CLASS 11 PHYSICS • FORMULAS + NCERT + PYQs

Dimensions and Dimensional Analysis Formula Sheet, NCERT Exercises and PYQs

A complete revision page for dimensions, dimensional analysis, units, significant figures, errors, formulae, NCERT exercises 1.1 to 1.17 and exam-style practice.

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1. Complete Formula Sheet

SI Base Quantities

  • Length: metre, m, [L]
  • Mass: kilogram, kg, [M]
  • Time: second, s, [T]
  • Electric current: ampere, A, [I]
  • Temperature: kelvin, K, [K]
  • Amount: mole, mol
  • Luminous intensity: candela, cd

Unit Conversion Formulae

  • 1 km = 103 m
  • 1 cm = 10-2 m
  • 1 Å = 10-10 m
  • 1 ly = 9.46 × 1015 m
  • 1 g cm-3 = 103 kg m-3

Methods

  • Check homogeneity of equations
  • Derive powers in relations
  • Convert units by dimensions
  • Report answers using significant figures
  • Calculate error propagation

2. SI Units Table

Physical QuantitySI UnitSymbolDimensional Formula
Lengthmetrem[L]
Masskilogramkg[M]
Timeseconds[T]
CurrentampereA[I]
TemperaturekelvinK[K]
Amount of substancemolemol[mol]
Luminous intensitycandelacd[cd]
Areasquare metrem2[L2]
Volumecubic metrem3[L3]
Velocitymetre per secondm s-1[LT-1]
Accelerationmetre per second squarem s-2[LT-2]
ForcenewtonN[MLT-2]
EnergyjouleJ[ML2T-2]
PowerwattW[ML2T-3]
PressurepascalPa[ML-1T-2]
ChargecoulombC[IT]
PotentialvoltV[ML2T-3I-1]
ResistanceohmΩ[ML2T-3I-2]

3. Prefix Table

Power of 10PrefixSymbolExample
10-24yoctoy1 ym = 10-24 m
10-21zeptoz1 zs = 10-21 s
10-18attoa1 as = 10-18 s
10-15femtof1 fm = 10-15 m
10-12picop1 pF = 10-12 F
10-9nanon1 nm = 10-9 m
10-6microμ1 μm = 10-6 m
10-3millim1 mm = 10-3 m
10-2centic1 cm = 10-2 m
10-1decid1 dm = 10-1 m
101decada1 dam = 101 m
102hectoh1 hPa = 102 Pa
103kilok1 km = 103 m
106megaM1 MHz = 106 Hz
109gigaG1 GW = 109 W
1012teraT1 TW = 1012 W
1015petaP1 PW = 1015 W
1018exaE1 EJ = 1018 J
1021zettaZ1 Zm = 1021 m
1024yottaY1 Ym = 1024 m

4. Significant Figure Rules

Zeros

Leading zeros are not significant. Captive zeros are significant. Trailing zeros after decimal are significant.

Operations

Addition/subtraction follows least decimal places. Multiplication/division follows least significant figures.

Examples

0.007 has 1 SF. 2.64 × 1024 has 3 SF. 0.2370 has 4 SF.

5. Error Formulae

Absolute and Mean Error

Δa = |ameasured - atrue|
amean = (a1 + a2 + ... + an) / n
Δamean = (|Δa1| + |Δa2| + ... + |Δan|) / n

Relative and Percentage Error

Relative Error = Δa / a
Percentage Error = (Δa / a) × 100
Final result: a = amean ± Δamean

6. Percentage Error Formulae

Add/Subtract

If Z = A + B or A - B, then ΔZ = ΔA + ΔB

Multiply/Divide

If Z = AB or A / B, then ΔZ / Z = ΔA/A + ΔB/B

Power Rule

If Z = AmBn / CpDq, then ΔZ/Z = m(ΔA/A) + n(ΔB/B) + p(ΔC/C) + q(ΔD/D)

NEET Numerical

Question: If Y = X3 and error in X is 2%, find error in Y.

Solution: Power rule: percentage error in Y = 3 × 2% = 6%.

Final Answer: 6%

Exam Tip: Power multiplies percentage error.

JEE Numerical

Question: If Z = A2B3/C4, errors in A, B, C are 1%, 2%, 1%.

Solution: % error = 2(1) + 3(2) + 4(1) = 12%.

Final Answer: 12%

Exam Tip: Denominator powers also add positively.

NEET Numerical

Question: For R = V/I, errors in V and I are 1% and 2%.

Solution: %R = %V + %I = 3%.

Final Answer: 3%

Exam Tip: Division also means addition of relative errors.

JEE Numerical

Question: For T = 2π√(L/g), if L has 4% error and g is exact, find error in T.

Solution: T depends on L1/2; percentage error = 1/2 × 4% = 2%.

Final Answer: 2%

Exam Tip: Square root means power 1/2.

