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Mechanics for JEE Physics: A Chapter-by-Chapter Guide from Kinematics to Oscillations
Mechanics in JEE physics is one chain of chapters: kinematics, then laws of motion, then work, energy and power, then system of particles and rotational motion, then gravitation, properties of solids and fluids, and oscillations. Each chapter reuses the tools of the one before it, so learn them in that order. Most students who find JEE physics mechanics hard have a weak early link, usually free-body diagrams or vectors, rather than a problem with rotation itself.
This guide is part of our JEE Physics home tutor in Gurgaon series. For each chapter, it gives the key ideas, the traps our tutors see most often, and worked examples we have checked step by step. It ends with a study-order table and a comparison of mechanics for JEE Main and JEE Advanced.
Why mechanics matters in JEE physics
Mechanics is the largest connected block of JEE physics, and almost all of it is taught in Class 11. In the official JEE (Main) 2026 syllabus, the mechanics chapters in this guide make up Units 2 to 7 and the oscillations part of Unit 10: Kinematics; Laws of Motion; Work, Energy and Power; Rotational Motion; Gravitation; Properties of Solids and Liquids; and Oscillations and Waves. Check the current year's syllabus, as NTA can revise it.
Mechanics also returns in Class 12: free-body diagrams in electrostatics, circular motion in magnetism, and energy conservation in the Bohr model. A student fluent in mechanics finds much of Class 12 physics easier.
Study order: the mechanics chain at a glance
This is the order we teach mechanics in, with what each chapter depends on.
| Step | Chapter | Depends on | Must be solid before moving on |
|---|---|---|---|
| 0 | Units, dimensions and vectors | Basic algebra and trigonometry | Resolving vectors into components; dot and cross products; dimensional checks |
| 1 | Kinematics | Vectors, basic calculus | Equations of motion, graphs, projectiles, relative velocity |
| 2 | Laws of motion | Kinematics, vectors | Free-body diagrams, constraint relations, friction, circular dynamics |
| 3 | Work, energy and power | Laws of motion | Work–energy theorem, conservation of energy, collisions, vertical circle |
| 4 | System of particles and rotational motion | All of the above | Centre of mass, torque, moment of inertia, angular momentum, rolling |
| 5 | Gravitation | Energy, circular motion | Variation of g, potential energy, orbits, escape velocity |
| 6 | Properties of solids and fluids | Laws of motion, energy | Stress and strain, pressure, Bernoulli, viscosity, surface tension |
| 7 | Oscillations | Laws of motion, energy, rotation | SHM equation, spring and pendulum systems, energy in SHM |
Step 0 is short, but weak vectors cause errors in every later chapter. Steps 1 to 4 are the core and deserve the most time; steps 5 to 7 are shorter but reuse every idea from the core. In the JEE (Main) 2026 syllabus, Unit 7 (Properties of Solids and Liquids) also includes heat, thermal expansion, calorimetry and heat transfer; this guide covers only its mechanics part (elasticity and fluids).
Kinematics
Kinematics describes motion without asking what causes it. For JEE, you need motion in a straight line and in a plane, projectiles, uniform circular motion and relative velocity.
Key ideas
- Equations of motion (v = u + at, s = ut + ½at2, v2 = u2 + 2as) hold only for constant acceleration. For variable acceleration, use v = dx/dt and a = dv/dt, or a = v dv/dx.
- Graphs: the slope of a position–time graph is velocity; the slope of a velocity–time graph is acceleration, and the area under it is displacement.
- Projectiles: treat horizontal and vertical motion separately. Time of flight T = 2u sinθ/g, maximum height H = u2sin2θ/2g, range R = u2sin 2θ/g, for level ground.
- Relative velocity: vAB = vA − vB. Rain-and-umbrella and river-crossing questions are relative-velocity questions.
Traps
- Using the equations of motion when acceleration is not constant.
- Confusing distance with displacement, and average speed with the magnitude of average velocity.
- Using the level-ground range formula for a projectile launched from a height or onto an incline.
Worked example: A ball is projected at 20 m/s at 30° to the horizontal from level ground. Find the time of flight, maximum height and range (take g = 10 m/s2).
