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NEET (UG) · revision cheat sheet · neetlogic.in

Physics

28 chapters · 86 topics · ~180 marks · 51 solved previous-year questions

Class 11

1. Units and Measurements

SI units, dimensions and dimensional analysis: Dimensional analysis can check an equation's consistency or derive how variables combine, but it can never fix a pure numerical constant like the ½ in KE = ½mv², and it can't distinguish between quantities that share dimensions, such as work and torque.
Significant figures and errors in measurement: For a quantity like Z = A^p B^q / C^r, the relative errors always add — even where the formula has division — because error propagation bounds the worst case rather than doing signed arithmetic.
Measuring instruments: vernier, screw gauge: A screw gauge's zero-error sign is the trap: a positive zero error is subtracted from the observed reading, but a negative zero error is subtracted as a negative number — which means it's effectively added back.

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier scale divisions (V.S.D.) are equal to its 9 main scale divisions (M.S.D.). The least division on the M.S. is 0.1 cm and the zero of the V.S. is at

x = 0.1

cm when the jaws of the Vernier callipers are closed. If the main scale reading for the diameter is

M = 5

cm and the number of the coinciding Vernier division is 8, the measured diameter after zero-error correction is:

Ans: 4.98 cm·1 VSD

= 0.9

MSD, so the least count is

0.1 \times 0.1 = 0.01

cm. Reading

= 5 + 8 \times 0.01 = 5.08

cm; the zero error is

+ 0.1

cm, so the corrected diameter is

5.08 - 0.10 = 4.98

cm.

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A force defined by

F = α t^{2} + β t

acts on a particle at a given time

t

. The factor which is dimensionless, if

α

and

β

are constants, is:

Ans:

\frac{α t}{β}

α t^{2}

and

β t

both have the dimensions of force, so

[α t] = [β]

. Hence

\frac{α t}{β}

is dimensionless.

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2. Motion in a Straight Line

Displacement, velocity, acceleration; graphs: Area under a v-t graph gives displacement, not distance — where the graph dips below the axis that area subtracts, so total distance needs the positive and negative areas summed separately, not net.
Equations of uniformly accelerated motion: The distance covered in the nth second, Sn = u + a(2n−1)/2, is a distinct formula from s = ut + ½at² — NEET often wants one specific second's distance, not the total displacement.
Relative velocity in one dimension: Relative velocity is a subtraction, v_AB = v_A − v_B, so assigned sign matters as much as speed — two objects add their speeds only when opposite directions are already opposite in sign.

In some appropriate units, the time (

t

) and position (

x

) relation of a moving particle is given by

t = x^{2} + x

. The acceleration of the particle is:

Ans:

- 2 (2 x + 1)^{- 3}

t = x^{2} + x \Rightarrow \frac{d t}{d x} = 2 x + 1 \Rightarrow v = \frac{1}{2 x + 1}

. Then

a = \frac{d v}{d t} = \frac{d v}{d x} v = - \frac{2}{( 2 x + 1 ) ^{2}} \cdot \frac{1}{2 x + 1} = - 2 (2 x + 1)^{- 3}

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Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every

T

min. A girl driving a scooty at a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period

T

of the bus service and the speed (assumed constant) of the buses.

Ans: 15 min, 120 km/h·Let bus speed be

v

. Same direction: buses catch her every 30 min, so

\frac{v T}{v - 60} = 30

; opposite:

\frac{v T}{v + 60} = 10

. Dividing,

\frac{v + 60}{v - 60} = 3 \Rightarrow v = 120

km/h, then

T = \frac{30 \times 60}{120} = 15

min.

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3. Motion in a Plane

Vectors: addition, resolution, products: The resultant of two vectors is maximum (A+B) when they're parallel and minimum (|A−B|) when antiparallel — it's never zero unless the two magnitudes are equal, a quick check for ruling out impossible resultants.
Projectile motion: Complementary projection angles (θ and 90°−θ) give the same range but different times of flight and maximum heights — same range never means everything else about the trajectory matches too.
Uniform circular motion and relative velocity: Uniform circular motion has constant speed but never constant velocity — the centripetal acceleration v²/r is always nonzero and points toward the centre, which is exactly why the motion counts as accelerated.

A particle moving with uniform speed in a circular path maintains:

Ans: Varying velocity and varying acceleration·In uniform circular motion the speed is constant but the direction of velocity keeps changing, so velocity varies; the centripetal acceleration has constant magnitude but continuously changing direction, so acceleration varies too.

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4. Laws of Motion

Newton's laws, inertia, impulse and momentum: Action–reaction pairs act on two different bodies, so they can never be added together to argue that nothing accelerates — a body stays at rest only when the net force on that one body is zero.
Friction: static, kinetic, rolling: Static friction isn't a fixed value — it self-adjusts to match the applied force up to its maximum, μsN — so the friction on a stationary block usually equals the applied force, not μsN itself.
Circular motion dynamics, banking of roads: tanθ = v²/rg is only the speed needing zero friction on a banked curve — with friction present, safe speed becomes a range between a v_min and v_max, both formulas picking up μ.
Connected bodies, pulleys and pseudo forces: A pseudo force exists only in a non-inertial, accelerating frame — solving the same problem from the ground frame must never include one, and mixing both approaches in one diagram is the most common error here.

A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (take

g = 9.8 m/ s^{2}

Ans: 21 N s·Speed just before impact

= 2 g \times 40 = 28

m/s downward; just after

= 2 g \times 10 = 14

m/s upward. Impulse

= m Δ v = 0.5 \times (14 + 28) = 21

N s.