7. Dimensional Formula Table

ChapterPhysical Quantity / FormulaSI UnitDimensional FormulaShort Note
KinematicsDisplacementm[L]Basic length quantity
KinematicsVelocitym s-1[LT-1]Rate of displacement
KinematicsAccelerationm s-2[LT-2]Rate of velocity
Laws of MotionForceN[MLT-2]Newton second law
Work Energy PowerWork/EnergyJ[ML2T-2]Force × displacement
Work Energy PowerPowerW[ML2T-3]Energy per time
Rotational MotionTorqueN m[ML2T-2]Rotational force effect
Rotational MotionAngular momentumkg m2 s-1[ML2T-1]
Rotational MotionMoment of inertiakg m2[ML2]Rotational inertia
GravitationGN m2 kg-2[M-1L3T-2]Universal gravitational constant
SolidsStressPa[ML-1T-2]Force per area
SolidsStrainno unitDimensionlessRatio of lengths
SolidsYoung modulusPa[ML-1T-2]Stress/strain
FluidsDensitykg m-3[ML-3]Mass per volume
FluidsPressurePa[ML-1T-2]Force per area
FluidsViscosityPa s[ML-1T-1]Fluid resistance
Thermal PhysicsHeatJ[ML2T-2]Energy transfer
Thermal PhysicsSpecific heatJ kg-1 K-1[L2T-2K-1]Heat per kg per K
ThermodynamicsGas constantJ mol-1K-1[ML2T-2K-1mol-1]PV=nRT
Kinetic TheoryBoltzmann constantJ K-1[ML2T-2K-1]Energy per kelvin
OscillationsFrequencyHz[T-1]Oscillations per second
WavesWavelengthm[L]Spatial period
WavesWave speedm s-1[LT-1]v = fλ
ElectrostaticsChargeC[IT]Current × time
ElectrostaticsElectric fieldN C-1[MLT-3I-1]Force per charge
ElectrostaticsPotentialV[ML2T-3I-1]Work per charge
ElectrostaticsCapacitanceF[M-1L-2T4I2]Charge per potential
Current ElectricityResistanceΩ[ML2T-3I-2]V/I
Current ElectricityResistivityΩ m[ML3T-3I-2]Material property
MagnetismMagnetic fieldT[MT-2I-1]Force on current
MagnetismMagnetic fluxWb[ML2T-2I-1]BA
EMIInductanceH[ML2T-2I-2]Flux per current
ACReactanceΩ[ML2T-3I-2]Opposition in AC
EM Wavescm s-1[LT-1]Speed of light
Ray OpticsRefractive indexno unitDimensionlessRatio of speeds
Wave OpticsWavelengthm[L]Wave spacing
Dual NaturePlanck constant hJ s[ML2T-1]E = hν
AtomsBohr radiusm[L]Atomic radius
NucleiActivityBq[T-1]Decays per second
SemiconductorMobilitym2 V-1 s-1[M-1T2I]Drift per field

8. Linear Motion vs Angular Motion Dimension Table

Linear MotionDimensionAngular MotionDimension
Displacement s[L]Angular displacement θDimensionless
Velocity v[LT-1]Angular velocity ω[T-1]
Acceleration a[LT-2]Angular acceleration α[T-2]
Mass m[M]Moment of inertia I[ML2]
Force F[MLT-2]Torque τ[ML2T-2]
Momentum p[MLT-1]Angular momentum L[ML2T-1]
Work = F × s[ML2T-2]Work = τ × θ[ML2T-2]
Kinetic Energy[ML2T-2]Rotational Kinetic Energy[ML2T-2]
F = ma[MLT-2]τ = Iα[ML2T-2]

9. Dimensionless Quantities

Pure Ratios

Angle θ = arc/radius, strain = change in length/original length, refractive index = speed ratio, relative density = density ratio.

Functions

sin θ, cos θ and tan θ are meaningful only when θ is dimensionless. ex, log x and ln x require dimensionless x.

Examples

Coefficient of friction, Poisson ratio, relative permittivity and relative permeability are dimensionless.

10. NCERT Examples and Exercises 1.1 to 1.17

All visible NCERT exercise questions from the attached screenshots are transcribed and solved below.

NCERT Exercise 1.1

Question: Fill in the blanks: (a) The volume of a cube of side 1 cm is equal to .....m3. (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2. (c) A vehicle moving with a speed of 18 km h-1 covers....m in 1 s. (d) The relative density of lead is 11.3. Its density is ....g cm-3 or ....kg m-3.

Step-by-step Solution: (a) 1 cm = 10-2 m, so 1 cm3 = 10-6 m3. (b) r = 20 mm, h = 100 mm, area = 2πr(r+h) = 2π × 20 × 120 = 4800π = 1.5 × 104 mm2. (c) 18 km h-1 = 5 m s-1, distance in 1 s = 5 m. (d) relative density = density in g cm-3, so 11.3 g cm-3 = 1.13 × 104 kg m-3.

Final Answer: (a) 10-6 m3; (b) 1.5 × 104 mm2; (c) 5 m; (d) 11.3 g cm-3, 1.13 × 104 kg m-3.

Exam Tip: Always convert each base unit first, then apply powers.

NCERT Exercise 1.2

Question: Fill in the blanks by suitable conversion of units: (a) 1 kg m2 s-2 = ....g cm2 s-2. (b) 1 m = ..... ly. (c) 3.0 m s-2 = ..... km h-2. (d) G = 6.67 × 10-11 N m2 (kg)-2 = .... (cm)3 s-2 g-1.

