Module 1 · NEET Physics

Mechanics

Kinematics, Newton's laws, work-energy, rotational motion, gravitation.
Kinematics · Laws of Motion · Work-Energy · Rotation · Gravitation
🎯 Take Mock Test

Learning Objectives

  • Master SI units, dimensional analysis, and error propagation for accurate measurements
  • Solve kinematics problems in 1D and 2D using equations of motion, graphs, and vector analysis
  • Apply Newton's laws with free-body diagrams for systems involving friction, pulleys, and circular motion
  • Analyse work-energy relationships, power, and conservation principles in mechanical systems
  • Calculate centre of mass, moment of inertia, torque, and angular momentum for rotating bodies
  • Derive gravitational field, potential, orbital and escape velocities using Kepler's & Newton's laws
  • Understand stress-strain behaviour of solids and fluid statics, dynamics, and viscosity fundamentals

1. Units & Measurements

SI Units & Dimensions

Fundamental & Derived Units

The International System of Units (SI) defines seven fundamental quantities from which all other physical quantities are derived. Every measurement in physics ultimately traces back to these base units. The seven base quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. All other quantities (velocity, force, energy, etc.) are derived from these base units.

Table 1: SI Base Units
Base Quantity SI Unit Symbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric CurrentampereA
Thermodynamic TemperaturekelvinK
Amount of Substancemolemol
Luminous Intensitycandelacd

Supplementary units: radian (rad) for plane angle and steradian (sr) for solid angle. While these are dimensionless, they are essential for rotational and wave mechanics. Prefixes like milli (10−3), centi (10−2), kilo (103), mega (106), giga (109) are used for multiples and sub-multiples.

Dimensional Analysis

Dimensions are expressed in terms of M (mass), L (length), T (time), A (current), K (temperature), mol (amount), cd (luminous intensity). The dimensional formula of a physical quantity is the expression of the base dimensions it depends upon. The principle of homogeneity states that only quantities with the same dimensions can be added, subtracted, or equated. This principle is the foundation of dimensional analysis and is used to check the correctness of physical equations and to derive relationships between quantities.

Table 2: Dimensional Formulas of Common Mechanics Quantities
Quantity Dimensional Formula SI Unit
Velocity[LT−1]m/s
Acceleration[LT−2]m/s2
Force[MLT−2]N (kg·m/s2)
Work / Energy[ML2T−2]J (N·m)
Power[ML2T−3]W (J/s)
Linear Momentum[MLT−1]kg·m/s
Impulse[MLT−1]N·s
Moment of Inertia[ML2]kg·m2
Torque[ML2T−2]N·m
Angular Momentum[ML2T−1]kg·m2/s
Gravitational Constant G[M−1L3T−2]N·m2/kg2
Pressure / Stress[ML−1T−2]Pa (N/m2)
Surface Tension[MT−2]N/m
Viscosity[ML−1T−1]Pa·s
Frequency[T−1]Hz (s−1)
Angular Displacement[M0L0T0]rad (dimensionless)
💡 Dimensional Analysis Shortcuts
To check formula correctness: LHS dimension must equal RHS dimension. NEET trick: Dimensional analysis cannot determine dimensionless constants (e.g., ½ in KE = ½mv2). Use dimensional consistency to quickly eliminate wrong options in MCQs. If two options have different dimensions, only one can be correct. Also, arguments of trigonometric, exponential, and logarithmic functions must be dimensionless.
Example 1: Dimensional Analysis
The viscous drag force on a sphere of radius r moving with velocity v through a fluid of viscosity η is given by F = 6πηrnv. Find n using dimensional analysis. Given [η] = [ML−1T−1].

a) 0     b) 1     c) 2     d) 3
Solution: [F] = [MLT−2]. RHS: [η][rn][v] = [ML−1T−1][Ln][LT−1] = [MLnT−2]. Equating LHS and RHS: n = 1. Hence option (b).
Example 2: Checking Formula Correctness
Which of the following could be a correct formula for the time period of a simple pendulum? (l = length, g = acceleration due to gravity)

a) T = 2π√(l/g)     b) T = 2π√(g/l)     c) T = 2πl/g     d) T = 2πg/l
Solution: [T] = 2π has no dimensions. [√(l/g)] = √(L/LT−2) = √(T2) = T. So option (a) has dimension of time. The others have dimensions of √(T−2) = T−1 (b), T2 (c), L−1T2 (d). Only (a) is dimensionally correct.

Significant Figures

Rules for counting significant figures: (1) All non-zero digits are significant. (2) Zeros between non-zero digits are significant. (3) Leading zeros are not significant. (4) Trailing zeros after decimal are significant. (5) In scientific notation N × 10m, all digits in N are significant. For addition/subtraction, round to the least precise decimal place. For multiplication/division, round to the least number of significant figures. Examples: 5200 (2 significant figures, unless written as 5200. which has 4), 0.0045 (2 significant figures), 3.1416 (5 significant figures). Use scientific notation to avoid ambiguity: 5.2 × 103 clearly has 2 significant figures.

Rounding Off

Rules for rounding off: (1) If the digit to be dropped is less than 5, the preceding digit is unchanged. (2) If the digit to be dropped is greater than 5, the preceding digit is increased by 1. (3) If the digit to be dropped is exactly 5, the preceding digit is increased by 1 if it is odd, unchanged if it is even (rule of even digit). Example: 3.45 rounds to 3.4 (4 is even), 3.35 rounds to 3.4 (3 is odd).

Error Analysis

Absolute Error = |Measured Value − True Value|. Mean Absolute Error = (Σ|Δxi|)/n. Relative Error = Mean Absolute Error / Mean Value. Percentage Error = Relative Error × 100%. For a quantity Z = ApBq / Cr, the maximum relative error is ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C).

💡 Error Propagation Rule
When quantities are multiplied/divided, add relative errors. When raised to powers, multiply the relative error by the power. For addition/subtraction, add absolute errors. This is a high-yield NEET concept tested every year. In NEET, percentage error questions are very common.
Example 3: Error Analysis
The resistance R = V/I where V = (100 ± 5) V and I = (10 ± 0.2) A. Find the percentage error in R.

a) 5%     b) 7%     c) 10%     d) 12%
Solution: ΔR/R = ΔV/V + ΔI/I = 5/100 + 0.2/10 = 0.05 + 0.02 = 0.07 = 7%. Option (b).

2. Kinematics

Motion in 1D

Equations of Motion (Constant Acceleration)

For motion with constant acceleration a, the three primary equations are:

  • v = u + at
  • s = ut + ½at2
  • v2 = u2 + 2as

where u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement. These are valid only for uniformly accelerated motion along a straight line. They can also be expressed in vector form for 2D motion: v = u + at, r = ut + ½at2.

Graphs in Kinematics

Graphical analysis is a powerful tool for solving kinematics problems. Three key graph types are used:

  • Displacement–Time (s–t) graph: Slope = velocity. A straight line indicates constant velocity; a curved line indicates acceleration. The area under the curve has no physical meaning.
  • Velocity–Time (v–t) graph: Slope = acceleration. Area under the graph = displacement (distance for speed–time). A horizontal line represents uniform velocity; a sloping line represents uniform acceleration.
  • Acceleration–Time (a–t) graph: Area under the graph = change in velocity. Area = vf − vi.

These graphs are interconvertible: differentiate s–t to get v–t; differentiate v–t to get a–t. Integrate a–t to get v–t; integrate v–t to get s–t. For v–t graphs: when the graph crosses the time axis, the body reverses direction. NEET problems frequently ask: "Which v–t graph corresponds to this s–t graph?" or vice versa.

Table: Kinematic Graph Characteristics
Graph Slope Area NEET Pattern
s vs tVelocityParabola for uniform acceleration
v vs tAccelerationDisplacementStraight line for uniform acceleration
a vs tChange in velocityHorizontal line for constant acceleration
💡 NEET Graph Shortcut
In v–t graphs, the displacement is the NET area (area above the axis minus area below the axis). For uniform acceleration, the s–t graph is a parabola. If acceleration is zero, v–t graph is horizontal; s–t graph is a straight line. A common NEET trick: the slope of a chord on the s–t graph gives average velocity; the slope of a tangent gives instantaneous velocity.
Table 3: Key Kinematic Variables
Variable Symbol SI Unit Nature
DisplacementsmVector
Velocityvm/sVector
Accelerationam/s2Vector
Speedvm/sScalar
DistancedmScalar

Derived formulas for motion under gravity: For free fall (a = g), the equations become v = u + gt, h = ut + ½gt2, v2 = u2 + 2gh. For upward projection, g is taken as negative (a = −g). Time of ascent = time of descent = u/g. Maximum height = u2/2g. For a body dropped from rest, distance travelled in nth second: Sn = u + a(2n − 1)/2 = g(2n − 1)/2. For vertically projected body, the speed at any height h is v = √(u2 − 2gh). The ratio of distances travelled in successive seconds is 1:3:5:7:... for a body starting from rest.

Table: Sign Conventions for Motion Under Gravity
Scenario u a s
Dropped from restu = 0+g+h (downward)
Thrown upward+u (up)−g+h (up), −h (down)
Thrown downward+u (down)+g+h (down)
Projected from height+u+g+h (down)

Graphical Analysis

Slope of s-t graph gives velocity. Slope of v-t graph gives acceleration. Area under v-t graph gives displacement (with sign). Area under a-t graph gives change in velocity. For non-uniform acceleration, the slope of the v-t graph at any point gives instantaneous acceleration. Key points for graph interpretation:

  • In a v-t graph, a horizontal line means constant velocity (a = 0)
  • A straight line with positive slope means constant positive acceleration
  • A curved line means non-uniform acceleration
  • The area between v-t graph and time axis = displacement; if the graph goes below the axis, that area represents negative displacement
  • The magnitude of displacement = area under speed-time graph (always positive)
💡 NEET Graph Shortcut
In v-t graphs: area above time-axis is positive displacement, below is negative. For uniformly accelerated motion, displacement = average velocity × time = ½(u + v)t. Use this to avoid solving quadratics. For s-t graphs, the slope at any point equals instantaneous velocity. In a-t graphs, the area between the graph and time-axis gives the change in velocity.
Example 4: 1D Kinematics
A particle starts from rest and moves with constant acceleration 2 m/s2 for 5 s, then continues with constant velocity for next 3 s, and finally decelerates uniformly to rest in 2 s. Find total distance travelled.

a) 45 m     b) 55 m     c) 65 m     d) 75 m
Solution: Phase 1 (acceleration): s1 = ½ × 2 × 52 = 25 m, v = 2 × 5 = 10 m/s. Phase 2 (constant v): s2 = 10 × 3 = 30 m. Phase 3 (deceleration): s3 = ½ × 10 × 2 = 10 m. Total = 25 + 30 + 10 = 65 m. Option (c).
Example 5: Free Fall
A stone is dropped from the top of a tower 80 m high. Find the time taken to reach the ground. (g = 10 m/s2)

a) 2 s     b) 3 s     c) 4 s     d) 5 s
Solution: s = ut + ½gt2. Here u = 0, s = 80 m, g = 10 m/s2. So 80 = 0 + ½ × 10 × t2. t2 = 80/5 = 16. t = 4 s. Option (c).
Example 6: Stopping Distance
A car moving at 72 km/h applies brakes and stops after travelling 40 m. Find the deceleration and time taken.

a) 4 m/s2, 5 s     b) 5 m/s2, 4 s     c) 5 m/s2, 5 s     d) 4 m/s2, 4 s
Solution: 72 km/h = 72 × 5/18 = 20 m/s. Using v2 = u2 + 2as: 0 = 400 + 2a(40). a = −400/80 = −5 m/s2. Using v = u + at: 0 = 20 + (−5)t. t = 4 s. Option (b).

Motion in 2D

Vectors in Kinematics

Position vector r = xᴋ + yᴌ. Velocity v = dr/dt = vxᴋ + vyᴌ. Acceleration a = dv/dt = axᴋ + ayᴌ. For uniform acceleration: r = r0 + ut + ½at2. Component-wise analysis is key to solving 2D problems. The motion in x and y directions are independent of each other.

Vector Addition and Resolution

Vectors follow the triangle and parallelogram laws of addition. For two vectors A and B with angle θ between them, the magnitude of their resultant R = √(A2 + B2 + 2AB cosθ), and the direction is given by tanφ = B sinθ/(A + B cosθ). Resolution of a vector into components: Ax = A cosθ, Ay = A sinθ (where θ is measured from the x-axis). Unit vectors along the axes: ᴋ, ᴌ, ᴍ. The dot product A·B = AB cosθ = AxBx + AyBy + AzBz. The cross product A × B = AB sinθ with direction given by the right-hand rule. Cross product magnitude equals the area of the parallelogram formed by the vectors. For NEET, focus on vector addition (co-planar vectors) and resolution into rectangular components using trigonometric functions.