Answer: T = 2 s, H = 5 m, R = 20√3 ≈ 34.6 m. The vertical component is u sin 30° = 10 m/s, and the horizontal component is u cos 30° = 10√3 m/s. Time of flight T = 2 × 10/10 = 2 s. Maximum height H = 102/(2 × 10) = 5 m. Range R = horizontal speed × T = 10√3 × 2 = 20√3 m. Check with the formula: u2sin 60°/g = 400 × (√3/2)/10 = 20√3.
Worked example: Rain falls vertically at 6 m/s. A man walks at 8 m/s. At what speed and angle does the rain appear to hit him, and how should he hold his umbrella?
Answer: 10 m/s, at about 53° to the vertical; tilt the umbrella forward by that angle. The velocity of rain relative to the man is vrain − vman: 6 m/s downwards plus 8 m/s backwards (opposite to his walking direction). Its magnitude is √(62 + 82) = 10 m/s, at tan−1(8/6) ≈ 53° from the vertical. The rain seems to come from in front of him, so he tilts the umbrella forward. Trap: adding the velocities instead of subtracting, which gives the right speed but the wrong direction.
Laws of motion
Laws of motion JEE questions are really questions about free-body diagrams. Newton's second law, F = ma, is easy to state; the skill is choosing the right body, drawing every force on it, and writing one equation per direction.
Key ideas
- Free-body diagram (FBD): isolate one body and draw only the forces acting on it: weight, normal reaction, tension, friction, spring force and any applied force.
- Constraint relations: bodies joined by a taut, inextensible string have related accelerations; with pulleys, the relation comes from the string length staying constant.
- Friction: static friction adjusts up to a limit μsN; kinetic friction is μkN once sliding starts.
- Circular dynamics: a net inward force of mv2/r is needed for circular motion. On a banked road with no friction, the safe speed satisfies tanθ = v2/rg.
- Momentum and impulse: F = dp/dt, impulse = change in momentum, and momentum is conserved when the net external force is zero.
Traps
- Writing static friction as μsN every time. It equals μsN only when the body is just about to slip.
- Adding "centripetal force" as an extra force on the FBD. It is not a new force; it is the name for the net inward force.
- Getting the direction of friction wrong in stacked-block problems. Friction opposes relative motion (or its tendency) between the surfaces in contact.
Worked example: A 5 kg block rests on a rough floor with μs = 0.5 and μk = 0.4. A horizontal force of 20 N is applied. Find the friction force (g = 10 m/s2).
Answer: 20 N, and the block does not move. The limiting static friction is μsN = 0.5 × 5 × 10 = 25 N. The applied force (20 N) is less than this, so the block stays at rest and static friction just balances the applied force: 20 N. Trap: answering 25 N (the maximum) or 20 N from μkN (kinetic friction applies only when the block slides).
Worked example (Advanced-style): A 2 kg block rests on a 4 kg block, which lies on a smooth floor. The coefficient of static friction between the blocks is 0.3. A horizontal force F is applied to the lower block. Find the largest F for which the blocks move together (g = 10 m/s2).
Answer: 18 N. The only horizontal force on the upper block is friction, which can be at most μmg = 0.3 × 2 × 10 = 6 N. So the upper block's acceleration can be at most 6/2 = 3 m/s2. If the blocks move together, the whole 6 kg system has this acceleration, so F = 6 × 3 = 18 N. Above 18 N, the lower block slides out from under the upper one. Trap: applying μ to the combined weight of both blocks.
From our tutors: in laws of motion, our rule for students is "no equation without a diagram". We ask them to draw a separate free-body diagram for each body, label every force, and say aloud which body exerts it. Most wrong answers we see in this chapter come from one missing or invented force, and the habit of naming the source of every force catches almost all of them.
Work, energy and power
The work–energy theorem says the total work done on a body equals its change in kinetic energy. Energy methods often solve in two lines what takes a page with forces, especially on curved paths.
Key ideas
- Work by a constant force is F · s; by a variable force it is ∫ F · ds. Power is the rate of doing work, P = F · v.
- Conservative forces (gravity, spring force) have potential energies; friction does not. Mechanical energy is conserved only when non-conservative forces do no work.
- Spring potential energy is ½kx2, where x is the extension or compression from natural length.
- Vertical circle: for a particle on a string, the minimum speed at the bottom to complete the circle is √(5gL), and the minimum speed at the top is √(gL).