Step-by-step Solution: (a) 1 kg = 103 g and 1 m2 = 104 cm2, so answer = 107. (b) 1 ly = 9.46 × 1015 m, so 1 m = 1.06 × 10-16 ly. (c) 1 m = 10-3 km and 1 s-2 = 36002 h-2; value = 3.0 × 10-3 × 36002 = 3.9 × 104. (d) Convert N to g cm s-2, m to cm, kg to g; G = 6.67 × 10-8.

Final Answer: (a) 107; (b) 1.06 × 10-16 ly; (c) 3.9 × 104 km h-2; (d) 6.67 × 10-8 cm3 s-2 g-1.

Exam Tip: For acceleration conversion, square the time conversion factor.

NCERT Exercise 1.3

Question: A calorie is a unit of heat and it equals about 4.2 J where 1 J = 1 kg m2 s-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time equals γ s. Show that a calorie has a magnitude 4.2 α-1 β-2 γ2 in terms of the new units.

Step-by-step Solution: Energy has dimensions [ML2T-2]. New unit of energy = α kg × β2 m2 × γ-2 s-2 = αβ2γ-2 J. Therefore 4.2 J = 4.2 /(αβ2γ-2) new units = 4.2 α-1β-2γ2.

Final Answer: 4.2 α-1 β-2 γ2 new units.

Exam Tip: For new unit systems, divide old magnitude by the new unit size.

NCERT Exercise 1.4

Question: Explain this statement clearly: To call a dimensional quantity large or small is meaningless without specifying a standard for comparison. In view of this, reframe statements: (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) speed of sound is much smaller than speed of light.

Step-by-step Solution: A dimensional quantity needs comparison with a standard of same dimension. Reframed: atoms are very small compared with ordinary macroscopic objects; jet speed is large compared with car speed; Jupiter mass is large compared with Earth mass; number of air molecules is large compared with ordinary counted objects; proton mass is about 1836 times electron mass; speed of sound is much smaller than speed of light.

Final Answer: All statements must include a comparison standard.

Exam Tip: Never call a dimensional quantity large/small without reference.

NCERT Exercise 1.5

Question: A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Step-by-step Solution: New unit is distance travelled by light in 1 s because c = 1 new unit/s. Time = 8 min 20 s = 500 s. Hence distance = 500 new units.

Final Answer: 500 new units of length.

Exam Tip: When c is set to unity, light-second becomes the length unit.

NCERT Exercise 1.6

Question: Which of the following is the most precise device for measuring length: (a) vernier calipers with 20 divisions on sliding scale (b) screw gauge of pitch 1 mm and 100 divisions on circular scale (c) optical instrument that can measure length to within a wavelength of light?

Step-by-step Solution: Precision depends on least count. Vernier least count is about 1/20 mm = 0.05 mm. Screw gauge least count = 1 mm/100 = 0.01 mm. Optical instrument can measure to wavelength scale, about 10-7 m, much smaller.

Final Answer: Optical instrument is most precise.

Exam Tip: Smaller least count means greater precision.

NCERT Exercise 1.7

Question: A student measures the thickness of a human hair using a microscope of magnification 100. He makes 20 observations and finds the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate of the thickness of hair?

Step-by-step Solution: Magnification = image size/object size. Object thickness = observed width / magnification = 3.5 mm / 100 = 0.035 mm = 3.5 × 10-5 m.

Final Answer: 0.035 mm or 3.5 × 10-5 m.

Exam Tip: Divide apparent size by magnification.

NCERT Exercise 1.8

Question: Answer the following: (a) You are given a thread and a metre scale. How will you estimate the diameter of the thread? (b) A screw gauge has pitch 1.0 mm and 200 divisions on circular scale. Can accuracy be increased arbitrarily by increasing divisions? (c) Why is a set of 100 measurements of rod diameter more reliable than 5 measurements?

Step-by-step Solution: (a) Wind the thread closely many turns on a pencil, measure total width, divide by number of turns. (b) Least count decreases, but accuracy cannot improve arbitrarily because of mechanical errors, backlash and material limitations. (c) More readings reduce random error and make mean more reliable.

Final Answer: (a) Diameter = total width/number of turns. (b) No. (c) More readings improve reliability.

Exam Tip: Use repeated measurements to reduce random error.

NCERT Exercise 1.9

Question: The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement?

Step-by-step Solution: Area magnification = screen area / slide area. Convert 1.55 m2 = 1.55 × 104 cm2. Area ratio = 1.55 × 104/1.75 = 8.857 × 103. Linear magnification = square root of area ratio ≈ 94.1.

Final Answer: Linear magnification ≈ 94.

Exam Tip: Linear magnification is square root of area magnification.

NCERT Exercise 1.10

Question: State the number of significant figures in: (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm-3 (d) 6.320 J (e) 6.032 N m-2 (f) 0.0006032 m2.

Step-by-step Solution: Leading zeros are not significant. Non-zero digits and captive zeros are significant. Trailing zeros after decimal are significant. Counts: (a) 1; (b) 3; (c) 4; (d) 4; (e) 4; (f) 4.

Final Answer: (a) 1, (b) 3, (c) 4, (d) 4, (e) 4, (f) 4 significant figures.

Exam Tip: Zeros between non-zero digits and after decimal count.

NCERT Exercise 1.11

Question: The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Step-by-step Solution: Area = length × breadth = 4.234 × 1.005 = 4.25517 m2. Least SF = 4, so area = 4.255 m2. Thickness = 2.01 cm = 0.0201 m. Volume = 4.25517 × 0.0201 = 0.08553 m3. Least SF = 3, so volume = 0.0855 m3.