Relative velocity in 2D: vAB = vA − vB (vector subtraction). For two bodies moving at angles, relative velocity magnitude = √(vA2 + vB2 − 2vAvBcosθ). The shortest distance between two moving bodies can be found using relative velocity concepts.

Projectile Motion

A projectile launched with initial velocity u at angle θ with horizontal follows a parabolic trajectory under constant gravitational acceleration g (ignoring air resistance). The horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to gravity. The motion can be analysed by treating the horizontal and vertical components independently.

  • Horizontal velocity: ux = u cosθ (constant throughout)
  • Vertical velocity: uy = u sinθ − gt (changes due to g)
  • Time of flight: T = 2u sinθ / g
  • Maximum height: H = u2 sin2θ / 2g
  • Range: R = u2 sin2θ / g
  • Equation of trajectory: y = x tanθ − gx2 / 2u2 cos2θ

Projectile from a height: If a projectile is fired from a height h above ground level, the equation of trajectory method is used. Time of flight is obtained by solving for y = −h in the vertical motion equation. Range = u cosθ × T. For horizontal projection (θ = 0) from height h: T = √(2h/g), R = u × √(2h/g). For a projectile fired horizontally from a moving platform (e.g., an aeroplane at altitude h with speed u), the bomb's initial horizontal velocity equals the plane's speed. The bombs fall in a parabolic path and are always vertically below the plane (neglecting air resistance). The distance from the target = u × √(2h/g).

Projectile on an inclined plane: When a projectile is fired up an incline (angle β) with velocity u at angle θ with the incline: range up the incline R = 2u2cosθsin(θ+β)/gcos2β. Time of flight T = 2u sin(θ+β)/(g cosβ). Maximum range on an inclined plane occurs at θ = 45° − β/2. For a projectile fired down an incline, the angle for maximum range is θ = 45° + β/2. These formulas are useful for NEET, but it is often faster to resolve the acceleration due to gravity along and perpendicular to the incline.

💡 Projectile Shortcuts for NEET
(1) Range is same for complementary angles θ and (90° − θ). (2) Maximum range at θ = 45°. (3) Rmax = u2/g. (4) H/R = tanθ/4. (5) At highest point, KE = ½m(u cosθ)2 and PE = ½m(u sinθ)2 so KE/PE = cot2θ. (6) For a given speed, the product of time of flight and range is maximum at θ = 45°. (7) If a projectile is fired from a height, solve using the equation of trajectory directly.
Example 7: Projectile Motion
A projectile is fired with velocity 20 m/s at 30° to horizontal. Find its range. (g = 10 m/s2)

a) 20√3 m     b) 30 m     c) 40 m     d) 20 m
Solution: R = u2 sin2θ / g = (20)2 × sin60° / 10 = 400 × (√3/2) / 10 = 20√3 m. Option (a).
Example 8: Projectile Height & Time
A ball is thrown at 30 m/s at 60° to horizontal. Find maximum height and time of flight. (g = 10 m/s2)

a) 33.75 m, 5.2 s     b) 22.5 m, 3 s     c) 45 m, 6 s     d) 33.75 m, 3 s
Solution: u sinθ = 30 × √3/2 = 25.98 m/s. H = (u sinθ)2/2g = (25.98)2/20 = 675/20 = 33.75 m. T = 2u sinθ/g = 2 × 25.98/10 = 5.2 s. Option (a).

Relative Motion

Velocity of A relative to B: vAB = vA − vB. For relative velocity in 2D, vector subtraction applies. In rain-man problems, the relative velocity of rain w.r.t. man determines the angle at which the man should hold his umbrella. In river-boat problems, the velocity of the boat relative to ground is the vector sum of boat velocity relative to water and water velocity relative to ground.

💡 Relative Motion Tips
(1) For crossing a river in minimum time, head straight perpendicular to the bank. (2) For minimum drift, the boat should head upstream at angle θ = sin−1(vr/vb). (3) In rain-man problems, tanθ = vman/vrain (where θ is the angle from the vertical). (4) For two bodies moving towards each other, relative velocity = sum of speeds.
Example: River-Boat Problem
A river flows at 3 m/s. A boat can travel at 5 m/s in still water. Find the time to cross a 200 m wide river if the boat heads straight perpendicular to the bank. Also find the drift.

a) 40 s, 120 m     b) 50 s, 100 m     c) 40 s, 100 m     d) 50 s, 120 m
Solution: Time = width / vb = 200/5 = 40 s. Drift = vr × t = 3 × 40 = 120 m. Option (a).

3. Laws of Motion

Newton's Laws

First Law (Law of Inertia): A body continues in its state of rest or uniform motion unless acted upon by an external unbalanced force. Inertia is the property of mass that resists change in motion. Greater mass means greater inertia. Examples: when a bus suddenly stops, passengers lurch forward (due to inertia of motion); when a bus suddenly starts, passengers fall backward (inertia of rest).

Second Law: The rate of change of momentum is directly proportional to the applied force and takes place in the direction of the force. Mathematically, F = dp/dt = d(mv)/dt = ma (for constant mass). The SI unit of force is newton (N) = kg·m/s2. The second law also gives the concept of impulse: Impulse J = FΔt = Δp = m(v − u). The impulse-momentum theorem is especially useful for problems involving varying forces and collisions where the time of contact is known.

Third Law: Every action has an equal and opposite reaction. FAB = −FBA. Action-reaction pairs act on different bodies, never on the same body. Examples: a gun recoils when a bullet is fired; a rocket propulsion works by expelling gases downward; a man walking on ground pushes the ground backward and the ground pushes him forward.

💡 Newton's Laws & Impulse Tips
(1) Impulse = area under F-t graph. (2) For a bouncing ball, change in momentum = m(v + u) (not m(v − u)) because direction reverses. (3) The reaction force from a wall when a ball hits it depends on the time of contact (impulse = F_avg × Δt). (4) In problems involving connected bodies, always draw FBDs separately for each body to avoid sign errors.

Analysis of Bouncing Ball

The motion of a ball bouncing off a floor is a rich problem combining free fall, impact, and coefficient of restitution. Key stages: (1) Ball dropped from height h: speeds up to v = √(2gh) just before impact. (2) During impact (Δt ≈ 0.01 s): large impulsive force acts upward; impulse J = m(v' + v) = m(v + ev). (3) Ball rebounds with speed v' = ev = e√(2gh) and rises to height h' = e2h. (4) The total time for n bounces: total time = √(2h/g)[1 + 2e + 2e2 + ...] = √(2h/g)(1 + e)/(1 − e). (5) The total distance travelled before stopping = h + 2e2h + 2e4h + ... = h(1 + e2)/(1 − e2). These geometric series results are frequently tested in NEET and other competitive exams.

Example: Bouncing Ball Height
A ball is dropped from 10 m and e = 0.6. Find the height after the second bounce.

a) 3.6 m     b) 1.296 m     c) 6.0 m     d) 0.78 m
Solution: After first bounce, h₁ = e²h = 0.36 × 10 = 3.6 m. After second bounce, h₂ = e²h₁ = 0.36 × 3.6 = 1.296 m. Option (b).
Example: Impulse-Momentum
A ball of mass 0.2 kg moving at 10 m/s strikes a wall and rebounds with the same speed. If contact time is 0.02 s, find the average force on the ball.

a) 100 N     b) 200 N     c) 300 N     d) 400 N
Solution: Change in momentum = m(v − (−u)) = 0.2(10 + 10) = 4 kg·m/s. F_avg = Δp/Δt = 4/0.02 = 200 N. Option (b).

Free Body Diagrams

A free body diagram (FBD) isolates a body and shows all forces acting on it. Steps: (1) Identify the system/body of interest. (2) Draw all external forces (gravity mg downwards, normal reaction N perpendicular to surface, tension T along the string away from the body, friction f opposing relative motion, applied forces). (3) Resolve forces into components along convenient axes (usually along the direction of acceleration and perpendicular to it). (4) Apply ΣF = ma along each axis. For equilibrium (a = 0), ΣF = 0 in all directions.

Connected Motion (Pulley Systems)

For systems of connected bodies, treat the entire system to find acceleration, then isolate individual bodies to find tension. For an Atwood machine (two masses m1 and m2 connected by a string over a frictionless, massless pulley): a = (m1 − m2)g / (m1 + m2) and T = 2m1m2g / (m1 + m2). For systems with multiple blocks and pulleys, write constraint equations relating the accelerations. The number of independent constraint equations equals the number of strings.

Mass on incline connected to hanging mass: If mass m1 is on a smooth incline at angle θ and mass m2 is hanging vertically, connected by a string over a pulley at the top: a = (m2g − m1g sinθ)/(m1 + m2). If the incline is rough with coefficient μ, subtract kinetic friction from the numerator: a = (m2g − m1g sinθ − μm1g cosθ)/(m1 + m2). Tension T = m2(g − a) = m1(a + g sinθ + μg cosθ). For multiple pulley systems (block and tackle), the mechanical advantage equals the number of rope segments supporting the load.

Example: Connected Bodies on Incline
A 4 kg block on a smooth 30° incline is connected by a string over a pulley to a 2 kg hanging mass. Find acceleration. (g = 10 m/s2)

a) 0 m/s2     b) 2 m/s2     c) 4 m/s2     d) 6 m/s2
Solution: For m1 = 4 kg on incline, m2 = 2 kg hanging. a = (m2g − m1g sinθ)/(m1 + m2) = (20 − 40 × 0.5)/6 = (20 − 20)/6 = 0 m/s2. The system is in equilibrium. Option (a).

Pseudo Forces (Non-Inertial Frames)

When analysing motion from a non-inertial (accelerated) frame of reference, a pseudo force Fp = −ma0 must be applied to each body of mass m, where a0 is the acceleration of the frame. This is useful for problems involving accelerating lifts, vehicles, or rotating reference frames. For an accelerating lift: apparent weight = m(g ± a). If lift accelerates upward, apparent weight = m(g + a) (heavier feeling). If lift accelerates downward, apparent weight = m(g − a) (lighter feeling). If the cable breaks (free fall), a = g, apparent weight = 0 (weightlessness).

Galilean Relativity: The laws of mechanics are the same in all inertial frames. The velocity of a body in frame S' moving with constant velocity v relative to frame S is v' = v − V. Acceleration is invariant (same in all inertial frames): a' = a. This is why a ball dropped from a moving train appears to fall vertically to an observer on the train, but follows a parabolic path to an observer on the ground.

Rotating frames and centrifugal force: In a rotating reference frame with angular velocity ω, pseudo forces experienced include: centrifugal force Fcf = mω2r (away from centre) and Coriolis force Fcor = −2m(ω × v) (deflects moving objects sideways). The Coriolis force explains the direction of cyclones (counterclockwise in Northern hemisphere, clockwise in Southern) and the deviation of long-range projectiles. Centrifugal force affects the effective value of g at different latitudes.

Example: Apparent Weight in Lift
A 60 kg person stands on a weighing scale in a lift. Find the scale reading when the lift descends with acceleration 2 m/s2. (g = 10 m/s2)

a) 480 N     b) 600 N     c) 720 N     d) 0 N
Solution: When lift descends with acceleration a, apparent weight = m(g − a) = 60(10 − 2) = 60 × 8 = 480 N. The scale reads 480 N. Option (a).

Lami's Theorem (Equilibrium of Concurrent Forces)

If three concurrent forces F1, F2, F3 acting at a point keep it in equilibrium, then each force is proportional to the sine of the angle between the other two forces: F1/sinα = F2/sinβ = F3/sinγ, where α is the angle between F2 and F3, β between F1 and F3, and γ between F1 and F2. This theorem is very useful for solving problems involving strings, springs, and walls in equilibrium without resolving into components.

Example: Lami's Theorem
A block of weight 100 N is suspended by two strings making angles 30° and 60° with the horizontal. Find tensions in the strings.

a) 50 N, 86.6 N     b) 86.6 N, 50 N     c) 100 N, 100 N     d) 50 N, 50 N
Solution: Angles between forces: between T1 and T2 = 90°, between T1 and W = 150° (180−30), between T2 and W = 120° (180−60). Using Lami: T1/sin120° = T2/sin150° = W/sin90° = 100/1 = 100. T1 = 100 sin120° = 100 × √3/2 = 86.6 N. T2 = 100 sin150° = 100 × 1/2 = 50 N. Option (b).
Example 9: Atwood Machine
Two masses 4 kg and 2 kg are connected by a string over a frictionless pulley. Find acceleration and tension. (g = 10 m/s2)

a) 3.33 m/s2, 26.7 N     b) 5 m/s2, 20 N     c) 3.33 m/s2, 20 N     d) 5 m/s2, 26.7 N
Solution: a = (4 − 2)g / (4 + 2) = 2 × 10 / 6 = 3.33 m/s2. T = 2m1m2g / (m1 + m2) = 2 × 4 × 2 × 10 / 6 = 26.7 N. Option (a).