- Collisions: momentum is conserved in all collisions (when external impulse is negligible); kinetic energy is conserved only in elastic collisions. The coefficient of restitution e compares the speed of separation to the speed of approach.
Traps
- Conserving kinetic energy in an inelastic collision.
- Forgetting the work done by friction when using energy conservation.
- Using the vertical-circle result for a rod. A light rigid rod can push as well as pull, so the minimum speed at the top is zero, and the minimum at the bottom is √(4gL).
Worked example: A 2 kg block moving at 4 m/s on a smooth floor hits a light spring of spring constant 800 N/m. Find the maximum compression.
Answer: 0.2 m. At maximum compression the block momentarily stops, so all its kinetic energy is stored in the spring: ½ × 2 × 42 = ½ × 800 × x2. That gives 16 = 400x2, so x2 = 0.04 and x = 0.2 m.
Worked example: A 2 kg ball moving at 6 m/s strikes a 1 kg ball at rest, and they stick together. Find their common velocity and the kinetic energy lost.
Answer: 4 m/s; 12 J lost. Momentum is conserved: 2 × 6 = (2 + 1)v, so v = 4 m/s. Kinetic energy before = ½ × 2 × 36 = 36 J. After = ½ × 3 × 16 = 24 J. The loss is 12 J, which goes into heat, sound and deformation. Trap: trying to conserve kinetic energy, which gives a different, wrong velocity.
Is mechanics where your child's JEE physics marks are slipping? Book a free JEE Physics demo class in Gurgaon. Our tutor will test the chain from free-body diagrams to energy and show you exactly where the gap is.
Book a Free JEE Physics Demo +91 92204 75088System of particles and rotational motion
Rotational motion JEE questions are the ones students fear most, but every linear idea has a rotational partner: force becomes torque, mass becomes moment of inertia, and momentum becomes angular momentum.
Key ideas
- Centre of mass: the external forces move the centre of mass as if all the mass were there. If the net external force is zero, the centre of mass does not accelerate; this is how "man walking on a boat" questions are solved.
- Torque and angular acceleration: τ = Iα about a fixed axis (or about the centre of mass).
- Moment of inertia: learn the standard values (ring MR2, disc ½MR2, solid sphere ⅖MR2, rod about its centre ML2/12) and the parallel and perpendicular axes theorems.
- Angular momentum: L = Iω for rotation about a fixed axis; it is conserved when the net external torque is zero.
- Rolling without slipping: v = Rω. The total kinetic energy is ½Mv2 + ½Iω2. On an incline, a = g sinθ / (1 + I/MR2).
Traps
- Taking torques about a point that is accelerating (other than the centre of mass) without accounting for it.
- Using the parallel axes theorem from an axis that is not through the centre of mass.
- Thinking friction does work in pure rolling. The contact point is momentarily at rest, so static friction does no work there.
- Assuming kinetic energy is conserved when angular momentum is. When a spinning skater pulls in her arms, L stays the same but kinetic energy increases, because she does work.
Worked example: A uniform rod of length 1.5 m is pivoted at one end and released from rest in a horizontal position. Find its angular acceleration just after release, and its angular speed as it passes the vertical (g = 10 m/s2).
Answer: α = 10 rad/s2; ω = √20 ≈ 4.47 rad/s. About the pivot, I = ML2/12 + M(L/2)2 = ML2/3 by the parallel axes theorem. Just after release, the weight acts at the centre, giving torque MgL/2. So α = (MgL/2)/(ML2/3) = 3g/2L = 30/3 = 10 rad/s2. At the vertical, the centre has fallen L/2, so MgL/2 = ½(ML2/3)ω2, giving ω2 = 3g/L = 20. Trap: using g/L, as if the rod were a point mass at its end.
Worked example: A solid sphere rolls without slipping down an incline of 30°. Find its acceleration (g = 10 m/s2). Which reaches the bottom first: a solid sphere, a disc or a ring, released together?
Answer: 25/7 ≈ 3.57 m/s2; the sphere wins, then the disc, then the ring. For a solid sphere, I/MR2 = 2/5, so a = g sinθ/(1 + 2/5) = (5/7) × 10 × 0.5 = 25/7 m/s2. The smaller I/MR2 is, the larger the acceleration: sphere 2/5, disc 1/2, ring 1. The result does not depend on mass or radius, only on shape.