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There are two inclined surfaces of equal length

L

and the same angle of inclination

4 5^{\circ}

with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down the rough surface as the smooth surface. The coefficient of kinetic friction (

μ_{k}

) between the object and the rough surface is close to:

Ans: 0.75·Smooth:

a_{1} = g sin 4 5^{\circ}

. Rough:

a_{2} = g (sin 4 5^{\circ} - μ cos 4 5^{\circ})

. Same distance,

t_{2} = 2 t_{1} \Rightarrow a_{2} = a_{1} /4

, so

1 - μ = \frac{1}{4} \Rightarrow μ = 0.75

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5. Work, Energy and Power

Work by constant and variable forces: Work is zero whenever force stays perpendicular to displacement at every instant — centripetal force on a circling body and the upward force on someone carrying a bag horizontally both do zero work.
Kinetic and potential energy, conservation: The work-energy theorem, W_net = ΔKE, holds even with friction present — it's total mechanical energy that's conserved only when every non-conservative force does zero work, and NEET tests which claim survives friction.
Collisions: elastic and inelastic: Momentum is conserved in every collision, elastic or not — only kinetic energy needs the collision to be elastic — and for two equal masses in a 1D elastic collision, the velocities simply swap.
Power and vertical circular motion: At the top of a vertical circle the minimum safe speed is √(gr), not zero — gravity alone supplies the whole centripetal force there, so tension or normal force drops to exactly zero, never negative.

The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m and car B stops after 1500 m. If

F_{A}

and

F_{B}

are the forces applied by the brakes on cars A and B respectively, then the ratio

\frac{F _{A}}{F _{B}}

is:

Ans:

\frac{2}{3}

·Work–energy:

F s = K E

\frac{F _{A}}{F _{B}} = \frac{K E _{A} / s _{A}}{K E _{B} / s _{B}} = \frac{100/1000}{225/1500} = \frac{0.1}{0.15} = \frac{2}{3}

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Two bodies A and B of the same mass undergo a completely inelastic one-dimensional collision. Body A moves with velocity

v_{1}

while body B is at rest before the collision. The velocity of the system after the collision is

v_{2}

. The ratio

v_{1} : v_{2}

is:

Ans: 2 : 1·Momentum conservation for a perfectly inelastic collision:

m v_{1} = 2 m v_{2} \Rightarrow v_{1} : v_{2} = 2 : 1

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6. System of Particles and Rotational Motion

Centre of mass and its motion: Internal forces — an explosion, an internal collision — never change the centre of mass's velocity or path, only external forces do; a shell that explodes mid-air has its COM land exactly where the intact shell would have.
Torque, angular momentum and their conservation: Angular momentum L = Iω is conserved whenever net external torque is zero, even as I itself changes — a skater pulling their arms in spins faster purely because I falls, with no external torque involved.
Moment of inertia, theorems of axes: The perpendicular axis theorem (Iz = Ix+Iy) applies only to flat, planar bodies — using it on a solid sphere or cylinder is a common invalid move — while the parallel axis theorem works for any rigid body.
Rolling motion and rotational kinetic energy: Racing shapes down an incline without slipping, acceleration depends only on the k²/R² ratio, never on mass or radius — so a solid sphere always beats a solid cylinder, which always beats a ring, at any size.

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

Ans: 108 days·No external torque, so

I ω

is conserved.

I \propto R^{2}

for a uniform sphere, so doubling

R

quadruples

I

and the period becomes

4 \times 27 = 108

days.

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A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of

6 0^{\circ}

with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take

g = 10 m/ s^{2}

Ans:

1003

N·Take torques about the foot. The wall pushes horizontally with

N

; the rod's weight acts at its midpoint. With the rod at

6 0^{\circ}

to the wall (i.e.

3 0^{\circ}

to the floor):

N \cdot L sin 3 0^{\circ} = m g \cdot \frac{L}{2} cos 3 0^{\circ} \Rightarrow N = m g \frac{c o s 3 0 ^{\circ}}{2 s i n 3 0 ^{\circ}} = 200 \times \frac{3 /2}{1} = 1003

N. Friction at the floor balances

N

1003

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7. Gravitation

Newton's law of gravitation, g and its variation: g falls off in both directions from the surface but not symmetrically: it drops as g(1−2h/R) with altitude, roughly twice as fast per unit distance as g(1−d/R) with depth, and reaches zero only at the centre.
Gravitational potential and potential energy: Gravitational potential energy is negative everywhere at finite distance, by the convention that it's zero at infinity — rising toward zero as a satellite escapes doesn't mean the value turns positive.
Kepler's laws, satellites, escape velocity: Escape velocity depends only on the planet's mass and radius, never on the projection angle or the escaping body's own mass. Kepler's third law, T² ∝ r³, likewise links period to orbital radius alone, independent of the orbiting body's mass.

The radius of the Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

Ans: 88 Earth days·Kepler's third law:

T^{2} \propto r^{3}

\frac{T _{M er c u r y}}{T _{M a r s}} = (\frac{1}{4})^{3/2} = \frac{1}{8}

, so

T_{M er c u r y} = \frac{687}{8} \approx 86

— nearest option 88 Earth days.

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A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

Ans: 27 N·

F \propto \frac{1}{( R + h ) ^{2}}

. With

h = \frac{R}{3}

R + h = \frac{4 R}{3}

, so

F = 48 \times \frac{9}{16} = 27

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8. Mechanical Properties of Solids

Stress, strain and Hooke's law: Hooke's law (stress ∝ strain) holds only up to the proportional limit — the elastic limit lies slightly beyond it, so a material can still spring back to shape over a small range where stress and strain aren't proportional.
Young's, bulk and shear moduli; Poisson's ratio: Young's modulus and shear modulus are defined only for solids — a liquid or gas can't sustain a shape-restoring shear or tensile stress, so only the bulk modulus applies to fluids.
Elastic potential energy: Elastic energy stored per unit volume of a stretched wire is ½ × stress × strain, which simplifies to ½Y(strain)² — the same ½ × force × extension logic as a spring, expressed through Young's modulus.