Friction

Friction is a contact force that opposes relative motion or tendency of relative motion between surfaces in contact.

Table 4: Types of Friction
Type Formula Condition
Static Frictionfs ≤ μsNBody at rest; self-adjusting up to limit
Limiting FrictionfL = μsNMaximum static friction just before motion
Kinetic Frictionfk = μkNBody is sliding
Rolling Frictionfr = μrNBody is rolling

Typically: μs > μk > μr. N is the normal reaction force. Friction is independent of the area of contact (to first approximation). The angle of repose α = tan−1s) is the maximum angle of an inclined plane before a block slips. For an inclined plane at angle θ:

  • If tanθ < μs: block remains at rest (friction = mg sinθ)
  • If tanθ = μs: block is on verge of slipping (limiting friction)
  • If tanθ > μs: block slides down with acceleration a = g(sinθ − μk cosθ)

For a block pushed up an incline with force F parallel to the incline: net acceleration depends on whether F exceeds the sum of mg sinθ and friction. For a block on a rough horizontal surface pulled at angle θ above horizontal: N = mg − F sinθ, so friction is reduced, making it easier to move the block.

💡 Friction NEET Tricks
(1) On an inclined plane, slipping begins when tanθ > μs. (2) For a block on a horizontal surface pushed at angle θ below horizontal, N = mg + F sinθ. (3) For a block on a horizontal surface pulled at angle θ above horizontal, N = mg − F sinθ. (4) Minimum force to move a block on horizontal surface is at angle θ = tan−1(μ) from horizontal. (5) The cone of friction is the cone with semi-vertical angle λ = tan−1s); if the resultant of all external forces lies inside this cone, the block will not move regardless of the magnitude of the force.
Example 10: Friction on Incline
A block of mass 10 kg rests on a horizontal surface with μs = 0.4 and μk = 0.3. A horizontal force F is applied. Find minimum F to just move the block. (g = 10 m/s2)

a) 30 N     b) 40 N     c) 50 N     d) 60 N
Solution: Limiting friction fs,max = μsN = 0.4 × 10 × 10 = 40 N. To just move the block, applied force must exceed 40 N. Minimum F = 40 N. Option (b).
Example: Inclined Plane with Friction
A 5 kg block slides down a 37° rough incline with μk = 0.25. Find acceleration. (g = 10 m/s2, sin37° = 0.6, cos37° = 0.8)

a) 2 m/s2     b) 4 m/s2     c) 6 m/s2     d) 8 m/s2
Solution: Component of gravity down the incline = mg sinθ = 50 × 0.6 = 30 N. Normal reaction N = mg cosθ = 50 × 0.8 = 40 N. Kinetic friction fk = μkN = 0.25 × 40 = 10 N. Net force = 30 − 10 = 20 N. a = F/m = 20/5 = 4 m/s2. Option (b).

Circular Motion

For uniform circular motion, the speed is constant but the velocity direction changes continuously, producing centripetal acceleration directed toward the centre. Centripetal acceleration ac = v2/r = ω2r. Centripetal force Fc = mv2/r = mω2r. Angular velocity ω = 2π/T = 2πf, where T is time period and f is frequency. The centripetal force is not a separate force but is provided by tension, gravity, friction, or normal reaction depending on the situation.

Example 11: Circular Motion
A car of mass 1000 kg moves on a circular track of radius 50 m at 20 m/s. What is the centripetal force required?

a) 4000 N     b) 8000 N     c) 10000 N     d) 20000 N
Solution: Fc = mv2/r = 1000 × (20)2 / 50 = 1000 × 400 / 50 = 8000 N. Option (b).
💡 Circular Motion Concepts
(1) On a banked road (angle θ), optimum speed v = √(rg tanθ) for no friction dependence; friction reduces banking requirement. (2) In vertical circular motion, Ttop = mv2/r − mg, Tbottom = mv2/r + mg; minimum speed at top = √(gr). (3) For a conical pendulum, T = 2π√(L cosθ/g). (4) Centrifugal force is a pseudo force observed in rotating frames. (5) In death well, the limiting speed to avoid falling is v = √(gr/μ).
Example: Banking of Roads
A circular road of radius 100 m is banked at an angle such that no friction is needed at 20 m/s. Find the banking angle. (g = 10 m/s2)

a) tan−1(0.2)     b) tan−1(0.4)     c) tan−1(0.6)     d) tan−1(0.8)
Solution: tanθ = v2/rg = 400/(100 × 10) = 0.4. θ = tan−1(0.4). Option (b).

Friction in Circular Motion

For a vehicle negotiating a flat circular curve, the centripetal force is provided entirely by friction. The maximum safe speed without skidding is vmax = √(μsrg), where μs is the coefficient of static friction between tyres and road. For a banked curve (angle θ), the optimum speed requiring no friction is vo = √(rg tanθ). If friction is also present, the maximum safe speed on a banked curve increases: vmax = √(rg(μs + tanθ)/(1 − μs tanθ)). In practice, roads are banked for the average expected speed, and friction handles deviations. In a "death well" (vertical circular wall), the minimum speed to avoid sliding down is v = √(rg/μ), where the normal reaction provides centripetal force and friction balances weight.

Example: Safe Speed on Curved Road
A circular road of radius 80 m has μs = 0.5 between tyres and road. Find maximum safe speed without skidding. (g = 10 m/s2)

a) 15 m/s     b) 20 m/s     c) 25 m/s     d) 40 m/s
Solution: vmax = √(μsrg) = √(0.5 × 80 × 10) = √400 = 20 m/s. Option (b).

Motion in a Vertical Circle

A body moving in a vertical circle experiences both centripetal force and varying gravitational force. The tension or normal reaction varies at different points. For a mass m tied to a string whirled in a vertical circle of radius R:

  • At topmost point (A): Minimum speed vA = √(gR) for string tautness. TA = mvA2/R − mg = 0 (at minimum speed).
  • At lowest point (B): vB = √(5gR) for completing the circle. TB = mvB2/R + mg = 6mg.
  • At horizontal position (C): vC = √(3gR). TC = mvC2/R = 3mg.

Critical condition: vB ≥ √(5gR) for complete vertical circle. If vB < √(2gR), the mass oscillates about the lowest point. If √(2gR) < vB < √(5gR), the mass leaves the circle between the horizontal and topmost positions and follows a parabolic path. For a bead on a rigid circular wire, the bead can complete the circle at any non-zero speed because the wire provides radial support (unlike a string which can only provide tension). The difference in tension between lowest and highest points is always 6mg, independent of speed, which is a useful NEET shortcut.

Table: Vertical Circle Speed and Tension Summary
Position Angle from Vertical Velocity (for complete circle) Tension
Lowest√(5gR)6mg
Horizontal90°√(3gR)3mg
Highest180°√(gR)0
Example: Vertical Circle Kinematics
A 0.2 kg stone tied to a 0.5 m string is whirled in a vertical circle. Find minimum speed at the lowest point to just complete the circle. (g = 10 m/s2)

a) 3 m/s     b) 4 m/s     c) 5 m/s     d) 6 m/s
Solution: For completing vertical circle, vB ≥ √(5gR) = √(5 × 10 × 0.5) = √25 = 5 m/s. Minimum speed = 5 m/s. Option (c).

4. Work, Energy & Power

Work

Work done by a constant force: W = F·s = Fs cosθ, where θ is the angle between force and displacement vectors. Work is a scalar quantity and its SI unit is joule (J).

  • Positive work: 0° ≤ θ < 90° (force aids motion; e.g., gravity when falling)
  • Zero work: θ = 90° (force perpendicular to displacement; e.g., centripetal force, tension in a string for horizontal circular motion)
  • Negative work: 90° < θ ≤ 180° (force opposes motion; e.g., friction, gravity when moving upward)

Work done by a variable force is given by the area under the force-displacement graph: W = ∫F·ds.

Work-Energy Theorem

Work done by net force equals change in kinetic energy: Wnet = ΔKE = KEf − KEi = ½m(vf2 − vi2). This is a powerful tool that avoids detailed kinematic analysis and can solve problems involving non-uniform acceleration.

Conservative vs Non-Conservative Forces

Conservative forces (gravity, spring force, electrostatic force): Work done is path-independent; work done around a closed path is zero; total mechanical energy is conserved. Δ(KE + PE) = 0. The work done by a conservative force equals the negative of the change in potential energy: Wc = −ΔPE.

Non-conservative forces (friction, air drag, viscous force): Work done depends on path; mechanical energy is not conserved. WNC = ΔKE + ΔPE = change in total mechanical energy.

Table 5: Forms of Energy in Mechanics
Energy Form Expression Notes
Kinetic Energy (translational)½mv2Due to linear motion of centre of mass
Kinetic Energy (rotational)½Iω2Due to rotation about an axis
Gravitational PE (near Earth)mghh measured from reference level; valid for h << R
Elastic PE (spring)½kx2k = spring constant, x = extension/compression
Gravitational PE (general)−GMm/rr from centre of mass; zero at infinity; always negative
Table 5b: Work Done by Different Forces
Force Work Done Remarks
Gravity (near Earth)W = mghPath-independent; depends only on vertical height
Spring ForceW = −½k(xf2 − xi2)Negative of change in spring PE
FrictionW = −fk × dAlways negative; depends on path length d
Centripetal ForceW = 0Always perpendicular to displacement
Tension (in string)W = 0For a fixed-length string, no work done
Variable Force (general)W = ∫F·dxArea under F-x graph

Power

Power is the rate of doing work: P = dW/dt = F · v (for constant force). SI unit: watt (W) = J/s. Average power = Total work / Total time. Instantaneous power = Fv cosθ. Power is also equal to the rate at which energy is transferred or converted.

💡 Work-Energy NEET Shortcuts
(1) Work done by gravity depends only on vertical height, not path. (2) When only conservative forces act, use: KEi + PEi = KEf + PEf. (3) For spring problems, total energy = ½kA2 (A = amplitude in SHM). (4) Work done by friction is always negative and equals fk × distance travelled. (5) For a body sliding down a rough incline, the loss in PE = gain in KE + work done against friction.
Example 12: Work-Energy Theorem
A body of mass 2 kg is dropped from height 20 m. Find its speed just before hitting the ground. (g = 10 m/s2)

a) 10 m/s     b) 15 m/s     c) 20 m/s     d) 25 m/s
Solution: Using energy conservation: mgh = ½mv2. v = √(2gh) = √(2 × 10 × 20) = √400 = 20 m/s. Option (c).
Example 13: Spring Energy
A spring of spring constant 200 N/m is compressed by 0.05 m. Find the work done in compressing it.

a) 0.25 J     b) 0.125 J     c) 0.5 J     d) 0.05 J
Solution: Work = ½kx2 = ½ × 200 × (0.05)2 = 100 × 0.0025 = 0.25 J. Option (a).

Work-Energy Theorem in Rotational Motion

The work-energy theorem extends to rotational motion: work done by torque = change in rotational kinetic energy. W = ∫τdθ = Δ(½Iω2). For a body undergoing both translation and rotation, total work done by forces = change in total KE: W = Δ(½mvcm2 + ½Icmω2). A classic problem: a yo-yo falling while unwinding string uses this principle to relate linear and angular motion. The power delivered by a torque P = τω.

💡 Rotational Work-Energy Tips
(1) For a rolling body, the work done by friction is zero (if no slipping), because the point of contact is instantaneously at rest. (2) In a perfectly inelastic collision between rotating bodies, angular momentum is conserved but KE is not. (3) The work done in rotating a body from rest to angular speed ω equals ½Iω2. (4) Torque and angular displacement must be in the same direction for positive work.
Example: Elastic Collision
A 2 kg ball moving at 3 m/s strikes a 1 kg ball at rest head-on elastically. Find velocities after collision.

a) 1 m/s, 4 m/s     b) 2 m/s, 3 m/s     c) 0 m/s, 6 m/s     d) 1 m/s, 5 m/s
Solution: v1 = (m1−m2)u1/(m1+m2) = (2−1)×3/3 = 1 m/s. v2 = 2m1u1/(m1+m2) = 4×3/3 = 4 m/s. Option (a).
Example: Power
A motor lifts 500 kg at constant speed of 0.5 m/s. Find power output of the motor. (g = 10 m/s2)

a) 2500 W     b) 1250 W     c) 250 W     d) 500 W
Solution: P = Fv cosθ = mgv = 500 × 10 × 0.5 = 2500 W. Option (a).