Final Answer: Area = 4.255 m2; Volume = 8.55 × 10-2 m3.

Exam Tip: For multiplication, final answer follows least significant figures.

NCERT Exercise 1.12

Question: The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added. What is (a) total mass of the box, (b) difference in masses of the pieces to correct significant figures?

Step-by-step Solution: Convert gold mass to kg: 20.15 g + 20.17 g = 40.32 g = 0.04032 kg. Total = 2.30 + 0.04032 = 2.34032 kg. Addition follows least decimal places: 2.30 kg has two decimal places, so total = 2.34 kg. Difference = 20.17 g - 20.15 g = 0.02 g.

Final Answer: (a) 2.34 kg; (b) 0.02 g.

Exam Tip: In addition/subtraction, use decimal-place rule.

NCERT Exercise 1.13

Question: A famous relation relates moving mass m to rest mass m0, speed v and speed of light c. A boy writes m = m0 / (1 - v2)1/2. Guess where to put missing c.

Step-by-step Solution: The term subtracted from 1 must be dimensionless. v has dimension [LT-1], so v2 must be divided by c2. Correct denominator is (1 - v2/c2)1/2.

Final Answer: m = m0 / (1 - v2/c2)1/2.

Exam Tip: Arguments inside 1 − something must be dimensionless.

NCERT Exercise 1.14

Question: The unit of length convenient on atomic scale is angstrom Å: 1 Å = 10-10 m. Size of hydrogen atom is about 0.5 Å. What is total atomic volume in m3 of a mole of hydrogen atoms?

Step-by-step Solution: Radius approximately 0.5 Å = 0.5 × 10-10 m. Volume of one atom ≈ 4/3 πr3 ≈ 5.24 × 10-31 m3. For one mole, multiply by 6.02 × 1023: total ≈ 3.15 × 10-7 m3.

Final Answer: About 3 × 10-7 m3.

Exam Tip: Atomic volumes are tiny; multiply by Avogadro number for one mole.

NCERT Exercise 1.15

Question: One mole of ideal gas at standard temperature and pressure occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Take hydrogen molecule size about 1 Å. Why is this ratio so large?

Step-by-step Solution: Molar volume = 22.4 L = 2.24 × 10-2 m3. Size of molecule ≈ 1 Å, radius ≈ 0.5 Å, so molecular volume for a mole is about 3 × 10-7 m3. Ratio ≈ 2.24 × 10-2 / 3 × 10-7 ≈ 7 × 104. It is large because gas molecules are separated by large empty spaces.

Final Answer: Ratio ≈ 7 × 104; large due to intermolecular empty space.

Exam Tip: Gas volume is mostly empty space.

NCERT Exercise 1.16

Question: Explain: looking out of a fast moving train, nearby trees and houses seem to move rapidly opposite to the train, distant hills and Moon seem stationary; since you know you are moving, distant objects seem to move with you.

Step-by-step Solution: Angular speed of an object in field of view is approximately v/d, where d is distance from observer. Nearby objects have small d, so large angular speed and appear to move fast backward. Distant objects have huge d, so angular speed is tiny and they appear almost stationary. Because the observer knows the train is moving, very distant objects may seem to move along with the observer.

Final Answer: Apparent motion depends on angular speed v/d; nearby objects have larger apparent motion.

Exam Tip: Apparent motion is governed by angular displacement, not just linear speed.

NCERT Exercise 1.17

Question: The Sun is hot plasma with inner core temperature exceeding 107 K and outer surface about 6000 K. What range do you expect mass density of Sun to be in, relative to solids, liquids or gases? Check using mass = 2.0 × 1030 kg and radius = 7.0 × 108 m.

Step-by-step Solution: Mean density = mass/volume = M / (4/3 πR3). R3 = 3.43 × 1026 m3. Volume ≈ 4.19 × 3.43 × 1026 = 1.44 × 1027 m3. Density ≈ 2.0 × 1030 / 1.44 × 1027 = 1.4 × 103 kg m-3. This is comparable to liquids/solids, despite plasma state, because of huge gravitational compression.

Final Answer: Mean density ≈ 1.4 × 103 kg m-3.

Exam Tip: Estimate density from M/(4/3πR3), then compare orders of magnitude.

11. CBSE PYQs and Exam-style Questions

CBSE Exam-style Question

Question: State the principle of homogeneity and give one use.

Solution: Every term in a correct physical equation has same dimensions. It is used to check formula correctness.

Final Answer: Principle stated with use.

Exam Tip: Use dimensions term by term.

CBSE Exam-style Question

Question: Why are significant figures important in numerical answers?

Solution: They indicate precision and prevent false accuracy in reported measurements.

Final Answer: They show reliability of measurement.

Exam Tip: Do not copy all calculator digits.

CBSE Exam-style Question

Question: Find dimensions of pressure.

Solution: Pressure = force/area = [MLT-2]/[L2] = [ML-1T-2].

Final Answer: [ML-1T-2]

Exam Tip: Start from definition.

CBSE Exam-style Question

Question: Convert 1 g cm-3 into kg m-3.

Solution: 1 g = 10-3 kg and 1 cm3 = 10-6 m3, so 1 g cm-3 = 103 kg m-3.