The maximum elongation of a steel wire of 1 m length, if the elastic limit of steel and its Young's modulus are respectively

8 \times 1 0^{8} N m^{- 2}

and

2 \times 1 0^{11} N m^{- 2}

, is:

Ans: 4 mm·At the elastic limit, strain

= \frac{stress}{Y} = \frac{8 \times 1 0 ^{8}}{2 \times 1 0 ^{11}} = 4 \times 1 0^{- 3}

. Elongation

= strain \times L = 4 \times 1 0^{- 3} \times 1

= 4

mm.

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9. Mechanical Properties of Fluids

Pressure, Pascal's law, buoyancy: Buoyant force depends only on the volume of fluid displaced and the fluid's density — once an object is fully submerged, pushing it deeper doesn't change the buoyant force, since the displaced volume stays the same.
Viscosity, Stokes' law, terminal velocity: Terminal velocity under Stokes' law scales as the square of the radius, not linearly — doubling a drop's radius quadruples its terminal speed, and NEET tests this exponent directly more often than the formula's derivation.
Bernoulli's principle and its applications: Bernoulli's equation assumes non-viscous, incompressible, streamline flow — the familiar 'pressure drops where speed increases' rule also silently assumes the same height, so applying it to turbulent or viscous flow is already outside its own assumptions.
Surface tension, capillarity, excess pressure: Excess pressure is 4T/r inside a soap bubble but only 2T/r inside a plain liquid drop or a submerged air bubble — the extra factor comes from the soap film having two free surfaces instead of one.

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If the surface tension of water is

0.07 N m^{- 1}

, then the excess force required to take it away from the surface is:

Ans: 19.8 mN·A disc lifted off a liquid surface is held by surface tension along its full circumference:

F = T \times 2 π r = 0.07 \times 2 π \times 0.045 \approx 1.98 \times 1 0^{- 2}

= 19.8

mN.

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Consider a water tank shown in the figure. It has one wall at

x = L

and can be taken to be very wide in the

z

direction. When filled with a liquid of surface tension

S

and density

ρ

, the liquid surface makes an angle

θ_{0}

(

θ_{0} ≪ 1

) with the

x

-axis at

x = L

. If

y (x)

is the height of the surface, then the equation for

y (x)

is (take

sin θ (x) \approx tan θ (x) \approx \frac{d y}{d x}

;

g

is the acceleration due to gravity):

Ans:

\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{S} y

·At height

y

the hydrostatic pressure

ρ g y

is balanced by the surface-tension pressure

S \times

(curvature). For a nearly flat surface the curvature is

\frac{d ^{2} y}{d x ^{2}}

, so

S \frac{d ^{2} y}{d x ^{2}} = ρ g y

, i.e.

\frac{d ^{2} y}{d x ^{2}} = \frac{ρ g}{S} y

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10. Thermal Properties of Matter

Thermal expansion and calorimetry: For an isotropic solid the linear, areal and volume expansion coefficients are in the ratio 1:2:3, so 'the' expansion coefficient without saying which one invites a factor-of-2-or-3 error; water alone contracts while warming from 0°C to 4°C.
Change of state and latent heat: Temperature stays flat on a heating curve during a phase change even while heat keeps flowing in — that heat is going entirely into latent heat, not into raising temperature, until the change is complete.
Heat transfer: conduction, convection, radiation: Radiation is the only one of the three heat-transfer modes that needs no medium at all — it's how the Sun's heat crosses vacuum — while conduction needs direct contact and convection needs actual bulk fluid movement.
Newton's law of cooling, Stefan's and Wien's laws: Newton's law of cooling is only an approximation valid for a small temperature excess over the surroundings; the always-true relation beneath it is Stefan's law, E ∝ T⁴, and Wien's law says a hotter body's peak wavelength shifts shorter.

A metallic bar of Young's modulus

0.5 \times 1 0^{11} N m^{- 2}

and coefficient of linear thermal expansion

1 0^{- 5}^{\circ} C^{- 1}

, length 1 m and area of cross-section

1 0^{- 3} m^{2}

, is heated from

0^{\circ}

C to

10 0^{\circ}

C without expansion or bending. The compressive force developed in it is:

Ans:

50 \times 1 0^{3}

N·Thermal strain prevented

= α Δ T = 1 0^{- 5} \times 100 = 1 0^{- 3}

. Force

= Y A α Δ T = 0.5 \times 1 0^{11} \times 1 0^{- 3} \times 1 0^{- 3} = 5 \times 1 0^{4}

= 50 \times 1 0^{3}

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Three identical heat-conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity

2 K

while that in the middle has thermal conductivity

K

. The left end of the combination is maintained at temperature

3 T

and the right end at

T

. The rods are thermally insulated from outside. In steady state, the temperature at the left junction is

T_{1}

and that at the right junction is

T_{2}

. The ratio

T_{1} / T_{2}

is:

Ans:

\frac{5}{3}

·In series the heat current is the same, so the temperature drop across each rod is

\propto \frac{1}{K}

: drops in the ratio

1 : 2 : 1

across a total drop of

2 T

, i.e.

0.5 T

T

0.5 T

. Hence

T_{1} = 2.5 T

T_{2} = 1.5 T

and

\frac{T _{1}}{T _{2}} = \frac{5}{3}

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11. Thermodynamics

First law, work and internal energy: Internal energy is a state function — for an ideal gas it depends on temperature alone — while heat and work are path-dependent, so two different processes between the same states can have very different Q and W despite identical ΔU.
Thermodynamic processes: isothermal, adiabatic, isobaric, isochoric: On a P-V diagram, an adiabatic curve through a point is always steeper than the isothermal curve through that same point, because γ > 1 — a fast way to identify which curve is which without reading labels.
Second law, heat engines, Carnot cycle: Carnot efficiency, η = 1 − T2/T1 with temperatures in kelvin, depends only on the two reservoir temperatures, never on the working substance — no gas swap raises a reversible engine's efficiency between fixed temperatures.

Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius

r_{A}

and

r_{B}

, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio

\frac{r _{A}}{r _{B}}

is equal to:

Ans:

\frac{3}{4}

·Same heat, same

Δ U

, so the work done (

P Δ V

) is the same:

π r_{A}^{2} \times 16 = π r_{B}^{2} \times 9 \Rightarrow \frac{r _{A}}{r _{B}} = \frac{3}{4}

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A thermodynamic system is taken through the cycle

ab c d a

shown in the

P

–

V

diagram. The work done by the gas along the path

b c

is:

Ans: Zero·Path

b c

runs vertically on the

P

–

V

diagram: the volume stays at 400 cm

^{3}

while the pressure rises. With

Δ V = 0

, the work done by the gas

\int P d V

is zero.

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12. Kinetic Theory

Ideal gas equation and kinetic interpretation of pressure: At a fixed temperature, rms speed depends only on molar mass — lighter molecules move faster, not slower, which is why hydrogen effuses faster than oxygen at the same temperature.
Degrees of freedom, law of equipartition, specific heats: Each degree of freedom contributes ½kT, so Cv = (f/2)R: monatomic gas has f=3 (Cv=3/2R), diatomic has f=5 at ordinary temperatures (Cv=5/2R), and only gains its two vibrational modes — becoming f=7 — at high temperature.
Mean free path: Mean free path depends on the number density of molecules, hence inversely on pressure at fixed temperature, not on molecular speed — doubling pressure at constant T halves it, but raising T at constant volume leaves it unchanged.

A container has two chambers of volumes

V_{1} = 2

litres and

V_{2} = 3

litres separated by a partition made of a thermal insulator. The chambers contain

n_{1} = 5

and

n_{2} = 4

moles of ideal gas at pressures

p_{1} = 1

atm and

p_{2} = 2

atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

Ans: 1.6 atm·Temperature is the same throughout (no heat exchange with outside, ideal gases). Final

P = \frac{n _{1} + n _{2}}{V _{1} + V _{2}} R T

and

R T = \frac{p _{1} V _{1}}{n _{1}} = \frac{p _{2} V _{2}}{n _{2}}

. Using

p_{1} V_{1} + p_{2} V_{2} = (n_{1} + n_{2}) R T

P = \frac{1 \times 2 + 2 \times 3}{5} = 1.6

atm.

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An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheres at a temperature of

2 7^{\circ}

C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given

R = \frac{100}{12} J mo l^{- 1} K^{- 1}

, molecular mass of

O_{2}

= 32, 1 atm pressure

= 1.01 \times 1 0^{5} N/ m^{2}

]

Ans: 0.116 kg·Final moles

n = \frac{P V}{R T} = \frac{12 \times 1.01 \times 1 0 ^{5} \times 0.030}{( 100/12 ) \times 300} \approx 14.5

mol (gauge 11 atm means absolute 12 atm). Withdrawn

\approx 18.2 - 14.5 = 3.7

mol

\approx 0.116

kg.

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13. Oscillations

Simple harmonic motion: equations, energy: Total energy in SHM is constant and proportional to amplitude squared, but keeps trading between KE (maximum at the mean position) and PE (maximum at the extremes) — and acceleration itself is never constant, since a = −ω²x.
Spring–mass system and simple pendulum: A spring-mass system's period, T = 2π√(m/k), doesn't involve g and works the same on the Moon, while a pendulum's period, T = 2π√(L/g), depends on effective g and simply stops swinging in a freely falling lift.
Damped, forced oscillations and resonance: Resonance occurs when the driving frequency matches the system's own natural frequency, producing a sharp buildup in amplitude — the textbook reason soldiers break step crossing a bridge rather than march across it in rhythm.

Two identical point masses P and Q, suspended from two separate massless springs of spring constants

k_{1}

and

k_{2}

respectively, oscillate vertically. If their maximum speeds are the same, the ratio

A_{Q} / A_{P}

of the amplitude

A_{Q}

of mass Q to the amplitude

A_{P}

of mass P is:

Ans:

\frac{k _{1}}{k _{2}}

·Maximum speed

= A ω

, equal for both:

A_{Q} ω_{Q} = A_{P} ω_{P}

. With

ω \propto k

for equal masses,

\frac{A _{Q}}{A _{P}} = \frac{ω _{P}}{ω _{Q}} = \frac{k _{1}}{k _{2}}

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If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is

\frac{x}{2}

times its original time period. Then the value of

x

is:

Ans: 2·

T = 2 π \frac{l}{g}

is independent of mass. Halving

l

gives

T^{'} = \frac{T}{2} = \frac{2}{2} T

, so

x = 2

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14. Waves

Transverse and longitudinal waves, speed of a wave: Wave speed is fixed by the medium's own properties, not by the source's frequency or the wave's amplitude — crossing into a new medium leaves frequency exactly as the source set it, while wavelength and speed both change.
Superposition, standing waves, beats: Beat frequency is simply the difference of the two frequencies, never their sum or average — and a standing wave's nodes and antinodes stay fixed in space, unlike a travelling wave where every point oscillates with equal amplitude.
Doppler effect: The Doppler formula's sign convention is the real trap: source and observer velocities are each positive when moving toward the other and negative when moving away, and an inconsistent choice is the most common source of an inverted answer.

A pipe open at both ends has a fundamental frequency

f

in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to:

Ans:

f

·Open pipe of length

L

f = \frac{v}{2 L}

. Dipped to half its length it becomes a closed pipe of length

\frac{L}{2}

f^{'} = \frac{v}{4 ( L /2 )} = \frac{v}{2 L} = f

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Class 12

15. Electric Charges and Fields

Coulomb's law and superposition: Force from multiple charges adds as vectors, not scalars — the net force on a charge comes from superposing each pairwise Coulomb force with its own direction, so just adding magnitudes fails whenever the charges aren't collinear.
Electric field, field lines, dipole: On a dipole's axial line the field is 2kp/r³, but on its equatorial line it's only kp/r³ and points opposite to the dipole moment — both fall off as 1/r³, but the factor of 2 and the direction are what NEET swaps.
Electric flux and Gauss's law with applications: Gauss's law counts only the charge enclosed by the surface — charges outside contribute zero net flux, though they still affect the field at points on it — and the field just outside a charged conductor is σ/ε0, twice an isolated sheet's σ/2ε0.