Mechanical Advantage, Velocity Ratio & Efficiency

Simple machines (levers, pulleys, inclined planes, screws, wedges, gears) help lift heavy loads with less effort. Key parameters: Mechanical Advantage (MA) = Load/Effort. Velocity Ratio (VR) = Distance moved by effort / Distance moved by load. Efficiency η = MA/VR = Work output / Work input = Useful work done / Total work done. For an ideal (frictionless) machine, MA = VR and η = 1 (100%). In real machines, friction reduces efficiency. For a lever: MA = Load arm / Effort arm (for first-class lever like see-saw or crowbar). For a single fixed pulley: MA = 1 (only changes direction). For a single movable pulley: MA = 2. For a system of n pulleys (block and tackle): MA = n (ideal). For an inclined plane: MA = Length/Height = 1/sinθ. The screw jack has a very high MA = 2πL/p, where L = lever arm length and p = pitch of the screw. For NEET, focus on lever and pulley problems with efficiency.

Example: Efficiency of Pulley System
A pulley system lifts a 500 N load by applying 125 N effort over 4 m distance. If the load rises 0.5 m, find the efficiency.

a) 50%     b) 75%     c) 80%     d) 90%
Solution: Work input = Effort × Effort distance = 125 × 4 = 500 J. Work output = Load × Load distance = 500 × 0.5 = 250 J. Efficiency = Wout/Win = 250/500 = 0.5 = 50%. Option (a).

Variable Mass Systems: Rocket Propulsion

For systems where mass changes with time (like a rocket ejecting fuel), Newton's second law is applied in its original form F = dp/dt = d(mv)/dt. The rocket equation (Tsiolkovsky equation) gives the velocity of a rocket at any instant: v = v0 + u ln(m0/m), where u is the exhaust velocity of gases relative to the rocket, m0 is the initial total mass, and m is the mass at that instant. The thrust force on the rocket = u(dm/dt). For a rocket launched vertically, net acceleration a = (u dm/dt)/m − g. The total impulse delivered = u(m0 − m). Multistage rockets use staging to drop empty fuel tanks, improving the mass ratio and achieving higher final velocities.

Example: Rocket Velocity
A rocket with initial mass 5000 kg ejects gas at 2 km/s relative to the rocket. If 80% of the mass is fuel, find the final velocity of the rocket (neglect gravity).

a) 1.6 km/s     b) 3.2 km/s     c) 4.0 km/s     d) 2.0 km/s
Solution: m0 = 5000 kg, m = 5000 − 0.8 × 5000 = 1000 kg. u = 2 km/s = 2000 m/s. v = u ln(m0/m) = 2000 × ln(5000/1000) = 2000 × ln5 = 2000 × 1.609 = 3218 m/s ≈ 3.2 km/s. Option (b).

5. Rotational Motion

Kinematics of Rotational Motion

Rotational motion is described by angular displacement (θ), angular velocity (ω = dθ/dt), and angular acceleration (α = dω/dt = d2θ/dt2). For constant angular acceleration, the equations are completely analogous to linear kinematics: ω = ω0 + αt, θ = ω0t + ½αt2, ω2 = ω02 + 2αθ. The relationship between linear and angular quantities for a point at distance r from the axis: s = rθ (arc length), v = rω (tangential speed), at = rα (tangential acceleration), ac = v2/r = rω2 (centripetal/radial acceleration). Total linear acceleration magnitude: a = √(at2 + ac2). The direction of the total acceleration is not along the radius unless α = 0 (uniform circular motion). The angular velocity vector direction is given by the right-hand thumb rule.

Table: Linear vs Angular Kinematics - Complete Analogy
Linear Quantity Angular Equivalent Relation
Displacement sθ (radians)s = rθ
Velocity vω (rad/s)v = rω
Acceleration aα (rad/s2)at = rα
Mass mMoment of inertia II = Σmr2
Force FTorque ττ = r × F
Momentum pAngular momentum LL = Iω
KE = ½mv2KE = ½Iω2
Example: Angular Kinematics
A wheel rotating at 600 rpm is uniformly accelerated to 1800 rpm in 10 s. Find angular acceleration and number of rotations in this time.

a) 4π rad/s2, 200 rotations     b) 2π rad/s2, 200 rotations     c) 4π rad/s2, 100 rotations     d) 2π rad/s2, 100 rotations
Solution: ω0 = 600 × 2π/60 = 20π rad/s. ω = 1800 × 2π/60 = 60π rad/s. α = (ω − ω0)/t = 40π/10 = 4π rad/s2. θ = ω0t + ½αt2 = 20π(10) + ½(4π)(100) = 200π + 200π = 400π rad. Rotations = θ/2π = 200. Option (a).

Centre of Mass

For a system of particles, the centre of mass (COM) is the weighted average of positions: RCM = (Σmiri) / (Σmi). For continuous bodies, RCM = (∫r dm) / M. Velocity of COM: VCM = (Σmivi) / M. Acceleration of COM: ACM = Fext / M. In the absence of external forces, COM remains at rest or moves with constant velocity. The COM of a system of particles can lie outside the material of the body (e.g., a ring, a hollow sphere, or a boomerang). COM of common bodies: uniform rod = mid-point; triangle = centroid at h/3 from base; semicircular ring = 2R/π from centre; semicircular disc = 4R/3π from centre; solid hemisphere = 3R/8 from centre; hollow hemisphere = R/2 from centre.

Example: Centre of Mass
Two particles of masses 2 kg and 3 kg are located at (1,2) m and (4,6) m respectively. Find the position of their centre of mass.

a) (2.8, 4.4) m     b) (2.5, 4.0) m     c) (3.0, 4.5) m     d) (2.8, 4.0) m
Solution: xCM = (2×1 + 3×4)/(2+3) = (2+12)/5 = 14/5 = 2.8 m. yCM = (2×2 + 3×6)/5 = (4+18)/5 = 22/5 = 4.4 m. Option (a).

Moment of Inertia

Moment of inertia I = Σmiri2 (discrete system) or I = ∫r2 dm (continuous body). It measures rotational inertia, analogous to mass in linear motion. SI unit: kg·m2. Two important theorems:

  • Perpendicular Axis Theorem (for planar bodies): Iz = Ix + Iy, where z is perpendicular to the plane.
  • Parallel Axis Theorem: I = ICM + Mh2, where h is the distance between the parallel axes.
Table 6: Moment of Inertia of Common Shapes (NEET Important)
Body Axis I
Thin RingThrough centre, perpendicular to planeMR2
DiscThrough centre, perpendicular to planeMR2/2
Solid SphereThrough centre (diametral)2MR2/5
Spherical ShellThrough centre (diametral)2MR2/3
Thin RodThrough centre, perpendicular to rodML2/12
Thin RodThrough end, perpendicular to rodML2/3
Hollow CylinderAxis of cylinderMR2
Solid CylinderAxis of cylinderMR2/2
Rectangular PlateThrough centre, perpendicular to planeM(a2+b2)/12

Torque & Angular Momentum

Torque: τ = r × F, magnitude τ = rF sinθ. Direction is given by the right-hand rule (perpendicular to plane of r and F). Torque is the rotational analogue of force. The net external torque about a point equals the rate of change of angular momentum: τext = dL/dt. For a system of particles, τext = Iα (where α = dω/dt is angular acceleration).

Angular momentum: L = r × p = m(r × v). For rotation about a fixed axis: L = Iω. The direction of L is along the axis of rotation (right-hand rule). Conservation of angular momentum: if τext = 0, then L = constant, i.e., I1ω1 = I2ω2. This explains why a spinning ice skater rotates faster when pulling arms in (decreasing I increases ω to keep L constant), why a diver tucks in to spin faster, and why a cat always lands on its feet.

Couple: A pair of equal, parallel, and opposite forces separated by distance d produces a torque (couple) of magnitude τ = Fd, independent of the point about which it is taken. A couple produces pure rotation without translation.

Equilibrium of rigid bodies: For a body to be in complete equilibrium, both net force and net torque must be zero: ΣF = 0 and Στ = 0 (about any point). This is used to solve problems involving ladders, beams, and see-saws.

Table: Translational vs Rotational Quantities
Translational Rotational Relation
Mass (m)Moment of Inertia (I)I = Σmr2
Force (F)Torque (τ)τ = r × F
Linear Momentum (p)Angular Momentum (L)L = r × p
F = maτ = Iαα = τ/I
KE = ½mv2KE = ½Iω2v = ωR
W = ∫F·dxW = ∫τdθP = τω
Impulse = ΔpAngular Impulse = ΔL∫τdt = ΔL
Example: Torque on a Beam
A uniform beam of length 6 m and mass 20 kg is pivoted at its centre. Where should a 30 kg child sit to balance a 40 kg child sitting at one end?

a) 0.5 m from centre     b) 1 m from centre     c) 1.5 m from centre     d) 2 m from centre
Solution: Taking torques about centre: 40g × 3 = 30g × d. So d = 40 × 3 / 30 = 4 m from centre on the other side. But the 40 kg child is at 3 m from centre, so 30 kg child at 4 m from centre would balance. However, the beam is only 6 m total (3 m each side). So the 30 kg child sits at 3 m from centre (at the opposite end). Wait: 40 × 3 = 120, so d = 120/30 = 4 m. Since max distance is 3 m, the beam cannot balance. This indicates the problem has no solution with these masses. The 30 kg child must sit at the opposite end (3 m) giving torque = 30g × 3 = 90g, while the 40 kg child gives 40g × 3 = 120g. The beam is unbalanced.

Rolling Motion

For pure rolling without slipping: v = ωR. Total KE = KEtrans + KErot = ½mv2 + ½Iω2 = ½mv2(1 + I/mR2). The friction in pure rolling is static and does no work (point of contact is instantaneously at rest). For rolling down an incline, acceleration a = g sinθ / (1 + I/mR2). The object with smaller I/mR2 reaches bottom first.

💡 Rotational Motion Shortcuts
(1) For rolling without slipping, vcm = ωR, acm = αR. (2) KErolling = ½mvcm2(1 + k2/R2) where k = radius of gyration. (3) Acceleration of a rolling body on an incline: a = g sinθ / (1 + k2/R2). (4) The body with smaller k2/R2 reaches bottom first (solid sphere > disc > hollow sphere > ring). (5) Angular momentum L = mvr about a point; use right-hand rule for direction. (6) Torque is the rate of change of angular momentum: τ = dL/dt.

Centre of Mass Motion & Collisions

Velocity of centre of mass: vcm = (m1v1 + m2v2 + ...)/(m1 + m2 + ...) = total momentum / total mass. In the absence of external forces, the velocity of the centre of mass remains constant (total momentum conserved). The centre of mass behaves as if all mass is concentrated there and all external forces act there. For two-body collisions, the centre of mass frame simplifies analysis significantly.

Collisions in Centre of Mass Frame: In the centre of mass frame, the total momentum is zero. For elastic collisions, the speeds of approach and separation are equal in the centre of mass frame: relative speed of approach = relative speed of separation. The coefficient of restitution e = relative speed of separation / relative speed of approach. For perfectly elastic collision, e = 1; for perfectly inelastic, e = 0; for partially elastic, 0 < e < 1.

Table: Collision Characteristics in 1D
Type KE Conserved e Value Final Velocities (equal masses)
Elastic (head-on)Yes1v1' = v2, v2' = v1 (exchange)
InelasticNo0 < e < 1Depends on e and masses
Perfectly InelasticNo (max loss)0v' = (m1v1 + m2v2)/(m1 + m2)
Example: Collision with Coefficient of Restitution
A ball is dropped from height 10 m onto a horizontal floor. If e = 0.5, find the height to which it rebounds after the first bounce. (g = 10 m/s2)

a) 2.5 m     b) 5 m     c) 1.25 m     d) 7.5 m
Solution: Speed just before impact v = √(2gh) = √(2 × 10 × 10) = √200 = 10√2 m/s. Speed just after impact v' = e × v = 0.5 × 10√2 = 5√2 m/s. Rebound height h' = v'2/2g = (50)/(20) = 2.5 m. Option (a).
Example 14: Moment of Inertia
Four particles each of mass m are placed at corners of a square of side a. Find MOI about an axis through centre perpendicular to the plane.

a) ma2     b) 2ma2     c) 3ma2     d) 4ma2
Solution: Each particle is at distance r = a/√2 from centre. I = Σmr2 = 4m(a/√2)2 = 4m(a2/2) = 2ma2. Option (b).
Example 15: Angular Momentum Conservation
A disc with I = 2 kg·m2 rotates at 20 rad/s. A ring with I = 1 kg·m2 is gently placed coaxially on it. Find new angular velocity.

a) 10 rad/s     b) 13.3 rad/s     c) 15 rad/s     d) 20 rad/s
Solution: Conservation of angular momentum: I1ω1 = (I1 + I22. 2 × 20 = (2 + 1) × ω2. ω2 = 40/3 = 13.3 rad/s. Option (b).
Example: Rolling Down an Incline
A solid sphere (I = 2MR2/5) rolls down a 30° incline of height 10 m without slipping. Find its speed at the bottom. (g = 10 m/s2)

a) 10 m/s     b) √(1000/7) m/s     c) √(200/7) m/s     d) √140 m/s
Solution: Energy conservation: mgh = ½mv2 + ½Iω2 = ½mv2 + ½(2MR2/5)(v/R)2 = ½mv2(1 + 2/5) = ½mv2(7/5). So v = √(10gh/7) = √(10 × 10 × 10 / 7) = √(1000/7) m/s. Option (b).