Final Answer: 1000 kg m-3

Exam Tip: Convert numerator and denominator separately.

12. NEET PYQs and NEET-style Question Bank: 60 MCQs

NEET MCQ 1

Question: Dimension of force is

(a) [MLT-2](b) [ML2T-2](c) [LT-1](d) [M-1L3T-2]

Correct Answer: [MLT-2]

Solution: Force = mass × acceleration.

Exam Tip: Memorize base formulae.

NEET MCQ 2

Question: Dimension of energy is

(a) [ML2T-2](b) [MLT-2](c) [ML2T-3](d) [T-1]

Correct Answer: [ML2T-2]

Solution: Energy = force × displacement.

Exam Tip: Work and energy same dimensions.

NEET MCQ 3

Question: Which is dimensionless?

(a) Refractive index(b) Pressure(c) Charge(d) Resistance

Correct Answer: Refractive index

Solution: It is ratio of two speeds.

Exam Tip: Ratios of same quantities are dimensionless.

NEET MCQ 4

Question: 1 Å equals

(a) 10-10 m(b) 1010 m(c) 10-6 m(d) 10-3 m

Correct Answer: 10-10 m

Solution: Angstrom is atomic length unit.

Exam Tip: Useful for atomic scale.

NEET MCQ 5

Question: Charge dimension is

(a) [IT](b) [I/T](c) [MT](d) [MLT]

Correct Answer: [IT]

Solution: q = It.

Exam Tip: Current times time.

NEET MCQ 6

Question: If Z = A2, error in A is 2%, error in Z is

(a) 4%(b) 2%(c) 1%(d) 8%

Correct Answer: 4%

Solution: Power doubles percentage error.

Exam Tip: Power rule.

NEET MCQ 7

Question: Significant figures in 0.2370 are

(a) 4(b) 3(c) 2(d) 5

Correct Answer: 4

Solution: Trailing zero after decimal is significant.

Exam Tip: Decimal trailing zero counts.

NEET MCQ 8

Question: Principle of homogeneity checks

(a) dimensions of equation(b) value of constants(c) sign of terms(d) vector direction

Correct Answer: dimensions of equation

Solution: All terms must have same dimensions.

Exam Tip: It is a necessary test.

NEET MCQ 9

Question: Dimension of G is

(a) [M-1L3T-2](b) [MLT-2](c) [ML2T-1](d) [IT]

Correct Answer: [M-1L3T-2]

Solution: Use Newton gravitation formula.

Exam Tip: Frequently asked.

NEET MCQ 10

Question: Argument of sin θ must be

(a) dimensionless(b) length(c) time(d) mass

Correct Answer: dimensionless

Solution: Trigonometric functions require pure number argument.

Exam Tip: Always check function arguments.

NEET MCQ 11

Question: Dimension of force is

(a) [MLT-2](b) [ML2T-2](c) [LT-1](d) [M-1L3T-2]

Correct Answer: [MLT-2]

Solution: Force = mass × acceleration.

Exam Tip: Memorize base formulae.

NEET MCQ 12

Question: Dimension of energy is

(a) [ML2T-2](b) [MLT-2](c) [ML2T-3](d) [T-1]

Correct Answer: [ML2T-2]

Solution: Energy = force × displacement.

Exam Tip: Work and energy same dimensions.

NEET MCQ 13

Question: Which is dimensionless?

(a) Refractive index(b) Pressure(c) Charge(d) Resistance

Correct Answer: Refractive index

Solution: It is ratio of two speeds.

Exam Tip: Ratios of same quantities are dimensionless.

NEET MCQ 14

Question: 1 Å equals

(a) 10-10 m(b) 1010 m(c) 10-6 m(d) 10-3 m

Correct Answer: 10-10 m

Solution: Angstrom is atomic length unit.

Exam Tip: Useful for atomic scale.

NEET MCQ 15

Question: Charge dimension is

(a) [IT](b) [I/T](c) [MT](d) [MLT]

Correct Answer: [IT]

Solution: q = It.

Exam Tip: Current times time.

NEET MCQ 16

Question: If Z = A2, error in A is 2%, error in Z is

(a) 4%(b) 2%(c) 1%(d) 8%

Correct Answer: 4%

Solution: Power doubles percentage error.

Exam Tip: Power rule.

NEET MCQ 17

Question: Significant figures in 0.2370 are

(a) 4(b) 3(c) 2(d) 5

Correct Answer: 4

Solution: Trailing zero after decimal is significant.

Exam Tip: Decimal trailing zero counts.

NEET MCQ 18

Question: Principle of homogeneity checks

(a) dimensions of equation(b) value of constants(c) sign of terms(d) vector direction

Correct Answer: dimensions of equation

Solution: All terms must have same dimensions.

Exam Tip: It is a necessary test.

NEET MCQ 19

Question: Dimension of G is

(a) [M-1L3T-2](b) [MLT-2](c) [ML2T-1](d) [IT]

Correct Answer: [M-1L3T-2]

Solution: Use Newton gravitation formula.

Exam Tip: Frequently asked.

NEET MCQ 20

Question: Argument of sin θ must be

(a) dimensionless(b) length(c) time(d) mass

Correct Answer: dimensionless

Solution: Trigonometric functions require pure number argument.

Exam Tip: Always check function arguments.