Two identical charged conducting spheres A and B have their centres separated by a certain distance. The charge on each sphere is

q

and the force of repulsion between them is

F

. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B, and finally removed from both. The new force of repulsion between spheres A and B (the radii of A and B are negligible compared to the distance of separation, so they can be considered point charges) is best given as:

Ans:

\frac{3 F}{8}

·Touching A: the third sphere and A share

q

, so A becomes

q /2

. Touching B: it carries

q /2

to B, which then holds

\frac{q /2 + q}{2} = \frac{3 q}{4}

. New force

\propto \frac{q}{2} \cdot \frac{3 q}{4} = \frac{3}{8} q^{2}

, i.e.

\frac{3 F}{8}

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16. Electrostatic Potential and Capacitance

Potential and potential energy of charges and dipoles: Potential from multiple charges adds as plain signed numbers, not vectors, much simpler than combining fields — and a dipole's potential is exactly zero at every point on its equatorial line, not just far away.
Conductors and dielectrics: A conductor in equilibrium is an equipotential throughout, but that doesn't mean its surface charge density is uniform — charge concentrates where curvature is sharpest, which is the entire physical basis for how a lightning rod works.
Capacitors: series, parallel, energy stored: In series, capacitors share the same charge, not voltage; in parallel they share the same voltage, not charge — the opposite pairing from resistor networks. Adding a dielectric to an isolated, constant-charge capacitor actually lowers its stored energy as C rises.

An electric dipole with dipole moment

5 \times 1 0^{- 6}

C m is aligned with the direction of a uniform electric field of magnitude

4 \times 1 0^{5}

N/C. The dipole is then rotated through an angle of

6 0^{\circ}

with respect to the electric field. The change in the potential energy of the dipole is:

Ans: 1.0 J·

U = - pE cos θ

Δ U = - pE (cos 6 0^{\circ} - cos 0^{\circ}) = pE (1 - \frac{1}{2}) = 5 \times 1 0^{- 6} \times 4 \times 1 0^{5} \times 0.5 = 1.0

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The plates of a parallel plate capacitor are separated by

d

. Two slabs of different dielectric constants

K_{1}

and

K_{2}

, with thicknesses

\frac{3}{8} d

and

\frac{d}{2}

respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If

K_{1} = 1.25 K_{2}

, the value of

K_{1}

is:

Ans: 2.66·Effective gap

= \frac{3 d /8}{K _{1}} + \frac{d /2}{K _{2}} + \frac{d}{8}

(air). Capacitance doubling means this equals

\frac{d}{2}

\frac{3}{8 K _{1}} + \frac{1}{2 K _{2}} = \frac{3}{8}

. With

K_{2} = 0.8 K_{1}

\frac{3}{8 K _{1}} + \frac{5}{8 K _{1}} = \frac{3}{8} \Rightarrow K_{1} = \frac{8}{3} \approx 2.66

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17. Current Electricity

Ohm's law, resistivity, temperature dependence: Resistivity is a property of the material only — stretching a wire to twice its length doubles R (since R = ρl/A) but leaves ρ itself unchanged, and resistivity rises with temperature for metals but falls for semiconductors.
Cells, EMF, internal resistance, combinations: Terminal voltage equals EMF only on open circuit — the moment current is drawn, V = E − Ir is strictly less than E. Cells in series suit a large external resistance; cells in parallel suit a small one, where less internal resistance matters most.
Kirchhoff's rules, Wheatstone bridge, potentiometer: A potentiometer's advantage over a voltmeter is that, at balance, it draws zero current from the cell being measured, while a voltmeter always draws some and reads slightly low. In a balanced Wheatstone bridge, only the galvanometer branch carries zero current.

A wire of resistance

R

is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of them together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:

Ans:

\frac{R}{16}

·Each piece is

\frac{R}{8}

. Four in parallel give

\frac{R}{32}

; two such sets in series give

\frac{R}{16}

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Two heaters A and B have power ratings of 1 kW and 2 kW, respectively. The two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:

Ans: 2 : 9·Series:

P_{s} = \frac{P _{1} P _{2}}{P _{1} + P _{2}} = \frac{2}{3}

kW. Parallel:

P_{p} = P_{1} + P_{2} = 3

kW. Ratio

= \frac{2/3}{3} = \frac{2}{9}

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18. Moving Charges and Magnetism

Force on a moving charge and on a current: The magnetic force on a moving charge is always perpendicular to its velocity, so it can never change the charge's speed or kinetic energy, only its direction — and it's exactly zero when velocity is parallel or antiparallel to B.
Biot–Savart and Ampère's laws: Ampère's law is exact for any current distribution but only practically solvable when the geometry is symmetric enough to pull B outside the integral — for anything less symmetric, Biot–Savart's element-by-element integration is what actually gets used.
Torque on a current loop, moving-coil galvanometer: Torque on a current loop is maximum when the loop's plane lies along B and zero when the plane is perpendicular to B — a galvanometer uses a radial field specifically to hold this torque near maximum through the whole swing, giving a linear scale.

An electron (mass

9 \times 1 0^{- 31}

kg and charge

1.6 \times 1 0^{- 19}

C) moving with speed

c /100

(

c

= speed of light) is injected into a magnetic field

B

of magnitude

9 \times 1 0^{- 4}

T perpendicular to its direction of motion. We wish to apply a uniform electric field

E

together with the magnetic field so that the electron does not deflect from its path. Then (speed of light

c = 3 \times 1 0^{8} m s^{- 1}

Ans:

E

is perpendicular to

B

and its magnitude is

27 \times 1 0^{2} V m^{- 1}

·For no deflection the electric force must cancel the magnetic force:

e E = e v B

, with

E

perpendicular to both

v

and

B

E = v B = 3 \times 1 0^{6} \times 9 \times 1 0^{- 4} = 2.7 \times 1 0^{3} = 27 \times 1 0^{2}

V/m.