6. Gravitation

Kepler's Laws

First Law (Law of Orbits): Each planet moves in an ellipse with the Sun at one focus. The semi-major axis a and eccentricity e define the orbit's shape. The eccentricity e = √(1 − b2/a2), where b is the semi-minor axis. For Earth's orbit, e ≈ 0.0167 (nearly circular).

Second Law (Law of Areas): The line joining the planet to the Sun sweeps equal areas in equal time intervals. dA/dt = L/2m = constant, where L is the angular momentum. This implies that the planet moves faster when closer to the Sun (perihelion) and slower when farther (aphelion). The areal velocity is constant: dA/dt = ½r2ω = constant.

Third Law (Law of Periods): The square of the orbital period is proportional to the cube of the semi-major axis. T2 ∝ a3. For circular orbits around a central mass M: T2 = (4π2/GM)r3. This is derived by equating centripetal force to gravitational force: GMm/r2 = mω2r = m(2π/T)2r.

Types of Satellites

Geostationary satellites: Orbit in the equatorial plane with T = 24 h, matching Earth's rotation. Orbital radius r ≈ 42000 km (altitude ≈ 36000 km). They appear stationary from Earth. Used for communication, broadcasting, weather monitoring.

Low Earth Orbit (LEO) satellites: Altitude 200-2000 km, T ≈ 90 min, v ≈ 7.9 km/s. Used for remote sensing, spy satellites, International Space Station.

Polar satellites: Orbit passing over the poles at low altitude (800 km). Used for mapping, environmental monitoring, reconnaissance.

Table: Satellite Orbit Comparison
Parameter LEO Satellite Geostationary Polar Satellite
Altitude200-2000 km~36000 km500-800 km
Orbital Period~90 min24 h~100 min
Orbital Speed~7.9 km/s~3.1 km/s~7.5 km/s
InclinationVaries (0-90°)0° (equatorial)~90°
Visibility from EarthMoves across skyFixed positionCovers poles
ApplicationsRemote sensing, ISSCommunication, TVWeather, mapping
💡 Satellite NEET Points
(1) The period of a satellite depends only on the orbital radius, NOT on the mass of the satellite. (2) For a given orbit, KE = −TE = GMm/2r, PE = −2KE = −GMm/r. (3) To shift a satellite to a higher orbit, energy must be supplied. (4) A satellite in orbit is in free fall; objects inside appear weightless. (5) Parking orbit (geostationary) is unique to Earth. (6) Binding energy of a satellite = |TE| = GMm/2r.

Universal Gravitation

Newton's law of universal gravitation: F = Gm1m2/r2, where G = 6.67 × 10−11 N·m2/kg2. Gravitational force is attractive, central (acts along the line joining the centres), and follows the inverse square law. It is the weakest of the four fundamental forces but dominates at astronomical scales. Gravitational force is conservative.

Gravitational Field of Continuous Distributions

For calculating the gravitational field due to extended objects, we use integration or known results. Key results for NEET:

  • Field inside a uniform spherical shell: E = 0 everywhere inside the shell. Potential V = constant = −GM/R (where R = radius of shell). This is a very important result: a spherical shell exerts no gravitational force on a particle inside it.
  • Field outside a spherical shell or solid sphere: E = GM/r2 (as if all mass is concentrated at the centre).
  • Field inside a uniform solid sphere: E = GMr/R3 (directly proportional to distance from centre). At centre, E = 0. Potential V = −GM(3R2 − r2)/2R3.
  • Field due to a uniform circular ring: On the axis at distance x from centre, E = GMx/(R2 + x2)3/2. Maximum field at x = R/√2.

These results are frequently tested in NEET in comparison-type questions (e.g., "the gravitational field inside a hollow sphere is...") and in problems requiring integration or application of Gauss's law for gravity analogously to electrostatics.

Gravitational Field & Potential

Gravitational field intensity E = F/m = GM/r2 (directed towards the mass causing the field). It is a vector quantity.

Gravitational potential V = W/m = −GM/r (scalar). It is the work done per unit mass in bringing a test mass from infinity to that point.

Gravitational potential energy U = −GMm/r. Relationship between field and potential: E = −dV/dr. The negative sign indicates that the field points in the direction of decreasing potential. The superposition principle applies: total field/potential is the vector/scalar sum of individual contributions.

Escape & Orbital Velocity

Escape Velocity: Minimum velocity required for an object to escape Earth's gravitational field (reach infinity with zero residual speed): ve = √(2GM/R) = √(2gR). For Earth: ve ≈ 11.2 km/s. It is independent of the mass of the escaping object and depends only on the mass and radius of the planet.

Orbital Velocity: Velocity required for a satellite to orbit at radius r: vo = √(GM/r). For a circular orbit just above Earth's surface, vo ≈ 7.9 km/s. Relation: ve = √2 × vo (for same r). vo decreases with increasing orbital radius.

Energy of a satellite: Total energy E = KE + PE = GMm/2r + (−GMm/r) = −GMm/2r (negative, indicating a bound state). To increase orbital radius, energy must be added to the satellite.

Table 7: Gravitational Parameters of Earth
Parameter Value Unit
Mass (M)5.97 × 1024kg
Radius (R)6.37 × 106m
g at surface9.8m/s2
Escape velocity11.2km/s
Orbital velocity (LEO)7.9km/s
Gravitational constant G6.67 × 10−11N·m2/kg2
💡 Collision NEET Tips
(1) In perfectly elastic head-on collision with equal masses, velocities are exchanged. (2) If a body collides with a much heavier body at rest, its velocity reverses direction (e = 1) or it moves with the heavy body (e = 0). (3) Energy loss in inelastic collision: ΔKE = ½(m1m2/(m1+m2))(u1−u2)2(1−e2). (4) In collision problems, always conserve momentum first, then apply the coefficient of restitution if needed. (5) For a collision between two bodies, the velocity of approach before collision equals the velocity of separation after collision only for perfectly elastic collisions (e = 1).
Example: Oblique Collision
A ball moving at 4 m/s strikes a stationary ball of equal mass. After elastic collision, one moves at 2 m/s at 60° to the original direction. Find the speed of the other ball.

a) 2√3 m/s     b) 2 m/s     c) 4 m/s     d) 3 m/s
Solution: Momentum conservation: x: 4m = 2m cos60° + mv2cosθ = m + mv2cosθ. y: 0 = 2m sin60° − mv2sinθ = m√3 − mv2sinθ. From y: v2sinθ = √3. From KE conservation (elastic, equal mass): 42 = 22 + v22. v22 = 16 − 4 = 12. v2 = 2√3 m/s. Option (a).
Example 16: Escape Velocity
If the mass of Earth becomes 4 times and radius becomes half, the escape velocity becomes:

a) doubled     b) halved     c) 4 times     d) 2√2 times
Solution: ve = √(2GM/R). New ve' = √(2G · 4M / (R/2)) = √(2GM/R × 8) = √8 × ve = 2√2 × ve. Option (d).
Example 17: g at Depth
At what depth below Earth's surface does g become half its surface value? (R = 6400 km)

a) 1600 km     b) 2400 km     c) 3200 km     d) 4800 km
Solution: g' = g(1 − d/R). Setting g' = g/2: g/2 = g(1 − d/R). 1/2 = 1 − d/R. d/R = 1/2. d = R/2 = 3200 km. Option (c).

7. Elasticity

Stress & Strain

Stress = Deforming Force / Area (SI: N/m2 or Pa). Stress is of three types: longitudinal (tensile/compressive), volumetric (bulk), and shear (tangential). Strain = Change in dimension / Original dimension (dimensionless). Within elastic limit, stress ∝ strain (Hooke's Law). The proportionality constant is the modulus of elasticity.

Elastic Moduli

Young's Modulus Y = Longitudinal Stress / Longitudinal Strain = (F/A) / (ΔL/L) = FL / AΔL. It measures resistance to length change. For a wire: ΔL = FL/AY. Spring constant of a wire: k = YA/L. For a composite wire made of two different materials joined end-to-end, the extension in each part is proportional to (L/Y) of that part. For parallel combination of wires, the net spring constant k = k1 + k2 and for series combination, 1/k = 1/k1 + 1/k2.

Bulk Modulus B = Volume Stress / Volume Strain = −ΔP / (ΔV/V). It measures resistance to volume change. Compressibility = 1/B. For gases, isothermal bulk modulus = P (for ideal gas at constant temperature), adiabatic bulk modulus = γP (where γ = Cp/Cv). For liquids, B is very large (liquids are nearly incompressible).

Shear Modulus (Rigidity Modulus) G = Shear Stress / Shear Strain = (F/A) / θ. It measures resistance to shape change. Only solids can sustain shear stress; fluids cannot. For a cubical block subjected to a tangential force on its top face, shear strain = x/h where x is the displacement of the top face and h is the height.

Table 8: Elastic Moduli Comparison
Modulus Definition Formula
Young's Modulus (Y)Longitudinal stress / Longitudinal strainFL / AΔL
Bulk Modulus (B)Volume stress / Volume strain−ΔP V / ΔV
Shear Modulus (G)Shear stress / Shear strainF / Aθ

Stress-Strain Curve & Elastic Hysteresis

The stress-strain curve for a ductile material (e.g., mild steel) shows distinct regions: (1) Proportionality limit (Hooke's law is valid up to this point; stress ∝ strain), (2) Elastic limit (beyond which permanent deformation occurs; the material will not return to original shape when load is removed), (3) Yield point (a large strain increase for a very small increase in stress; the material appears to flow), (4) Ultimate tensile strength (the maximum stress the material can withstand; necking begins), (5) Breaking point (fracture occurs, also called the fracture point). For brittle materials (e.g., cast iron, glass), there is no significant plastic region; fracture occurs near the elastic limit. For elastomers (e.g., rubber), there is a large elastic region with no well-defined yield point.

Elastic hysteresis: The area between the loading and unloading stress-strain curves represents the energy dissipated as heat within the material. This is called elastic hysteresis. Materials with high hysteresis (e.g., rubber) are used in shock absorbers and vibration dampers. The elastic after-effect is the delay in regaining original shape after unloading.

Poisson's Ratio

When a body is stretched longitudinally, it contracts laterally (and vice versa). Poisson's ratio σ (or μ) = lateral strain / longitudinal strain = −(ΔD/D) / (ΔL/L). The negative sign compensates for the fact that lateral strain is opposite in sign to longitudinal strain. For most materials, σ is between 0 and 0.5. Cork has σ ≈ 0 (no lateral contraction), rubber has σ ≈ 0.5 (nearly incompressible), and steel has σ ≈ 0.3. The theoretical limits are −1 < σ < 0.5. Poisson's ratio relates the three elastic moduli: Y = 3B(1 − 2σ) and Y = 2G(1 + σ). These relations are useful in NEET to find one modulus given the others.

Elastic Potential Energy

When a wire is stretched, work done is stored as elastic potential energy. The energy stored in a stretched wire = ½ × Stress × Strain × Volume = ½ F × ΔL (since volume = A·L). Elastic energy per unit volume = ½ × Stress × Strain = ½ Y (Strain)2. For a spring: U = ½kx2. This energy is recoverable when the material is unloaded (within elastic limit). The energy density concept is important in NEET problems asking for the energy stored in a stretched wire or rod.

Applications: Bending of Beams & Torsion

Bending of beams: When a beam is loaded, it bends. The top surface contracts (compression) and the bottom surface extends (tension). The neutral axis (at the centre of the beam) experiences no strain. The bending moment M = YIG/R, where IG is the geometric moment of inertia of the cross-section and R is the radius of curvature. For a rectangular beam, IG = bd3/12. The depression of a cantilever (beam fixed at one end and loaded at the other) δ = WL3/3YIG. This explains why I-beams are stronger (most material is far from the neutral axis, maximising IG).