Rolling without slipping of rings, cylinders and spheres is named in the JEE Advanced 2026 syllabus. The JEE Main 2026 syllabus lists rigid-body rotation, the equations of rotational motion and rolling friction, but does not use the phrase "rolling without slipping". We teach rolling to all JEE students anyway, because it follows directly from those listed ideas.
Gravitation
Gravitation applies Newton's law, F = GMm/r2, to planets, satellites and the variation of g. It is short and relies on energy conservation and circular motion.
Key ideas
- Variation of g: at height h, gh = g/(1 + h/R)2 (approximately g(1 − 2h/R) for small h); at depth d, gd = g(1 − d/R).
- Potential energy: U = −GMm/r, taking zero at infinity. Near the surface, mgh is only an approximation.
- Orbits: orbital speed vo = √(GM/r); for a satellite in a circular orbit, kinetic energy = GMm/2r, potential energy = −GMm/r, and total energy = −GMm/2r.
- Escape velocity: ve = √(2GM/R) = √(2gR). Near the surface, ve = √2 × orbital speed.
- Kepler's laws: especially T2 ∝ r3 for circular orbits.
Traps
- Using mgh for heights comparable to the Earth's radius.
- Using the small-height approximation g(1 − 2h/R) when h is not small.
Worked example: Compare the value of g at a height R/2 above the Earth's surface with its value at a depth R/2 below it, where R is the Earth's radius.
Answer: 4g/9 at height, g/2 at depth. At height h = R/2: gh = g/(1 + 1/2)2 = g/(9/4) = 4g/9 ≈ 0.44g. At depth d = R/2: gd = g(1 − 1/2) = g/2. Trap: using the approximation g(1 − 2h/R), which gives 0 here and is clearly wrong because h is not small.
Worked example: A satellite of mass m moves in a circular orbit of radius r around a planet of mass M. How much energy is needed to move it to a circular orbit of radius 2r?
Answer: GMm/4r. The total energy in a circular orbit is −GMm/2r. In the new orbit it is −GMm/4r. The energy needed is the difference: −GMm/4r − (−GMm/2r) = GMm/4r. Trap: using only the change in potential energy (GMm/2r), which ignores the fact that the satellite slows down in the higher orbit.
Properties of solids and fluids
Properties of solids and liquids is Unit 7 of the JEE (Main) 2026 syllabus, and its mechanics topics also appear in the JEE (Advanced) 2026 syllabus: elasticity, fluid statics and flow, viscosity and surface tension. Most questions are short applications of a few laws.
Key ideas
- Elasticity: stress = F/A, strain = ΔL/L, and Young's modulus Y = stress/strain, so extension ΔL = FL/AY. Bulk modulus and modulus of rigidity are the volume and shape equivalents.
- Fluid statics: pressure at depth h is P0 + ρgh; Pascal's law explains the hydraulic lift; buoyant force equals the weight of fluid displaced.
- Fluid flow: the continuity equation A1v1 = A2v2 and Bernoulli's principle, P + ½ρv2 + ρgh = constant along a streamline. Torricelli's result for efflux speed is v = √(2gh).
- Viscosity: Stokes' law, F = 6πηrv, and terminal velocity, which is proportional to r2 for a small sphere falling through a fluid.
- Surface tension: excess pressure is 2T/r inside a drop and 4T/r inside a soap bubble (two surfaces); capillary rise h = 2T cosθ/(rρg).
Traps
- Using 2T/r for a soap bubble, which has two surfaces.
- Applying Bernoulli's principle between points not on the same streamline, or to viscous flow.
- Forgetting atmospheric pressure when a question asks for absolute pressure.
Worked example: Water stands 5 m deep in a large open tank. A small hole is made in the side near the bottom. Find the speed at which water comes out (g = 10 m/s2).
Answer: 10 m/s. Apply Bernoulli's principle between the open top surface and the hole. Both are at atmospheric pressure, and the top surface moves very slowly because the tank is large. So ρgh = ½ρv2, giving v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s.
Worked example: Find the excess pressure inside a soap bubble of radius 1 cm, if the surface tension of the soap solution is 0.03 N/m.
Answer: 12 Pa. A soap bubble has two surfaces, so the excess pressure is 4T/r = 4 × 0.03 / 0.01 = 12 Pa. For a liquid drop of the same radius, with one surface, it would be 2T/r = 6 Pa.