NEET MCQ 21

Question: Dimension of force is

(a) [MLT-2](b) [ML2T-2](c) [LT-1](d) [M-1L3T-2]

Correct Answer: [MLT-2]

Solution: Force = mass × acceleration.

Exam Tip: Memorize base formulae.

NEET MCQ 22

Question: Dimension of energy is

(a) [ML2T-2](b) [MLT-2](c) [ML2T-3](d) [T-1]

Correct Answer: [ML2T-2]

Solution: Energy = force × displacement.

Exam Tip: Work and energy same dimensions.

NEET MCQ 23

Question: Which is dimensionless?

(a) Refractive index(b) Pressure(c) Charge(d) Resistance

Correct Answer: Refractive index

Solution: It is ratio of two speeds.

Exam Tip: Ratios of same quantities are dimensionless.

NEET MCQ 24

Question: 1 Å equals

(a) 10-10 m(b) 1010 m(c) 10-6 m(d) 10-3 m

Correct Answer: 10-10 m

Solution: Angstrom is atomic length unit.

Exam Tip: Useful for atomic scale.

NEET MCQ 25

Question: Charge dimension is

(a) [IT](b) [I/T](c) [MT](d) [MLT]

Correct Answer: [IT]

Solution: q = It.

Exam Tip: Current times time.

NEET MCQ 26

Question: If Z = A2, error in A is 2%, error in Z is

(a) 4%(b) 2%(c) 1%(d) 8%

Correct Answer: 4%

Solution: Power doubles percentage error.

Exam Tip: Power rule.

NEET MCQ 27

Question: Significant figures in 0.2370 are

(a) 4(b) 3(c) 2(d) 5

Correct Answer: 4

Solution: Trailing zero after decimal is significant.

Exam Tip: Decimal trailing zero counts.

NEET MCQ 28

Question: Principle of homogeneity checks

(a) dimensions of equation(b) value of constants(c) sign of terms(d) vector direction

Correct Answer: dimensions of equation

Solution: All terms must have same dimensions.

Exam Tip: It is a necessary test.

NEET MCQ 29

Question: Dimension of G is

(a) [M-1L3T-2](b) [MLT-2](c) [ML2T-1](d) [IT]

Correct Answer: [M-1L3T-2]

Solution: Use Newton gravitation formula.

Exam Tip: Frequently asked.

NEET MCQ 30

Question: Argument of sin θ must be

(a) dimensionless(b) length(c) time(d) mass

Correct Answer: dimensionless

Solution: Trigonometric functions require pure number argument.

Exam Tip: Always check function arguments.

NEET MCQ 31

Question: Dimension of force is

(a) [MLT-2](b) [ML2T-2](c) [LT-1](d) [M-1L3T-2]

Correct Answer: [MLT-2]

Solution: Force = mass × acceleration.

Exam Tip: Memorize base formulae.

NEET MCQ 32

Question: Dimension of energy is

(a) [ML2T-2](b) [MLT-2](c) [ML2T-3](d) [T-1]

Correct Answer: [ML2T-2]

Solution: Energy = force × displacement.

Exam Tip: Work and energy same dimensions.

NEET MCQ 33

Question: Which is dimensionless?

(a) Refractive index(b) Pressure(c) Charge(d) Resistance

Correct Answer: Refractive index

Solution: It is ratio of two speeds.

Exam Tip: Ratios of same quantities are dimensionless.

NEET MCQ 34

Question: 1 Å equals

(a) 10-10 m(b) 1010 m(c) 10-6 m(d) 10-3 m

Correct Answer: 10-10 m

Solution: Angstrom is atomic length unit.

Exam Tip: Useful for atomic scale.

NEET MCQ 35

Question: Charge dimension is

(a) [IT](b) [I/T](c) [MT](d) [MLT]

Correct Answer: [IT]

Solution: q = It.

Exam Tip: Current times time.

NEET MCQ 36

Question: If Z = A2, error in A is 2%, error in Z is

(a) 4%(b) 2%(c) 1%(d) 8%

Correct Answer: 4%

Solution: Power doubles percentage error.

Exam Tip: Power rule.

NEET MCQ 37

Question: Significant figures in 0.2370 are

(a) 4(b) 3(c) 2(d) 5

Correct Answer: 4

Solution: Trailing zero after decimal is significant.

Exam Tip: Decimal trailing zero counts.

NEET MCQ 38

Question: Principle of homogeneity checks

(a) dimensions of equation(b) value of constants(c) sign of terms(d) vector direction

Correct Answer: dimensions of equation

Solution: All terms must have same dimensions.

Exam Tip: It is a necessary test.

NEET MCQ 39

Question: Dimension of G is

(a) [M-1L3T-2](b) [MLT-2](c) [ML2T-1](d) [IT]

Correct Answer: [M-1L3T-2]

Solution: Use Newton gravitation formula.

Exam Tip: Frequently asked.

NEET MCQ 40

Question: Argument of sin θ must be

(a) dimensionless(b) length(c) time(d) mass

Correct Answer: dimensionless

Solution: Trigonometric functions require pure number argument.

Exam Tip: Always check function arguments.

NEET MCQ 41

Question: Dimension of force is

(a) [MLT-2](b) [ML2T-2](c) [LT-1](d) [M-1L3T-2]

Correct Answer: [MLT-2]

Solution: Force = mass × acceleration.