Torsion of a cylinder: When a cylinder is twisted, shear stress develops. The torque required to twist a solid cylinder of radius R through angle φ over length L is τ = (πGR4φ)/(2L). The quantity (πR4/2) is called the polar moment of inertia. Torsion pendulums and drive shafts use this principle.

Table 10: Comparison of Material Behaviour Under Stress
Material Type Ductile (e.g., Mild Steel) Brittle (e.g., Glass) Elastomer (e.g., Rubber)
Plastic RegionSignificant (large strain before fracture)NegligibleVery small
FractureAfter neckingSudden, at elastic limitGradual tearing
Energy AbsorptionHigh (large area under curve)LowVery high (hysteresis)
Young's ModulusHigh (~2 × 1011 Pa)ModerateVery low

Elasticity in Structural Design

Understanding elasticity is crucial for designing safe structures. Factor of safety = Ultimate stress / Working stress. For bridges and buildings, a high factor of safety (3–5) is used. The choice of material depends on the application: steel has high Young's modulus (stiffness) for beams and columns; rubber has low Young's modulus for shock absorbers and seals; concrete has high compressive strength but low tensile strength, so steel reinforcement (RCC) is used. The elastic limit must never be exceeded in normal use; permanent deformation would compromise structural integrity. The bending of beams, deflection of bridges, and twisting of drive shafts all rely on elasticity principles. The maximum safe load a wire can support without exceeding the elastic limit is given by Fmax = Stresselastic limit × Area.

💡 Elasticity Key Points
(1) Young's modulus depends only on material, not on dimensions of the object. (2) For a wire: ΔL = FL/AY, i.e., extension ∝ length ∝ load ∝ 1/area. (3) Spring constant k = YA/L for a wire. (4) Elastic potential energy per unit volume = ½ × Stress × Strain = ½Y (Strain)2. (5) Poisson's ratio σ = Lateral strain / Longitudinal strain (dimensionless, typically between −1 and 0.5). (6) For a given material, the breaking stress is fixed; a thicker wire can support a larger load. (7) The area under the stress-strain curve gives the toughness of the material.
Example 18: Young's Modulus
A steel wire of length 2 m and cross-sectional area 1 mm2 is stretched by 1 mm under a 100 N load. Find Young's modulus.

a) 2 × 1011 Pa     b) 2 × 1010 Pa     c) 1 × 1011 Pa     d) 1 × 1010 Pa
Solution: Y = FL/AΔL = 100 × 2 / (1 × 10−6 × 1 × 10−3) = 200 / 10−9 = 2 × 1011 Pa. Option (a).
Example: Bulk Modulus
A 2 L steel sphere is submerged to a depth where pressure increases by 2 × 105 Pa. Bulk modulus of steel = 1.6 × 1011 Pa. Find the volume change.

a) 1.5 mm3     b) 2.5 mm3     c) 3.0 mm3     d) 4.0 mm3
Solution: B = −ΔP V / ΔV. ΔV = −ΔP V / B = (2 × 105)(0.002) / (1.6 × 1011) = 4 × 102 / 1.6 × 1011 = 2.5 × 10−9 m3 = 2.5 mm3. Option (b).
Example: Shear Modulus
A cubical block of side 0.1 m and G = 8 × 1010 Pa is subjected to a shear force of 2 × 107 N on its top face. Find the shear strain.

a) 0.0125     b) 0.025     c) 0.05     d) 0.1
Solution: Shear stress = F/A = 2 × 107 / (0.1 × 0.1) = 2 × 109 Pa. Shear strain = Shear stress / G = 2 × 109 / 8 × 1010 = 0.025. Option (b).

9. Oscillations (Simple Harmonic Motion)

SHM Basics

Simple Harmonic Motion is a periodic oscillatory motion where the restoring force is directly proportional to the displacement from the mean position and directed opposite to it. Mathematically, F = −kx, where k is the force constant. This leads to the differential equation d2x/dt2 = −ω2x, whose solution is x = A sin(ωt + φ) or x = A cos(ωt + φ').

Parameters of SHM

  • Amplitude (A): Maximum displacement from mean position
  • Angular frequency (ω): ω = 2π/T = 2πf
  • Time period (T): T = 2π/ω = 2π√(m/k) for spring-mass system
  • Frequency (f): f = 1/T = ω/2π
  • Phase (φ): Initial phase determines the starting position

Velocity & Acceleration in SHM

Velocity: v = ±ω√(A2 − x2). Maximum velocity (at mean position x = 0): vmax = ωA. Acceleration: a = −ω2x. Maximum acceleration (at extreme positions x = ±A): amax = ω2A.

Energy in SHM

Total mechanical energy in SHM is constant: E = ½kA2 = ½mω2A2. Kinetic energy varies as KE = ½mω2(A2 − x2). Potential energy varies as PE = ½mω2x2. At mean position (x = 0): KE = E, PE = 0. At extreme positions (x = ±A): KE = 0, PE = E.

Table: Comparison at Different Positions in SHM
Position x v a KE PE
Mean0±ωA (max)0max0
Extreme±A0±ω2A (max)0max
At x = A/2±A/2±√3ωA/2±ω2A/275% of E25% of E

Simple Pendulum

For a simple pendulum with small amplitude (sinθ ≈ θ): T = 2π√(L/g). The period is independent of the mass of the bob and the amplitude (for small amplitudes). For a compound pendulum: T = 2π√(I/mgd).

Spring-Mass Systems

For a horizontal spring: T = 2π√(m/k). For a vertical spring: same formula; the equilibrium extension due to gravity does not affect the period. For springs in series: 1/keq = 1/k1 + 1/k2. For springs in parallel: keq = k1 + k2.

💡 SHM NEET Shortcuts
(1) IN SHM, vavg over one complete cycle = 0, but speedavg = 2ωA/π. (2) KE and PE are equal when x = A/√2. (3) The total energy is proportional to the square of the amplitude. (4) For a liquid column oscillating in a U-tube, T = 2π√(L/g). (5) For a ball oscillating in a concave bowl of radius R, T = 2π√((R−r)/g).
Example: SHM Energy
A particle of mass 0.5 kg executes SHM of amplitude 0.1 m and angular frequency 10 rad/s. Find total energy.

a) 1.25 J     b) 2.5 J     c) 0.5 J     d) 0.25 J
Solution: E = ½mω2A2 = 0.5 × 0.5 × 102 × 0.12 = 0.25 × 100 × 0.01 = 0.25 J. Option (d).
Example: Simple Pendulum
A simple pendulum has a period of 2 s at a place. If its length is doubled, what will be the new period?

a) 2√2 s     b) 4 s     c) 1 s     d) √2 s
Solution: T ∝ √L. T2/T1 = √(L2/L1) = √2. T2 = 2√2 s. Option (a).
Example: Spring-Mass System
A spring of force constant 400 N/m is attached to a 4 kg mass. Find the time period of oscillation.

a) 0.2π s     b) 0.4π s     c) 0.5π s     d) 0.1π s
Solution: T = 2π√(m/k) = 2π√(4/400) = 2π√(0.01) = 2π × 0.1 = 0.2π s. Option (a).

Damped & Forced Oscillations

Damped oscillations: When energy is dissipated (due to friction, air resistance, etc.), the amplitude of oscillation decreases over time. The damping force is proportional to velocity: Fd = −bv. The equation becomes m(d2x/dt2) + b(dx/dt) + kx = 0. For light damping, x = A0e(−bt/2m) sin(ω't + φ), where ω' = √(ω02 − (b/2m)2). The amplitude decays exponentially.

Forced oscillations: When an external periodic force F0 sin(ωt) is applied to an oscillator, it oscillates at the driving frequency ω. At resonance (ω = ω0), the amplitude becomes maximum. This is used in tuning circuits, bridges (Tacoma Narrows Bridge collapse), and musical instruments. The sharpness of resonance is measured by the quality factor Q = ω0/Δω, where Δω is the half-width of the resonance curve. A high Q indicates low damping and sharp resonance.

Other Oscillatory Systems

Torsional Pendulum: A disc suspended by a wire executes torsional oscillations. The restoring torque is τ = −Cθ, where C is the torsional constant. Time period T = 2π√(I/C). This is independent of amplitude for small twists. Used in torsion pendulums for timing and in the Cavendish experiment to measure G.

Physical Pendulum (Compound Pendulum): Any rigid body pivoted at a point not through its centre of mass oscillates as a pendulum. T = 2π√(I/mgd), where I = moment of inertia about the pivot, d = distance from pivot to centre of mass. The equivalent length of a simple pendulum Leq = I/md. For a uniform rod of length L pivoted at one end, I = mL2/3, d = L/2, so T = 2π√(2L/3g).

LC Oscillations: In an LC circuit, charge oscillates with ω = 1/√(LC). This is a mechanical analogy of SHM where charge q ←→ displacement x, current I ←→ velocity v, inductance L ←→ mass m, 1/C ←→ spring constant k. NEET sometimes asks SHM analogy questions.

Example: Compound Pendulum
A uniform rod of length 1 m is pivoted at one end. Find its period of small oscillation. (g = 10 m/s2)

a) 1.62 s     b) 2.00 s     c) 2.57 s     d) 1.00 s
Solution: For a rod pivoted at end, I = mL2/3, d = L/2. T = 2π√(I/mgd) = 2π√((mL2/3)/(mgL/2)) = 2π√(2L/3g) = 2π√(2 × 1 / 30) = 2π√(0.0667) = 2π × 0.258 = 1.62 s. Option (a).

8. Fluid Mechanics

Pressure & Density

Pressure P = F/A. SI unit: pascal (Pa) = N/m2. 1 atm = 1.013 × 105 Pa = 760 mm of Hg = 101.3 kPa. In a static fluid, pressure at depth h: P = P0 + ρgh, where P0 is atmospheric pressure at the surface. Pressure is same at all points at the same depth in a static fluid, irrespective of shape of the container (hydrostatic paradox). Density ρ = m/V (kg/m3). Relative density (specific gravity) = ρsubstance / ρwater. A barometer measures atmospheric pressure (using a mercury column), while a manometer measures gauge pressure (pressure relative to atmosphere). Absolute pressure = Gauge pressure + Atmospheric pressure.

Table 9: Fluid Properties & Their Densities
Substance Density (kg/m3) Relative Density
Water (4°C)10001.00
Mercury1360013.6
Sea Water10251.025
Air (STP)1.290.00129
Ice9170.917
Aluminium27002.70
Gold1930019.3

Pascal's Law

Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. Hydraulic lift: F2/F1 = A2/A1. This provides mechanical advantage: a small force on a small-area piston can lift a heavy load on a large-area piston. Hydraulic brakes and hydraulic jacks work on this principle.

Archimedes' Principle

The buoyant force on a body immersed in a fluid equals the weight of the displaced fluid: Fb = ρfluid Vimmersed g. A body floats if its average density is less than the fluid density. Apparent weight = Actual weight − Buoyant force. Fraction of volume immersed = ρbody / ρfluid. For a floating body, weight = buoyant force. A body sinks if its density exceeds that of the fluid.

Bernoulli's Theorem & Applications

For an ideal incompressible fluid in streamline (non-turbulent) flow: P + ½ρv2 + ρgh = constant along a streamline. This is a statement of energy conservation for fluid flow. The constant is the total pressure. Important applications:

  • Venturimeter: Measures flow speed using pressure difference at constriction. The flow rate Q = A1A2√(2gh)/(A12 − A22).
  • Atomiser / Sprayer: Perfume spray works by creating low pressure at the constriction, drawing liquid up from the reservoir.
  • Aerofoil lift: Airplane wing shape causes higher velocity above the wing (longer path) and lower pressure, creating lift. F = (Pbelow − Pabove)A.
  • Magnus effect: Spinning ball in cricket/football curves due to pressure difference caused by unequal airspeeds on opposite sides.
  • Torricelli's theorem: Efflux velocity from an orifice at depth h: v = √(2gh). This is a special case of Bernoulli with P1 = P2 = Patm and v1 ≈ 0 at the free surface.
  • Blood flow and aneurysm: In a constricted artery, velocity is higher and pressure is lower (Venturi effect). An aneurysm (ballooning of artery) occurs where pressure is higher after the constriction.
  • Pitot tube: Measures fluid flow velocity by comparing stagnation pressure and static pressure: v = √(2(P0 − P)/ρ).

Torricelli's theorem in detail: The speed of efflux of a liquid from an orifice in a tank at depth h below the free surface is v = √(2gh). The horizontal range of the efflux stream hitting the ground below the tank bottom: if the orifice is at height y above the ground and the liquid level is H above the ground (so h = H − y), then time to fall t = √(2y/g) and range R = vt = √(2(H−y) × 2y) = 2√(y(H−y)). Maximum range occurs when y = H/2 (i.e., the orifice is at half the total height).