Oscillations
Simple harmonic motion (SHM) is motion in which the restoring force is proportional to the displacement and opposite to it: F = −kx. It combines laws of motion, energy and, for pendulums and rigid bodies, rotation. The JEE (Main) 2026 syllabus places oscillations and waves together in Unit 10; this guide covers the oscillations part.
Key ideas
- SHM equation: x = A sin(ωt + φ), with ω = √(k/m) and period T = 2π/ω.
- Velocity and energy: v = ω√(A2 − x2); maximum speed ωA at the mean position; total energy ½kA2, shared between kinetic and potential energy.
- Springs: in parallel, k = k1 + k2; in series, 1/k = 1/k1 + 1/k2.
- Simple pendulum: T = 2π√(L/g) for small oscillations. In a lift accelerating upwards at a, replace g by (g + a); downwards, by (g − a).
- The method: to find the period of any system, displace it slightly, write the restoring force (or torque), and bring it to the form a = −ω2x.
Traps
- Thinking the period of a spring–mass system depends on amplitude or on g. For an ideal spring it depends only on m and k.
- Assuming kinetic and potential energy are equal at half the amplitude. At x = A/2, potential energy is ½k(A/2)2, only one-quarter of the total, so kinetic energy is three-quarters.
Worked example: A 0.5 kg mass on a spring of constant 50 N/m oscillates with amplitude 0.1 m. Find the period, the maximum speed and the total energy.
Answer: T = π/5 ≈ 0.63 s; vmax = 1 m/s; E = 0.25 J. ω = √(k/m) = √(50/0.5) = √100 = 10 rad/s, so T = 2π/10 = π/5 s. Maximum speed = ωA = 10 × 0.1 = 1 m/s. Total energy = ½kA2 = ½ × 50 × 0.01 = 0.25 J. Check: ½mvmax2 = ½ × 0.5 × 1 = 0.25 J.
From our tutors: we teach one method for every SHM question: displace the system by a small x, write the net restoring force or torque, and compare with a = −ω2x. Students who memorise a separate formula for each system (spring, pendulum, floating block, liquid in a U-tube) get stuck when JEE gives an unfamiliar set-up. Students who learn the method can derive the period of a system they have never seen.
Mechanics for JEE Main vs JEE Advanced
Both exams test the same mechanics chain, but they differ in syllabus detail and in how deep the questions go. This comparison is based on the JEE (Main) 2026 syllabus and the physics syllabus in the JEE (Advanced) 2026 Information Brochure. Always check the current year's documents.
| Chapter | JEE Main 2026 syllabus (examples) | JEE Advanced 2026 syllabus (examples) | Typical depth difference |
|---|---|---|---|
| Kinematics | Motion in a straight line and in a plane, graphs, projectile motion, uniform circular motion, relative velocity | Kinematics in one and two dimensions (Cartesian coordinates only), projectiles, uniform circular motion, relative velocity | Advanced questions more often use calculus and variable acceleration |
| Laws of motion | Newton's laws, momentum and impulse, equilibrium of concurrent forces, static and kinetic friction, rolling friction, vehicle on level and banked roads | Newton's laws; inertial and uniformly accelerated frames of reference; static and dynamic friction | Advanced adds work in accelerated frames (pseudo forces) and more multi-body constraint problems |
| Work, energy and power | Work–energy theorem, spring potential energy, vertical circle, elastic and inelastic collisions in one and two dimensions | Kinetic and potential energy, work and power, conservation of momentum and energy, elastic and inelastic collisions | Advanced problems chain energy and momentum across several stages |
| Rotation | Centre of mass, torque, angular momentum and its conservation, moment of inertia, parallel and perpendicular axes, equilibrium of rigid bodies, equations of rotational motion | Also names rolling without slipping of rings, cylinders and spheres, and collision of point masses with rigid bodies | Rotation is where the gap between the two exams is widest |
| Gravitation | Variation of g, Kepler's laws, potential and potential energy, escape velocity, satellite speed, period and energy | Gravitational potential and field, Kepler's law, geostationary orbits, motion of planets and satellites in circular orbits, escape velocity | Broadly similar; Advanced combines it with energy and angular momentum |
| Solids and fluids | Elasticity (Young's, bulk and rigidity moduli), Pascal's law, viscosity, Stokes' law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli, surface tension | Hooke's law, Young's modulus, pressure, Pascal's law, buoyancy, surface tension, viscosity (Poiseuille's equation excluded), Stokes' law, terminal velocity, continuity equation, Bernoulli | Similar lists; Advanced questions combine fluids with other mechanics |
| Oscillations | SHM equation, phase, spring oscillations, energy in SHM, simple pendulum with derivation of its period | Linear and angular SHM; also forced and damped oscillation (in one dimension) and resonance | Advanced includes angular SHM of rigid bodies |
Question style differs as well. According to the JEE (Main) 2026 Information Bulletin, physics in each JEE Main paper has 20 multiple-choice and 5 numerical-value questions, marked +4 for a correct answer and −1 for a wrong answer in both sections. So mechanics for JEE Mains rewards accuracy at speed. JEE Advanced mechanics questions more often join chapters, such as a collision that sets a rigid body rotating. Its marking scheme is given in each paper's instructions.