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A model for the quantised motion of an electron in a uniform magnetic field

B

states that the flux passing through the orbit of the electron is

n (h / e)

, where

n

is an integer,

h

is Planck's constant and

e

is the magnitude of the electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (

m

is the mass of the electron):

Ans:

\frac{h e}{2 π m}

·Flux quantisation:

B π r^{2} = \frac{h}{e}

for the lowest state. In the field,

m v = e B r

. Magnetic moment

μ = \frac{e v r}{2} = \frac{e ^{2} B r ^{2}}{2 m} = \frac{e ^{2} B}{2 m} \cdot \frac{h}{e π B} = \frac{h e}{2 π m}

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19. Magnetism and Matter

Bar magnet and magnetic field lines: Magnetic field lines always form closed loops with no start or end point, since isolated magnetic poles don't exist — unlike electric field lines, which begin and end on charges. A bar magnet's field falls off as 1/r³, its axial value twice the equatorial one.
Earth's magnetism: Dip angle is 0° at the magnetic equator, where the field is purely horizontal, and 90° at the magnetic poles, where it's purely vertical — and Earth's actual magnetic south pole sits near the geographic north.
Dia-, para- and ferromagnetism: Diamagnetism is present in every material, including paramagnetic and ferromagnetic ones — it's just too weak to notice once a stronger attractive effect is present. Above its Curie temperature, a ferromagnetic material loses its domains and simply becomes paramagnetic.

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is

9.8 \times 1 0^{- 6} kg m^{2}

. If the magnitude of the magnetic moment of the needle is

x \times 1 0^{- 5} A m^{2}

, then the value of

x

is:

Ans:

1280 π^{2}

·Oscillation period

T = \frac{5}{20} = 0.25

s. For a magnetic needle,

T = 2 π \frac{I}{M B}

, so

M = \frac{4 π ^{2} I}{T ^{2} B} = \frac{4 π ^{2} \times 9.8 \times 1 0 ^{- 6}}{0.0625 \times 0.049} = 1280 π^{2} \times 1 0^{- 5}

A m^{2}

. Hence

x = 1280 π^{2}

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Match List I with List II.

List I (Material)	List II (Susceptibility $χ$ )
A. Diamagnetic	I. $χ = 0$
B. Ferromagnetic	II. $- 1 \leq χ < 0$
C. Paramagnetic	III. $χ ≫ 1$
D. Non-magnetic	IV. $0 < χ < ε$ (a small positive number)

Choose the correct answer from the options given below:

Ans: A-II, B-III, C-IV, D-I·Diamagnetic: small negative susceptibility,

- 1 \leq χ < 0

(II). Ferromagnetic:

χ ≫ 1

(III). Paramagnetic: small positive,

0 < χ < ε

(IV). Non-magnetic:

χ = 0

(I).

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20. Electromagnetic Induction

Faraday's and Lenz's laws: Lenz's law says the induced effect opposes the change in flux, not the flux itself — if flux through a loop is decreasing, the induced current flows to try to maintain it, not to fight its existing direction.
Motional EMF and eddy currents: Eddy currents are the same phenomenon whether nuisance or feature: laminated transformer cores exist purely to suppress them, while electromagnetic braking and induction furnaces deliberately rely on them — NEET likes sorting a list into wanted versus unwanted.
Self and mutual inductance: Self-inductance is fixed by a coil's geometry and turns alone, never by the current through it — what an inductor actually opposes is a change in current, not current itself, so a steady current flows through it perfectly freely.

A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to: A. hold the sheet there if it is magnetic. B. hold the sheet there if it is non-magnetic. C. move the sheet away from the pole with uniform velocity if it is conducting. D. move the sheet away from the pole with uniform velocity if it is both non-conducting and non-polar. Choose the correct statement(s) from the options given below:

Ans: A and C only·A magnetic sheet is attracted and needs a force to hold it (A). Moving a conductor away from a pole induces eddy currents that oppose the motion, so a force is needed to keep it moving uniformly (C). A non-magnetic sheet feels no force (B false); a non-conducting, non-polar sheet has no induced currents (D false).

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AB is a part of an electrical circuit (see figure). The potential difference

V_{A} - V_{B}

, at the instant when the current

i = 2

A and is increasing at a rate of 1 A/s, is:

Ans: 10 volt·Walk from A to B adding potential drops: across the inductor

L \frac{d i}{d t} = 1 \times 1 = 1

V (current increasing); across the cell, entering at the positive plate, a drop of 5 V; across the resistor

i R = 2 \times 2 = 4

V_{A} - V_{B} = 1 + 5 + 4 = 10

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21. Alternating Current

AC through R, L, C and LCR circuits: In a pure inductor, current lags voltage by 90°; in a pure capacitor, it leads by 90° — remembered as 'ELI the ICE man' (in L, voltage E leads current I; in C, current I leads voltage E).
Resonance, power and power factor: At series-LCR resonance, impedance is at its minimum (just R) and current is maximum, with the circuit behaving as purely resistive — don't confuse this with parallel resonance, where impedance peaks instead. A pure L or C dissipates zero average power over a cycle.
Transformers: An ideal transformer conserves power, not voltage or current individually — a step-up transformer that raises voltage by some factor drops current by that exact same factor, and transformers work only on AC, since they need a changing flux.