Example: Torricelli's Theorem
A water tank has a small hole at a depth of 5 m below the water surface. Find the efflux speed. (g = 10 m/s2)

a) 5 m/s     b) 10 m/s     c) 15 m/s     d) 20 m/s
Solution: v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s. Option (b).

Viscosity

Viscous force between adjacent layers of fluid: F = ηA(dv/dx), where η is the coefficient of viscosity (SI: Pa·s). Stokes' law: viscous drag force on a sphere moving through a viscous fluid = 6πηrv. Terminal velocity of a sphere falling under gravity in a viscous fluid: vt = 2r2(ρ − σ)g / 9η, where ρ = density of sphere, σ = density of fluid.

Streamline & Turbulent Flow

In streamline (laminar) flow, fluid particles follow smooth paths that do not cross each other, with the velocity at any point being constant in time. In turbulent flow, the fluid undergoes irregular fluctuations and mixing with eddies. The critical velocity vc is the speed at which the transition from laminar to turbulent occurs. Reynolds number Re = ρvD/η, where D is the diameter. Flow is laminar if Re < 1000, turbulent if Re > 2000, and transitional in between. Higher viscosity promotes laminar flow; higher velocity and larger diameter promote turbulent flow. The equation of continuity and Bernoulli's equation apply only to streamline flow of ideal (non-viscous, incompressible) fluids.

Example: Reynolds Number
Water (ρ = 1000 kg/m3, η = 0.001 Pa·s) flows through a pipe of diameter 0.02 m at 1.5 m/s. Find the Reynolds number and flow regime.

a) 30000, turbulent     b) 15000, turbulent     c) 3000, transitional     d) 30000, laminar
Solution: Re = ρvD/η = 1000 × 1.5 × 0.02 / 0.001 = 30/0.001 = 30000. Since Re > 2000, flow is turbulent. Option (a).

Surface Tension

Surface tension S = F/L (N/m). It arises due to cohesive forces between liquid molecules at the surface, creating a tendency to minimise surface area. The surface of a liquid behaves like a stretched elastic membrane under tension. Excess pressure inside a liquid drop = 2S/R. Inside a soap bubble (two surfaces) = 4S/R. Inside a liquid jet = S/R.

Capillary rise/fall: h = 2S cosθ / ρgr, where θ is the contact angle, r is the radius of the capillary tube. For water (θ ≈ 0°, cosθ ≈ 1), h = 2S/ρgr (capillary rise). This phenomenon causes water to rise in a narrow tube, in soil, and in filter paper. For mercury (θ ≈ 140°, cosθ is negative), capillary depression occurs. Factors affecting capillary rise: h ∝ T, h ∝ 1/ρ, h ∝ 1/r. In a capillary tube of insufficient length (L < h), the liquid will not overflow; instead the meniscus radius adjusts so that the liquid rises to the full height of the tube and the meniscus becomes hemispherical (contact angle changes).

Surface energy = S × ΔA, where ΔA is the change in surface area. This is the work required to increase the surface area by ΔA at constant temperature. For a liquid film with two surfaces (like a soap bubble), total work = 2S × ΔA. The surface tension decreases with increasing temperature; it becomes zero at the critical temperature. The surface tension of water at 20°C = 0.0727 N/m. Addition of soap reduces the surface tension of water, which helps in cleaning by improving wetting. The angle of contact determines whether a liquid wets the surface (θ < 90°) or does not wet the surface (θ > 90°).

Example: Capillary Rise
A capillary tube of radius 0.5 mm is dipped in water (ρ = 1000 kg/m3, S = 0.072 N/m). Find the height of water rise. (g = 10 m/s2, cosθ = 1)

a) 2.88 cm     b) 5.76 cm     c) 1.44 cm     d) 28.8 cm
Solution: h = 2S cosθ / ρgr = 2 × 0.072 × 1 / (1000 × 10 × 0.5 × 10−3) = 0.144 / 5 = 0.0288 m = 2.88 cm. Option (a).
Example: Excess Pressure in a Bubble
Find the excess pressure inside a soap bubble of radius 2 cm if surface tension is 0.03 N/m.

a) 3 Pa     b) 6 Pa     c) 12 Pa     d) 0.75 Pa
Solution: For a soap bubble with two surfaces, excess pressure ΔP = 4S/R = 4 × 0.03 / 0.02 = 0.12/0.02 = 6 Pa. Option (b).
💡 Fluid Mechanics NEET Must-Know
(1) Equation of continuity: A1v1 = A2v2 (mass conservation for incompressible flow). (2) For efflux through an orifice, the horizontal range x = 2√h(H−h) where h = depth of hole, H = total height of water column. Maximum range at h = H/2. (3) In a liquid at rest, pressure is same at all points on a horizontal plane. (4) For a body floating in a liquid, buoyant force = weight of body. (5) The pressure at a point in a static fluid is the same in all directions (Pascal's law). (6) For a body immersed completely, the apparent loss in weight equals weight of displaced fluid. (7) The surface tension of soap solution is less than that of pure water, which allows soap bubbles to form more easily. (8) The terminal velocity of a sphere is proportional to the square of its radius, so larger particles settle faster in sedimentation.
Example 19: Bernoulli's Principle
Water flows through a pipe of radius 0.1 m at 2 m/s. If the pipe narrows to 0.05 m radius, what is the speed in the narrow section?

a) 4 m/s     b) 6 m/s     c) 8 m/s     d) 10 m/s
Solution: By equation of continuity: A1v1 = A2v2. A = πr2. π(0.1)2 × 2 = π(0.05)2 × v2. v2 = (0.01 × 2) / 0.0025 = 0.02/0.0025 = 8 m/s. Option (c).
Example 20: Buoyancy
A block of density 600 kg/m3 and volume 0.001 m3 is placed in water (ρ = 1000 kg/m3). What fraction of its volume remains above the water surface?

a) 0.2     b) 0.4     c) 0.6     d) 0.8
Solution: For floating, Weight = Buoyant force. ρbVg = ρwVimmg. Vimm/V = ρbw = 600/1000 = 0.6. Fraction above = 1 − 0.6 = 0.4. Option (b).
Example: Viscous Drag & Terminal Velocity
A spherical ball of radius 2 mm and density 8000 kg/m3 falls through a liquid (η = 1 Pa·s, σ = 1000 kg/m3). Find terminal velocity. (g = 10 m/s2)

a) 0.024 m/s     b) 0.062 m/s     c) 0.124 m/s     d) 0.248 m/s
Solution: vt = 2r2(ρ − σ)g / 9η = 2 × (0.002)2 × (8000 − 1000) × 10 / (9 × 1) = 2 × 4 × 10−6 × 7000 × 10 / 9 = 0.062 m/s. Option (b).
Example: Pascal's Law (Hydraulic Lift)
In a hydraulic lift, the smaller piston has area 0.01 m2 and the larger piston has area 0.5 m2. A force of 100 N is applied on the smaller piston. Find the load lifted.

a) 2000 N     b) 5000 N     c) 10000 N     d) 2500 N
Solution: Pressure is same: F1/A1 = F2/A2. F2 = F1 × A2/A1 = 100 × 0.5/0.01 = 100 × 50 = 5000 N. Option (b).

📋 Formula Sheet

▶ 1. Units & Measurements
QuantityFormulaUnitDimensions
Densityρ = m/Vkg/m³[ML⁻³]
Speedv = s/tm/s[LT⁻¹]
Accelerationa = v/tm/s²[LT⁻²]
ForceF = maN[MLT⁻²]
WorkW = F·sJ[ML²T⁻²]
EnergyKE = ½mv²J[ML²T⁻²]
PowerP = W/tW[ML²T⁻³]
PressureP = F/APa[ML⁻¹T⁻²]
Momentump = mvkg·m/s[MLT⁻¹]
ImpulseJ = F·ΔtN·s[MLT⁻¹]

NEET Tip: Dimensional analysis can check formula correctness. Only quantities with same dimensions can be added/subtracted.

▶ 2. Kinematics

Equations of Motion (constant acceleration):

v = u + at

s = ut + ½at²

v² = u² + 2as

sn = u + a(2n−1)/2 (distance in nth second)


Projectile Motion:

Time of flight: T = 2u sinθ / g

Maximum height: H = u² sin²θ / 2g

Range: R = u² sin2θ / g

Maximum range at θ = 45° → Rmax = u²/g

Equation of trajectory: y = x tanθ − gx²/(2u² cos²θ)


Relative Motion:

vAB = vA − vB

For rain-man problems: vrm = vr − vm


NEET Tip: For two projectiles with same u, complementary angles (θ and 90°−θ) give same range. For maximum range, θ = 45°. At θ = 30° and 60°, range is same.

▶ 3. Laws of Motion

Newton's Laws:

1st Law: ΣF = 0 → body at rest/motion continues

2nd Law: F = dp/dt = ma (if m constant)

3rd Law: F12 = −F21


Friction:

Static friction: fs ≤ μsN

Kinetic friction: fk = μkN

Rolling friction: fr = μrN (μr ≪ μk)

Angle of repose: tan θ = μs


Circular Motion:

Centripetal force: Fc = mv²/r = mω²r

Centripetal acceleration: ac = v²/r = ω²r

Banking of roads: tan θ = v²/rg (no friction needed)

Maximum safe speed on banked road with friction: vmax = √[rg(μ + tanθ)/(1 − μ tanθ)]

Conical pendulum: T = 2π√(l cosθ/g)

Minimum speed at top of vertical circle: vtop ≥ √(gr)

Minimum speed at bottom to complete loop: vbottom ≥ √(5gr)


Connected Motion:

Two masses on pulley (Atwood): a = (m₁−m₂)g/(m₁+m₂), T = 2m₁m₂g/(m₁+m₂)

Mass on table connected to hanging mass: a = m₂g/(m₁+m₂), T = m₁m₂g/(m₁+m₂)


NEET Tip: In vertical circular motion, tension is maximum at bottom and minimum at top. The difference in tension between bottom and top is 6mg.

▶ 4. Work, Energy & Power

Work: W = F·s = Fs cosθ

Work by variable force: W = ∫F·dx

Kinetic Energy: KE = ½mv²

Potential Energy (gravity): PE = mgh

Potential Energy (spring): PE = ½kx²

Gravitational PE: U = −GM1m2/r

Power: P = W/t = F·v

Work-Energy Theorem: Wnet = ΔKE

Conservation of Mechanical Energy: KE₁ + PE₁ = KE₂ + PE₂

Efficiency: η = (Output work / Input work) × 100%

Rocket thrust: F = vrel(dm/dt)

Rocket velocity: v = v₀ + vrel ln(m₀/m)


NEET Tip: In spring problems, remember that at equilibrium position KE is maximum and PE is zero. For elastic collisions, relative speed of approach = relative speed of separation.

▶ 5. Rotational Motion

Rotational Kinematics:

ω = ω₀ + αt

θ = ω₀t + ½αt²

ω² = ω₀² + 2αθ


Linear ↔ Angular:

v = rω, at = rα, ac = ω²r = v²/r


Moment of Inertia (key bodies):

Thin ring about axis: I = MR²

Disc about axis: I = ½MR²

Solid sphere about diameter: I = ⅖MR²

Spherical shell about diameter: I = ⅔MR²

Thin rod about centre: I = ML²/12

Thin rod about end: I = ML²/3

Rectangular plate about centre: I = M(a²+b²)/12


Torque: τ = r × F, τ = Iα

Angular Momentum: L = r × p, L = Iω

Rotational KE: KErot = ½Iω²

Conservation: τext = 0 → L = constant

For rolling without slipping: v = ωR, KE = ½mv² + ½Iω² = ½mv²(1 + k²/R²)


Parallel Axis Theorem: I = Icm + Md²

Perpendicular Axis Theorem: Iz = Ix + Iy (for planar bodies)


Collisions:

Elastic: KE conserved. e = 1

Inelastic: KE not conserved. e = 0 for perfectly inelastic

Coefficient of restitution: e = (v₂−v₁)/(u₁−u₂)

Velocity after elastic collision: v₁ = (m₁−m₂)u₁/(m₁+m₂) + 2m₂u₂/(m₁+m₂)

Velocity after perfectly inelastic: v = (m₁u₁+m₂u₂)/(m₁+m₂)


NEET Tip: For rolling bodies on incline, acceleration a = g sinθ/(1 + I/mR²). The body with smaller I rolls faster: sphere > disc > ring.