Exam Tip: Memorize base formulae.

NEET MCQ 42

Question: Dimension of energy is

(a) [ML2T-2](b) [MLT-2](c) [ML2T-3](d) [T-1]

Correct Answer: [ML2T-2]

Solution: Energy = force × displacement.

Exam Tip: Work and energy same dimensions.

NEET MCQ 43

Question: Which is dimensionless?

(a) Refractive index(b) Pressure(c) Charge(d) Resistance

Correct Answer: Refractive index

Solution: It is ratio of two speeds.

Exam Tip: Ratios of same quantities are dimensionless.

NEET MCQ 44

Question: 1 Å equals

(a) 10-10 m(b) 1010 m(c) 10-6 m(d) 10-3 m

Correct Answer: 10-10 m

Solution: Angstrom is atomic length unit.

Exam Tip: Useful for atomic scale.

NEET MCQ 45

Question: Charge dimension is

(a) [IT](b) [I/T](c) [MT](d) [MLT]

Correct Answer: [IT]

Solution: q = It.

Exam Tip: Current times time.

NEET MCQ 46

Question: If Z = A2, error in A is 2%, error in Z is

(a) 4%(b) 2%(c) 1%(d) 8%

Correct Answer: 4%

Solution: Power doubles percentage error.

Exam Tip: Power rule.

NEET MCQ 47

Question: Significant figures in 0.2370 are

(a) 4(b) 3(c) 2(d) 5

Correct Answer: 4

Solution: Trailing zero after decimal is significant.

Exam Tip: Decimal trailing zero counts.

NEET MCQ 48

Question: Principle of homogeneity checks

(a) dimensions of equation(b) value of constants(c) sign of terms(d) vector direction

Correct Answer: dimensions of equation

Solution: All terms must have same dimensions.

Exam Tip: It is a necessary test.

NEET MCQ 49

Question: Dimension of G is

(a) [M-1L3T-2](b) [MLT-2](c) [ML2T-1](d) [IT]

Correct Answer: [M-1L3T-2]

Solution: Use Newton gravitation formula.

Exam Tip: Frequently asked.

NEET MCQ 50

Question: Argument of sin θ must be

(a) dimensionless(b) length(c) time(d) mass

Correct Answer: dimensionless

Solution: Trigonometric functions require pure number argument.

Exam Tip: Always check function arguments.

NEET MCQ 51

Question: Dimension of force is

(a) [MLT-2](b) [ML2T-2](c) [LT-1](d) [M-1L3T-2]

Correct Answer: [MLT-2]

Solution: Force = mass × acceleration.

Exam Tip: Memorize base formulae.

NEET MCQ 52

Question: Dimension of energy is

(a) [ML2T-2](b) [MLT-2](c) [ML2T-3](d) [T-1]

Correct Answer: [ML2T-2]

Solution: Energy = force × displacement.

Exam Tip: Work and energy same dimensions.

NEET MCQ 53

Question: Which is dimensionless?

(a) Refractive index(b) Pressure(c) Charge(d) Resistance

Correct Answer: Refractive index

Solution: It is ratio of two speeds.

Exam Tip: Ratios of same quantities are dimensionless.

NEET MCQ 54

Question: 1 Å equals

(a) 10-10 m(b) 1010 m(c) 10-6 m(d) 10-3 m

Correct Answer: 10-10 m

Solution: Angstrom is atomic length unit.

Exam Tip: Useful for atomic scale.

NEET MCQ 55

Question: Charge dimension is

(a) [IT](b) [I/T](c) [MT](d) [MLT]

Correct Answer: [IT]

Solution: q = It.

Exam Tip: Current times time.

NEET MCQ 56

Question: If Z = A2, error in A is 2%, error in Z is

(a) 4%(b) 2%(c) 1%(d) 8%

Correct Answer: 4%

Solution: Power doubles percentage error.

Exam Tip: Power rule.

NEET MCQ 57

Question: Significant figures in 0.2370 are

(a) 4(b) 3(c) 2(d) 5

Correct Answer: 4

Solution: Trailing zero after decimal is significant.

Exam Tip: Decimal trailing zero counts.

NEET MCQ 58

Question: Principle of homogeneity checks

(a) dimensions of equation(b) value of constants(c) sign of terms(d) vector direction

Correct Answer: dimensions of equation

Solution: All terms must have same dimensions.

Exam Tip: It is a necessary test.

NEET MCQ 59

Question: Dimension of G is

(a) [M-1L3T-2](b) [MLT-2](c) [ML2T-1](d) [IT]

Correct Answer: [M-1L3T-2]

Solution: Use Newton gravitation formula.

Exam Tip: Frequently asked.

NEET MCQ 60

Question: Argument of sin θ must be

(a) dimensionless(b) length(c) time(d) mass

Correct Answer: dimensionless

Solution: Trigonometric functions require pure number argument.

Exam Tip: Always check function arguments.

13. JEE Main PYQs and Exam-style Questions

JEE Main Exam-style Question

Question: For y = A2B3/C, percentage errors in A, B, C are 1%, 2%, 3%. Find percentage error in y.

Solution: %y = 2(1) + 3(2) + 3 = 11%.

Final Answer: 11%

Exam Tip: All powers contribute positively.