To an AC power supply of 220 V at 50 Hz, a resistor of

20Ω, a

capacitor of reactance

25Ω

and an inductor of reactance

45Ω

are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage are, respectively:

Ans: 7.8 A and 45°·

Z = R^{2} + (X_{L} - X_{C})^{2} = 2 0^{2} + 2 0^{2} = 202 \approx 28.3

Ω

, so

I = \frac{220}{28.3} \approx 7.8

tan ϕ = \frac{X _{L} - X _{C}}{R} = 1 \Rightarrow ϕ = 4 5^{\circ}

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A 10 μ F

capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (take

π = 3.14

Ans: 0.93 A·

X_{C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 1 0 ^{- 5}} \approx 318

Ω

. Peak current

= \frac{V _{0}}{X _{C}} = \frac{210 2}{318} \approx 0.93

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22. Electromagnetic Waves

Displacement current and Maxwell's equations (qualitative): Displacement current isn't an actual flow of charge — it's the term Maxwell added so Ampère's law gives a consistent answer for any surface, and the textbook example is a charging capacitor's gap, where conduction current is zero but displacement current isn't.
Electromagnetic spectrum: Every part of the electromagnetic spectrum, from gamma rays to radio waves, travels at exactly the same speed c in vacuum — only wavelength and frequency change across it, contrary to a common trap implying higher-energy radiation 'travels faster.'

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to the displacement current is:

Ans: Non-zero everywhere, with a maximum at the imaginary cylindrical surface connecting the peripheries of the plates·The growing electric field between the plates is a displacement current, and by Ampère–Maxwell it produces a magnetic field that grows with distance from the axis up to the plate edge and then falls as

1/ r

outside — non-zero everywhere, maximum at the rim.

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The electric field in a plane electromagnetic wave is given by

E_{z} = 60 cos (5 x + 1.5 \times 1 0^{9} t)

V/m. Then the expression for the corresponding magnetic field is (here the subscripts denote the direction of the field):

Ans:

B_{y} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t)

T·

B_{0} = \frac{E _{0}}{c} = \frac{60}{3 \times 1 0 ^{8}} = 2 \times 1 0^{- 7}

T. The wave travels along

- x

(phase

5 x + ω t

); with

E

along

z

E \times B

must point along

- x

, which requires

B

along

+ y

. So

B_{y} = 2 \times 1 0^{- 7} cos (5 x + 1.5 \times 1 0^{9} t)

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23. Ray Optics and Optical Instruments

Reflection and refraction, mirror and lens formulae: The biggest source of errors here is sign convention: a concave mirror has negative f, a convex mirror positive f — a convex mirror's image is always virtual and upright regardless of object distance, and getting the sign backwards flips every later answer.
Total internal reflection and prisms: Total internal reflection needs two conditions together: light going from a denser to a rarer medium, and the angle of incidence exceeding the critical angle (sin C = 1/n) — higher refractive index gives a smaller critical angle, which is why diamond traps light so well.
Optical instruments: microscope and telescope: A microscope's magnifying power rises as its lenses' focal lengths get shorter, but a telescope's, fo/fe, rises as the OBJECTIVE's focal length gets longer — opposite dependence, which is why microscope objectives are tiny and telescope objectives are built as large as possible.

A microscope has an objective of focal length 2 cm, an eyepiece of focal length 4 cm and a tube length of 40 cm. If the distance of distinct vision of the eye is 25 cm, the magnification of the microscope is:

Ans: 125·

m = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}}) = \frac{40}{2} (1 + \frac{25}{4}) = 20 \times 7.25 = 145

; with the commonly used approximation

m \approx \frac{L}{f _{o}} \cdot \frac{D}{f _{e}} = 20 \times 6.25 = 125

, which is the marked answer.

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In a certain camera, a combination of four similar thin convex lenses is arranged axially in contact. Then the power of the combination and the total magnification, in comparison to the power (

p

) and magnification (

m

) of each lens, will be, respectively:

Ans:

4 p

and

m^{4}

·Powers of thin lenses in contact add:

P_{t o t a l} = 4 p

. Magnifications multiply:

m_{t o t a l} = m^{4}

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24. Wave Optics

Huygens' principle: Huygens' wave theory correctly predicted that light slows down in a denser medium, v = c/n — the opposite of Newton's corpuscular prediction — a difference that later measurement of light's speed in water settled in the wave theory's favour.
Interference and Young's double slit: Bright and dark fringes in Young's double slit are evenly and identically spaced, β = λD/d, not narrower at the dark fringes — and the whole pattern needs coherent sources, which is why two ordinary, independent lamps never show a stable pattern.
Diffraction and polarisation: Polarisation only happens for transverse waves, and the fact that light can be polarised is itself the proof light is transverse. In single-slit diffraction, a sinθ = nλ locates the minima, not the maxima — the reverse of double-slit's condition for maxima.

An unpolarised light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then:

Ans: The reflected light is completely polarised and the angle of reflection is close to 60°·Brewster's angle:

tan θ_{B} = μ = 1.73 \Rightarrow θ_{B} = 6 0^{\circ}

. The reflected beam (at

6 0^{\circ}

) is completely polarised; the refracted beam, at

3 0^{\circ}

, is only partially polarised.

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The intensity of transmitted light when a polaroid sheet is placed between two crossed polaroids at

22. 5^{\circ}

from the polarisation axis of one of the polaroids is (

I_{0}

is the intensity of polarised light after passing through the first polaroid):

Ans:

\frac{I _{0}}{8}

·First polaroid →

I_{0}

. The middle sheet at

22. 5^{\circ}

I_{0} cos^{2} 22. 5^{\circ}

. The last, crossed with the first, is at

67. 5^{\circ}

to the middle sheet:

I_{0} cos^{2} 22. 5^{\circ} cos^{2} 67. 5^{\circ} = I_{0} (\frac{s i n 4 5 ^{\circ}}{2})^{2} = \frac{I _{0}}{8}

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25. Dual Nature of Radiation and Matter

Photoelectric effect and Einstein's equation: Increasing light intensity increases the number of photoelectrons, not their maximum kinetic energy — that depends only on frequency, KEmax = hν − φ — and below the threshold frequency, no electrons come out no matter how bright the light is.
Matter waves and de Broglie wavelength: Every moving object has a de Broglie wavelength, λ = h/mv, not just electrons — it's simply too small to matter macroscopically. For an electron accelerated through potential V, this gives λ = h/√(2meV), a frequently tested formula in its own right.

