbon-12 is written with 12 above and 6 below the C | 150 |
| Most electrons in shell n: 2n² | K (n = 1) 2, L (n = 2) 8, M (n = 3) 18; the outermost shell holds at most 8; shells fill from K outwards | 151 |
| Valency | the combining capacity of an atom: how many electrons it gains, loses or shares to complete its octet (sodium 2, 8, 1 has valency 1) | 153 |
| Atomic masses are in u | the unified atomic mass unit (earlier written amu) | 154 |
| Weighted average atomic mass | (mass₁ × abundance₁) + (mass₂ × abundance₂), with the abundances as fractions; chlorine: 35 × 0.75 + 37 × 0.25 = 35.5 u | 155 |
More practice: Journey Inside the Atom questions with answers.
Chemistry, Chapter 9: Atomic foundations of matter
| Law or formula | What it means | Page |
|---|---|---|
| Law of conservation of mass | the reactants and the products of any chemical reaction have the same total mass | 165-166 |
| Law of constant proportions | a compound always has its elements in the same mass ratio; in water, hydrogen to oxygen is 1 : 8 | 167 |
| Writing a formula | write the symbols, put each valency under its symbol, then cross them over as subscripts (a 1 is not written) | 175 |
| Molecular mass | add the atomic masses of all atoms in one molecule; H₂O: 2 × 1 + 16 = 18 u | 179 |
| Formula unit mass | the same sum for an ionic compound, which forms no molecules; Na₂O: 2 × 23 + 16 = 62 u | 179 |
Valencies to know (Table 9.1, pages 174-175), sorted here by valency:
| Valency | Ions |
|---|---|
| 1 | positive: Na⁺, K⁺, Li⁺, Ag⁺, Cu⁺ (cuprous), NH₄⁺; negative: Cl⁻, F⁻, Br⁻, I⁻, OH⁻, NO₃⁻, HCO₃⁻ |
| 2 | positive: Mg²⁺, Ca²⁺, Ba²⁺, Zn²⁺, Fe²⁺ (ferrous), Cu²⁺ (cupric); negative: O²⁻, S²⁻, SO₄²⁻, CO₃²⁻ |
| 3 | Al³⁺, Fe³⁺ (ferric) |
More practice: Atomic Foundations of Matter questions with answers.
Numbers from the cell and Earth chapters
| Key number | What it means | Page |
|---|---|---|
| 0.1 mm | the smallest gap the unaided eye can resolve, from about 25 cm | 9 |
| 1 mm = 1000 µm | micrometres are the unit for cell sizes | 10 |
| Cell size ≈ field of view (µm) ÷ number of cells across it | the estimate used with a microscope | 10 |
| Total magnification = eyepiece power × objective power | a 10X eyepiece with a 10X objective gives 100X | 10 |
| Typical cell diameter | prokaryotic 1 to 10 µm, eukaryotic 10 to 100 µm (Table 2.2) | 15 |
| Solar constant ≈ 1.4 kW m⁻² | sunlight per square metre at the top of the atmosphere; about 1 kW m⁻² reaches the ground on a clear day | 254-255 |
| Energy = intensity × area × time | in joules when the intensity is in W m⁻² (J s⁻¹ m⁻²), area in m² and time in s | 255 |
| Air: nitrogen 78%, oxygen 21% | with small amounts of other gases such as argon, CO₂ and water vapour | 257 |
| Troposphere 0 to 12 km, stratosphere 12 to 50 km | in the troposphere it gets about 6.5 °C colder per km; the ozone layer is in the stratosphere; outer space starts near 100 km | 258 |
Chapter 2 and Chapter 13 questions: Cell questions with answers and Earth as a System questions with answers.
8 worked questions (our own)
Question 1 (kinematic equations). A metro train leaves a station from rest and speeds up at a steady 0.8 m s⁻² for 25 s. (a) What is its speed then, in m s⁻¹ and in km h⁻¹? (b) How far has it gone? (c) Near the next station it brakes from that speed and stops in 160 m. Find its acceleration while braking.
Answer. (a) v = u + at = 0 + 0.8 × 25 = 20 m s⁻¹, which is 20 × 18/5 = 72 km h⁻¹. (b) s = ut + ½at² = 0 + ½ × 0.8 × 25² = 250 m. (c) v² = u² + 2as gives 0 = 20² + 2 × a × 160, so a = −400/320 = −1.25 m s⁻². The minus sign means the acceleration points against the motion: the train is slowing down.
Question 2 (sound). A bat sends out an ultrasonic squeak of frequency 50 kHz. Use 340 m s⁻¹ for sound in air. (a) Find the wavelength in millimetres. (b) The echo from a moth comes back 0.02 s later. How far away is the moth? (c) Could a person hear this squeak?
Answer. (a) λ = v/ν = 340/50,000 = 0.0068 m = 6.8 mm. (b) The sound goes there and back, so distance = v × t/2 = 340 × 0.02/2 = 3.4 m. (c) No: 50 kHz is above 20 kHz, the top of human hearing, so it is ultrasound.
Question 3 (concentration). (a) 12 g of glucose is dissolved in 228 g of water. Find the concentration as % m/m. (b) A nurse makes 250 mL of hospital saline (0.9% m/v). How much salt does she need? (c) A 500 mL bottle of hand sanitiser holds 350 mL of alcohol. What is its % v/v?