A mechanics preparation plan
This plan suits most students studying mechanics alongside Class 11 school work.
- Fix vectors first. Before kinematics, make sure resolving forces and velocities into components is automatic.
- Learn each chapter from NCERT, then a concept book. Our guide to the best books for JEE physics explains how NCERT, H.C. Verma and a problem bank fit together.
- Draw before you calculate. A free-body diagram for every laws-of-motion and rotation problem; a sketch of the path for every kinematics problem.
- Keep a one-page result sheet per chapter, with the standard results and when they apply. Our JEE physics formula sheet collects them in one place.
- Solve chapter-wise previous year questions straight after each chapter, and keep an error log that records why each mistake happened.
- Mix chapters once rotation is done: energy with rotation, friction with circular motion, SHM with rigid bodies.
- Revise the chain in order in Class 12 and before the exam, so that mechanics does not fade while electricity and magnetism take over.
Many mechanics marks are lost to arithmetic, units and sign errors rather than to physics. Our guide to common JEE physics numerical mistakes shows how to catch them. Students in Gurgaon and across Gurugram often start Class 11 with heavy school schedules, so plan mechanics around school tests: it is better to finish four chapters properly than to rush seven.
Frequently asked questions
Which chapters count as mechanics in JEE physics?
Kinematics; laws of motion; work, energy and power; system of particles and rotational motion; gravitation; the mechanics part of properties of solids and liquids; and oscillations. Units, dimensions and vectors come first as the base.
Is mechanics important for JEE Mains?
Yes. Mechanics covers several units of the JEE (Main) 2026 physics syllabus, and its methods, such as free-body diagrams, energy conservation and circular motion, are used in Class 12 chapters too. A weak mechanics base affects the whole physics paper.
Why is rotational motion so hard for JEE students?
Usually because the earlier links are weak. Rotation uses free-body diagrams, energy, momentum and vectors together. When those are solid, rotation becomes a set of parallels with linear motion: torque for force, moment of inertia for mass, and angular momentum for momentum.
How should I practise laws of motion for JEE?
Draw a separate free-body diagram for each body, write constraint relations for strings and pulleys, and check whether friction is static or kinetic before using a formula. Then solve chapter-wise previous year questions and log every mistake.
Is properties of solids and liquids in the JEE Main 2026 syllabus?
Yes. It is Unit 7 of the JEE (Main) 2026 physics syllabus, covering elasticity, fluid pressure, viscosity, Bernoulli's principle and surface tension, along with heat, thermal expansion, calorimetry and heat transfer. Check the current syllabus each year.
How long does it take to complete mechanics for JEE?
There is no fixed figure; it depends on the student's starting point and school schedule. Most students learn mechanics through Class 11, spending the most time on kinematics, laws of motion, energy and rotation.
Can a home tutor help with JEE mechanics?
Yes. A one-to-one tutor can watch how a student draws free-body diagrams and sets up equations, which is where most mechanics errors start. Our JEE Physics tutors in Gurgaon teach at home or online and plan each session around the student's own mistakes.
Want mechanics taught in the right order, with a tutor who checks every link from vectors to oscillations? Book a free JEE Physics demo class with Ajay Vatsyayan Classes, Saraswati kunj II, Wazirabad, Sector 52, Gurugram, Haryana 122003. Male and female tutors are available, at home or online.
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