The de Broglie wavelength of an electron orbiting in the

n = 2

state of the hydrogen atom is close to (given Bohr radius = 0.052 nm):

Ans: 0.67 nm·In the

n

th orbit the circumference equals

n

de Broglie wavelengths:

2 π r_{n} = nλ

. With

r_{2} = 4 a_{0} = 0.208

nm:

λ = \frac{2 π \times 0.208}{2} \approx 0.65

\approx 0.67

nm.

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A photon and an electron (mass

m

) have the same energy

E

. The ratio

λ_{photon} / λ_{electron}

of their de Broglie wavelengths is (

c

is the speed of light):

Ans:

\frac{2 m c ^{2}}{E}

·Photon:

λ_{p} = \frac{h c}{E}

. Electron:

λ_{e} = \frac{h}{2 m E}

. Ratio

= \frac{h c}{E} \cdot \frac{2 m E}{h} = c \frac{2 m}{E} = \frac{2 m c ^{2}}{E}

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26. Atoms

Rutherford and Bohr models: Rutherford's model correctly located the nucleus but couldn't explain atomic stability — classically an accelerating orbiting electron should radiate energy and spiral inward. Bohr fixed exactly this by allowing only orbits with quantised angular momentum, L = nh/2π.
Hydrogen spectrum and energy levels: Only the Balmer series lands substantially in the visible range, since it ends at n=2 — Lyman (ending at n=1) is entirely ultraviolet and the rest are infrared. Energy levels En = −13.6/n² eV are always negative, approaching zero only at ionisation.

A particle of mass

m

is moving around the origin with a constant force

F

pulling it towards the origin. If the Bohr model is used to describe its motion, the radius of the

n

th orbit and the particle's speed

v

in the orbit depend on

n

as:

Ans:

r \propto n^{2/3}; v \propto n^{1/3}

·Constant force

F = \frac{m v ^{2}}{r}

and Bohr quantisation

m v r = \frac{nh}{2 π}

. From the first,

v^{2} \propto r

; substituting,

r^{3/2} \propto n \Rightarrow r \propto n^{2/3}

and

v \propto r^{1/2} \propto n^{1/3}

NEET 2025 · open →

Match List I with List II.

List I (Spectral lines of hydrogen for transitions from)	List II (Wavelength, nm)
A. $n_{2} = 3$ to $n_{1} = 2$	I. 410.2
B. $n_{2} = 4$ to $n_{1} = 2$	II. 434.1
C. $n_{2} = 5$ to $n_{1} = 2$	III. 656.3
D. $n_{2} = 6$ to $n_{1} = 2$	IV. 486.1

Choose the correct answer from the options given below:

Ans: A-III, B-IV, C-II, D-I·Balmer wavelengths follow

\frac{1}{λ} \propto (\frac{1}{4} - \frac{1}{n ^{2}})

: the smallest jump (

n = 3 \to 2

) gives the longest wavelength, 656.3 nm

(H - α)

, then 486.1, 434.1 and 410.2 nm as

n

rises. Hence A-III, B-IV, C-II, D-I.

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27. Nuclei

Nuclear size, mass defect, binding energy: Binding energy per nucleon peaks near mass number 56 (iron) and falls on both sides — exactly why fusion of light nuclei and fission of heavy nuclei both release energy, moving toward that peak. Nuclear density comes out roughly the same for every nucleus.
Radioactivity and decay law: A radioactive decay constant is fixed and completely unaffected by temperature, pressure or chemical combination, unlike an ordinary reaction rate — so half-life stays exactly the same regardless of how much substance remains or what conditions it's kept in.
Fission and fusion: Commercial nuclear power runs on fission, splitting heavy nuclei like uranium — controlled fusion, despite powering the Sun, isn't commercially viable because forcing light nuclei close enough to fuse means overcoming their Coulomb repulsion at extreme temperature and pressure.

_{82}^{290} X α Y e^{+} Z e^{-} P e^{-} Q

In the nuclear emission stated above, the mass number and atomic number of the product

Q

, respectively, are:

Ans: 286, 81·An

α -

emission lowers

A

by 4 and

Z

by 2:

(286, 80)

. Each

β -

emission leaves

A

unchanged; a positron lowers

Z

by 1 and an electron raises it by 1. Net over

e^{+}, e^{-}, e^{-}

Z = 80 - 1 + 1 + 1 = 81

. So

Q

(286, 81)

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28. Semiconductor Electronics

Energy bands, intrinsic and extrinsic semiconductors: Doping never leaves a semiconductor charged — an n-type material has extra free electrons as majority carriers, but the donor ion left behind balances the charge exactly, so the material as a whole stays electrically neutral.
p-n junction diode, rectifiers, Zener diode: A Zener diode is the one diode deliberately operated in reverse breakdown, using its near-constant voltage there for regulation — every ordinary diode is designed to avoid that same region. A full-wave rectifier conducts on both halves of the input cycle; a half-wave rectifier, only one.
Logic gates: NAND and NOR are each called a universal gate because either one alone, used repeatedly, can build every other basic gate — AND, OR and NOT included — which is why real circuits are so often built almost entirely from just one of them.

A logic circuit provides the output

Y

as per the following truth table:

A	B	Y
0	0	1
0	1	0
1	0	1
1	1	0

The expression for the output

Y

is:

Ans:

\overset{ˉ}{B}

·The output is 1 exactly when

B = 0

, regardless of

A

— the table is that of

\overset{ˉ}{B}

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The output (

Y

) of the given logic implementation is similar to the output of a/an _____ gate.

Ans: NOR·The top gate is a NOR:

\overline{A + B}

. The bottom gate is a NAND:

\overline{A \cdot B}

. Their AND gives

Y = \overset{ˉ}{A} \overset{ˉ}{B} \cdot (\overset{ˉ}{A} + \overset{ˉ}{B}) = \overset{ˉ}{A} \overset{ˉ}{B}

, which is exactly

\overline{A + B}

— a NOR gate.

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