Answer. (a) Mass of solution = 12 + 228 = 240 g, so % m/m = 12/240 × 100 = 5%. (b) 0.9 g in every 100 mL, so 0.9 × 250/100 = 2.25 g. (c) 350/500 × 100 = 70% v/v.
Question 4 (force and motion). A car of mass 1,200 kg is moving at 20 m s⁻¹. The brakes give a net backward force of 6,000 N. Find (a) the acceleration, (b) the time to stop, (c) the stopping distance.
Answer. (a) a = F/m = 6,000/1,200 = 5 m s⁻², against the motion, so a = −5 m s⁻². (b) v = u + at gives 0 = 20 − 5t, so t = 4 s. (c) v² = u² + 2as gives 0 = 400 − 10s, so s = 40 m. Check with s = ut + ½at² = 20 × 4 − ½ × 5 × 16 = 80 − 40 = 40 m.
Question 5 (work and power). A lift carries a load of 400 kg up 24 m in 20 s at a steady speed. Take g = 10 m s⁻². Find (a) the work done against gravity, (b) the power, in watts and in horsepower.
Answer. (a) The upward force equals the weight, mg = 400 × 10 = 4,000 N, so W = F × s = 4,000 × 24 = 96,000 J. (b) P = W/t = 96,000/20 = 4,800 W, and 4,800/746 ≈ 6.4 hp.
Question 6 (conservation of energy). A swing seat of mass 4 kg is pulled back until it is 0.45 m above its lowest point, then let go. Ignore air resistance and take g = 10 m s⁻². Find (a) the potential energy of the seat at the moment it is released, measured from its lowest point, (b) its speed as it passes through the lowest point.
Answer. (a) U = mgh = 4 × 10 × 0.45 = 18 J. (b) At the lowest point all of it has become kinetic energy: ½ × 4 × v² = 18, so v² = 9 and v = 3 m s⁻¹. The mass cancels out (v² = 2gh), so a heavier seat let go from the same height would pass the lowest point at the same speed.
Question 7 (lever). In a wheelbarrow the wheel is the fulcrum. A load of bricks weighing 600 N sits 0.4 m from the wheel, and the handles are 1.2 m from the wheel. (a) What lifting force is needed at the handles? (b) What is the mechanical advantage? (c) Does the wheelbarrow save work?
Answer. (a) Effort × effort arm = load × load arm: F × 1.2 = 600 × 0.4, so F = 200 N. (b) MA = load/effort = 600/200 = 3 (the same as 1.2/0.4). (c) No. The handles move three times as far as the load, so the work stays the same; only the force needed is smaller.
Question 8 (atom and formula). A calcium atom has atomic number 20 and mass number 40. (a) How many protons, neutrons and electrons does it have? (b) Write its electron distribution and valency. (c) Using Table 9.1, write the formula of calcium hydrogencarbonate and of ammonium sulfate, and find their formula unit masses (Ca = 40 u, H = 1 u, C = 12 u, O = 16 u, N = 14 u, S = 32 u).
Answer. (a) 20 protons, 20 electrons, 40 − 20 = 20 neutrons. (b) 2, 8, 8, 2: two outer electrons to lose, so valency 2 and the ion Ca²⁺. (c) Ca²⁺ has valency 2 and HCO₃⁻ valency 1, so crossing over gives Ca(HCO₃)₂: 40 + 2 × (1 + 12 + 3 × 16) = 40 + 2 × 61 = 162 u. NH₄⁺ has valency 1 and SO₄²⁻ valency 2, so crossing over gives (NH₄)₂SO₄: 2 × (14 + 4 × 1) + 32 + 4 × 16 = 36 + 96 = 132 u.
Three mistakes that cost marks
- Mixing km h⁻¹ and m s⁻¹ in one equation. Change every speed to m s⁻¹ (multiply km h⁻¹ by 5/18) before you use v = u + at, ½mv² or v = λν.
- Dividing an echo distance by nothing. The time you hear is for the sound to go and come back, so the distance to the wall or object is v × t/2, not v × t.
- Using the mass of the solvent for % m/m. Divide by the mass of the whole solution (solute + solvent). In Question 3, 12/228 would give 5.26% instead of 5%.
Practise more
- Free practice, marked instantly: the practice page (pick Class 9) and printable Class 9 worksheets.
- Questions on every chapter, with marks: Class 9 Science important questions 2026-27.
- Old chapter names and where they went in the new book: Class 9 old chapter names 2026-27.
Where CMDLearn fits
CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.
Sources
All opened and read on 5 November 2026. NCERT page numbers are printed page numbers; the CBSE page number is a PDF page.
- NCERT, Exploration, Textbook of Science for Grade 9 (First Edition, April 2026), Chapters 2, 4 to 10 and 13, the pages named in each table: ncert.nic.in/textbook/pdf/iesc102.pdf, ncert.nic.in/textbook/pdf/iesc104.pdf (and so on to iesc110.pdf), and ncert.nic.in/textbook/pdf/iesc113.pdf
- CBSE, Science (Subject Code 086), Class IX (2026-27) curriculum, course structure (PDF page 4): cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/ScienceSt_SecP1_2026-27.pdf
The formulas, laws and numbers are the textbook's; the layout, the notes and the 8 questions are our own, and every answer was re-computed by script. This page is general information; follow your teacher, the textbook and CBSE's own notices.
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