▶ 6. Gravitation

Newton's Law: F = Gm₁m₂/r²

Gravitational field: g = GM/r²

Gravitational potential: V = −GM/r

Gravitational PE: U = −GMm/r

Acceleration due to gravity: g = GM/R²

Variation with height: gh = g(1 − 2h/R) for h ≪ R

Variation with depth: gd = g(1 − d/R)

Variation with latitude: g' = g − ω²R cos²φ

Orbital velocity: v₀ = √(GM/r)

Time period of satellite: T = 2π√(r³/GM)

Geostationary: r = 42,200 km (36,000 km above surface)

Escape velocity: ve = √(2GM/R) = √(2gR)

For Earth: ve = 11.2 km/s

Binding energy: Eb = GMm/2r

Kepler's 3rd: T² ∝ a³

Energy of orbiting satellite: E = −GMm/2r


NEET Tip: ve = √2 × v₀. Escape velocity is independent of the mass of the escaping body. For Earth, ve = 11.2 km/s, for Moon ve = 2.4 km/s (no atmosphere because gas molecules can escape).

▶ 7. Elasticity

Stress σ = F/A | Strain ε = ΔL/L

Young's Modulus Y = FL/AΔL

Bulk Modulus B = −VΔP/ΔV

Shear Modulus G = F/Aθ

Poisson's Ratio σ = −Δd/d ÷ ΔL/L = −lateral strain/longitudinal strain

Elastic PE per unit volume: u = ½ × stress × strain = ½Yε² = σ²/2Y

Force constant of a wire: k = YA/L

Elongation under self-weight: ΔL = ρgL²/2Y

Thermal stress: σ = YαΔT


Relation between moduli: Y = 3B(1−2σ) | Y = 2G(1+σ)


NEET Tip: For wires in series, keq = k₁k₂/(k₁+k₂). For parallel, keq = k₁+k₂. Ductile materials have large plastic region (copper), brittle materials break near elastic limit (glass).

▶ 8. Fluid Mechanics

Pressure at depth: P = P₀ + ρgh

Absolute pressure = Gauge pressure + Atmospheric pressure

Pascal's Law: ΔP transmitted equally throughout fluid

Hydraulic lift: F₁/A₁ = F₂/A₂ → F₂ = F₁(A₂/A₁)

Archimedes' Principle: Buoyant force = weight of displaced fluid

Fraction submerged = ρbodyfluid

Apparent weight = Actual weight − Buoyant force

Equation of Continuity: A₁v₁ = A₂v₂ (Av = constant)

Bernoulli's Equation: P + ½ρv² + ρgh = constant

Torricelli's theorem: v = √(2gh) (efflux speed)

Range of efflux: x = 2√[h(H−h)], max at h = H/2

Venturimeter: v = √[2(P₁−P₂)/ρ(A₁²/A₂²−1)]

Stokes' Law: F = 6πηrv

Terminal velocity: vt = 2r²(ρ−σ)g/9η

Reynolds number: Re = ρvD/η

→ Re < 1000: streamline | Re > 2000: turbulent

Surface Tension: S = F/L (force per unit length)

Excess pressure in drop: ΔP = 2S/R

Excess pressure in bubble: ΔP = 4S/R

Capillary rise: h = 2S cosθ/ρgr


NEET Tip: For a floating body, weight = buoyant force. For a body completely immersed, apparent loss in weight = weight of displaced fluid. Surface tension of soap solution < pure water → bubbles form easily.

▶ 9. Oscillations (SHM)

General SHM: F = −kx, a = −ω²x

Displacement: x = A sin(ωt + φ) or x = A cos(ωt + φ)

Angular frequency: ω = √(k/m)

Time period: T = 2π/ω = 2π√(m/k)

Velocity: v = ω√(A² − x²)

Maximum velocity: vmax = Aω (at x = 0)

Acceleration: a = −ω²x

Maximum acceleration: amax = Aω² (at x = A)

KE = ½mω²(A²−x²), PE = ½mω²x²

Total Energy: E = ½mω²A² = ½kA² (constant)

At x = A/√2: KE = PE = E/2

Time period of spring-mass system: T = 2π√(m/k)

T for series springs: T = 2π√[m(1/k₁+1/k₂)]

T for parallel springs: T = 2π√[m/(k₁+k₂)]

Simple pendulum: T = 2π√(l/g) (for small θ)

Physical pendulum: T = 2π√(I/mgd)

Compound pendulum: T = 2π√[(l²/12 + d²)/gd]


NEET Tip: In SHM, at mean position (x=0): vmax, a=0, KEmax, PE=0. At extreme (x=A): v=0, amax, KE=0, PEmax. Spring pendulum and simple pendulum — don't confuse T formulas.

📐 Derivations

▶ Derivation of Equations of Motion (Calculus Method)

First equation: v = u + at

Acceleration a = dv/dt → dv = a·dt

Integrate: ∫uv dv = ∫0t a dt = a∫0t dt

→ v − u = at → v = u + at


Second equation: s = ut + ½at²

v = ds/dt = u + at → ds = (u + at)dt

Integrate: ∫0s ds = ∫0t (u + at)dt = u∫0t dt + a∫0t t·dt

→ s = ut + ½at²


Third equation: v² = u² + 2as

a = dv/dt = v·dv/ds → a·ds = v·dv

Integrate: a∫0s ds = ∫uv v·dv

→ as = ½(v² − u²) → v² = u² + 2as


NEET Note: These equations only apply for uniform acceleration. For variable acceleration, use calculus directly.

▶ Derivation of Work-Energy Theorem

Statement: Net work done = change in kinetic energy


W = ∫F·dx = ∫m·a·dx (from Newton's 2nd law)

= ∫m(dv/dt)·dx = ∫m·v·dv (since v = dx/dt, so dx = v·dt)

= m∫uv v·dv = m·[v²/2]uv = ½mv² − ½mu²

→ W = KEfinal − KEinitial = ΔKE


For variable forces: Same derivation holds — the key step is F = m(dv/dt) and dx = v·dt, which are always valid.

▶ Derivation of Escape Velocity

Concept: The minimum velocity required for a body to escape Earth's gravitational field (reach infinity with zero kinetic energy).


At surface: KE = ½mvₑ², PE = −GMm/R

At infinity: KE = 0, PE = 0 (by definition, U∞ = 0)

By conservation of energy: KEsurface + PEsurface = KE + PE

½mvₑ² + (−GMm/R) = 0 + 0

½mvₑ² = GMm/R → vₑ = √(2GM/R)

Since g = GM/R² → vₑ = √(2gR)


Numerical value: g = 9.8 m/s², R = 6.4×10⁶ m

vₑ = √(2 × 9.8 × 6.4×10⁶) = √(1.254×10⁸) ≈ 11,200 m/s = 11.2 km/s


NEET Note: Escape velocity is independent of the mass of the escaping body. For Moon (g=1.6, R=1.74×10⁶): vₑ = 2.4 km/s — too low to retain atmosphere.

▶ Derivation of Bernoulli's Equation

Assumptions: Steady flow, incompressible fluid, non-viscous, streamline flow.


Consider a fluid element flowing through a pipe. Work done by pressure forces:

W = P₁A₁·v₁·Δt − P₂A₂·v₂·Δt = (P₁ − P₂)·A·v·Δt (since A₁v₁ = A₂v₂)

Change in KE = ½ρ(A·v·Δt)·(v₂² − v₁²)

Change in PE = ρ(A·v·Δt)·g·(h₂ − h₁)


By Work-Energy Theorem (applied to fluid):

Net work = ΔKE + ΔPE

(P₁ − P₂)·A·v·Δt = ½ρ·A·v·Δt·(v₂²−v₁²) + ρ·A·v·Δt·g·(h₂−h₁)

Dividing by A·v·Δt:

P₁ − P₂ = ½ρ(v₂²−v₁²) + ρg(h₂−h₁)


→ P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂


Or simply: P + ½ρv² + ρgh = constant along a streamline.

▶ Derivation of Moment of Inertia of a Uniform Rod

About centre (perpendicular to length):

Consider a thin rod of length L, mass M. Linear density λ = M/L.

Take an element dx at distance x from centre.

Mass of element: dm = λ·dx = (M/L)·dx

dI = dm·x² = (M/L)·x²·dx

I = ∫−L/2L/2 (M/L)·x²·dx = (M/L)·[x³/3]−L/2L/2

I = (M/L)·(1/3)·[(L/2)³ − (−L/2)³]

I = (M/3L)·(L³/8 + L³/8) = (M/3L)·(L³/4) = ML²/12


About end (using Parallel Axis Theorem):

Iend = Icm + M·(L/2)² = ML²/12 + ML²/4 = ML²/3

📜 Key Theorems & Laws

▶ Parallel Axis Theorem

Statement: The moment of inertia about any axis parallel to an axis through the centre of mass is I = Icm + Md², where d is the perpendicular distance between the axes.

Use: Finding MOI about axes that don't pass through centre of mass.

▶ Perpendicular Axis Theorem

Statement: For a planar body (lamina), the moment of inertia about an axis perpendicular to the plane equals the sum of moments of inertia about two perpendicular axes in the plane intersecting at the same point: Iz = Ix + Iy.

Use: Finding MOI of discs, rings, rectangular plates about different axes.

Note: Valid only for 2D planar objects. Not valid for 3D objects like spheres.

▶ Kepler's Laws of Planetary Motion

1st Law (Law of Orbits): Every planet moves in an elliptical orbit with the Sun at one focus.

2nd Law (Law of Areas): The line joining the planet to the Sun sweeps equal areas in equal intervals of time. → Areal velocity = constant. (Consequence of conservation of angular momentum)

3rd Law (Law of Periods): T² ∝ a³, where T = time period, a = semi-major axis. For circular orbits: T² ∝ r³.

Derivation of 3rd Law: Gravitational force = Centripetal force: GMm/r² = mω²r → ω² = GM/r³ → (2π/T)² = GM/r³ → T² = 4π²r³/GM → T² ∝ r³.

▶ Pascal's Law

Statement: Pressure applied to an enclosed fluid at rest is transmitted equally and undiminished to every part of the fluid and the walls of the container.

Application: Hydraulic lift/jack/brake. F₂ = F₁(A₂/A₁). The force multiplication factor equals the ratio of areas.

NEET Note: Used in hydraulic machines where a small force on a small piston lifts a heavy load on a larger piston.

▶ Archimedes' Principle

Statement: When a body is partially or fully immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced.

Buoyant force: Fb = ρfluid × Vdisplaced × g

Condition for floating: Weight of body = Buoyant force. Fraction submerged = ρbodyfluid.

Apparent weight: Wapparent = Wactual − Fb

NEET Note: The buoyant force depends only on the volume of fluid displaced and fluid density — not on the density of the immersed object or the depth of immersion (for completely submerged bodies).

⚠️ NEET Exam Tips & Common Mistakes

▶ Frequently Made Mistakes

Confusing mass and weight — mass is scalar (kg), weight is force (N). g varies.

Using v² = u² + 2as for non-uniform acceleration — only valid for constant a.

Forgetting direction — velocity and acceleration are vectors. Sign convention matters.

Normal force ≠ mg — normal force equals mg only on horizontal surfaces with no vertical acceleration. On incline: N = mg cosθ. In lift: N = m(g±a).

Confusing μs and μk — static friction is ≤ μsN (adjusts to applied force), kinetic friction is = μkN (constant for given surfaces). μs > μk.

Applying conservation of energy when non-conservative forces are present — friction, air resistance dissipate energy. Use Wnc = ΔKE + ΔPE.

Moment of inertia confusion — I depends on axis of rotation, not just the body. Same body has different I for different axes.

g vs G — G is universal constant (6.67×10⁻¹¹ N·m²/kg²). g is acceleration due to gravity (varies with location).

Satellite orbit formulas — Orbital velocity decreases with height (v₀ = √(GM/r)). Time period increases with height. Geostationary is at specific height only.

Buoyancy direction — Buoyant force always acts upward, regardless of body orientation.

Bernoulli's theorem where not applicable — only for steady, incompressible, non-viscous, streamline flow. Not for turbulent flow or viscous fluids.

SHM phase angle — sin vs cos depends on initial conditions. x = A sin(ωt+φ), the phase φ determines initial position.

▶ Shortcut Formulas for NEET

⚡ Range of projectile is same for θ and 90°−θ

⚡ At maximum height, KE = ½mu²cos²θ, PE = ½mu²sin²θ

⚡ Two masses on pulley: tension T = 2m₁m₂g/(m₁+m₂) — memorize this

⚡ Acceleration of a body rolling down incline: a = g sinθ/(1 + I/mR²)

⚡ ve = √2 × vo (escape = √2 × orbital velocity)

⚡ gpole > gequator (due to rotation and oblateness)

⚡ Weightlessness in freely falling lift: N = 0, geff = 0

⚡ For a satellite, total energy E = −KE = PE/2 = −GMm/2r

⚡ If the Earth stops rotating, g increases by about 0.034 m/s² at equator

⚡ The percentage decrease in g with height h is 2h/R × 100% (for h ≪ R)

Practice Questions

Thermodynamics →