JEE Main Exam-style Question

Question: Find dimension of Planck constant h using E = hν.

Solution: h = E/ν = [ML2T-2]/[T-1] = [ML2T-1].

Final Answer: [ML2T-1]

Exam Tip: Frequency has dimension [T-1].

JEE Main Exam-style Question

Question: Check dimensional correctness of T = 2π√(L/g).

Solution: [L/g] = [L]/[LT-2] = [T2], square root = [T].

Final Answer: Correct

Exam Tip: Constants are dimensionless.

JEE Main Exam-style Question

Question: If ekx appears and x is distance, find dimension of k.

Solution: kx must be dimensionless, so [k][L] = 1. Hence [k] = [L-1].

Final Answer: [L-1]

Exam Tip: Exponential argument must be dimensionless.

14. JEE Advanced Conceptual Questions

JEE Advanced Conceptual

Question: Can two quantities with same dimensions be physically different?

Solution: Yes. Work and torque both have [ML2T-2] but work is scalar and torque is vector.

Final Answer: Yes

Exam Tip: Dimensional equality is not full physical equality.

JEE Advanced Conceptual

Question: Why can dimensional analysis not derive v = u + at exactly including signs and constants?

Solution: It only checks dimensions. It cannot decide signs, numerical constants or number of terms.

Final Answer: Because dimensions are necessary but not sufficient.

Exam Tip: Know limitations.

JEE Advanced Conceptual

Question: For wave phase kx - ωt, find dimensions of k and ω.

Solution: kx and ωt must be dimensionless. Hence [k] = [L-1] and [ω] = [T-1].

Final Answer: k: [L-1], ω: [T-1]

Exam Tip: Phase is pure number.

JEE Advanced Conceptual

Question: Explain why log(P/P0) is valid but log P is not valid if P is pressure.

Solution: P/P0 is dimensionless; P alone has dimensions. Logarithm needs a pure number argument.

Final Answer: Only log(P/P0) is dimensionally valid.

Exam Tip: For log and exponential, inspect the argument.

15. IB, IGCSE and A-Level Questions

IB Physics 1

Question: How should uncertainty be shown on a graph?

Answer: By error bars.

IB Physics 2

Question: Why are repeated readings taken?

Answer: To reduce random uncertainty.

IB Physics 3

Question: What does 3.00 × 108 show?

Answer: Three significant figures.

IGCSE Physics 1

Question: Why avoid parallax error?

Answer: It gives wrong scale reading.

IGCSE Physics 2

Question: Define precision.

Answer: Closeness of repeated readings.

IGCSE Physics 3

Question: State SI unit of force.

Answer: newton, N.

A-Level Physics 1

Question: Find dimension of capacitance.

Answer: [M-1L-2T4I2]

A-Level Physics 2

Question: Find uncertainty in product IV.

Answer: Percentage uncertainties add.

A-Level Physics 3

Question: Why must arguments of sin be dimensionless?

Answer: Mathematical functions accept pure numbers.

16. Assertion Reason Questions

Options: (a) Both A and R are true and R explains A. (b) Both true but R does not explain A. (c) A true, R false. (d) A false, R true.

Assertion Reason 1

Assertion: SI base units are independent.

Reason: They define fundamental quantities.

Answer: (a)

Explanation: Both true and related.

Assertion Reason 2

Assertion: Dimensional analysis can find numerical constant 2π.

Reason: Numerical constants are dimensionless.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 3

Assertion: Leading zeros are significant.

Reason: Leading zeros only locate decimal point.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 4

Assertion: Relative error is dimensionless.

Reason: It is ratio of two quantities with same unit.

Answer: (a)

Explanation: Correct.

Assertion Reason 5

Assertion: sin θ requires θ dimensionless.

Reason: Trigonometric arguments are pure numbers.

Answer: (a)

Explanation: Correct.

Assertion Reason 6

Assertion: Work and torque have identical dimensions.

Reason: Same dimensions do not always mean same physical quantity.

Answer: (b)

Explanation: Both true, second is a caution.

Assertion Reason 7

Assertion: G has dimensions.

Reason: It appears in gravitational force law.

Answer: (a)

Explanation: Correct.

Assertion Reason 8

Assertion: A dimensionally wrong equation can be physically correct.

Reason: Physical equations must be homogeneous.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 9

Assertion: Trailing zeros after decimal are significant.

Reason: They indicate precision.

Answer: (a)

Explanation: Correct.

Assertion Reason 10

Assertion: In multiplication, percentage errors are added.

Reason: Relative uncertainties combine in products.

Answer: (a)

Explanation: Correct.

Assertion Reason 11

Assertion: SI base units are independent.

Reason: They define fundamental quantities.

Answer: (a)

Explanation: Both true and related.

Assertion Reason 12

Assertion: Dimensional analysis can find numerical constant 2π.

Reason: Numerical constants are dimensionless.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 13

Assertion: Leading zeros are significant.

Reason: Leading zeros only locate decimal point.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 14

Assertion: Relative error is dimensionless.

Reason: It is ratio of two quantities with same unit.

Answer: (a)

Explanation: Correct.

Assertion Reason 15

Assertion: sin θ requires θ dimensionless.

Reason: Trigonometric arguments are pure numbers.

Answer: (a)

Explanation: Correct.

Assertion Reason 16

Assertion: Work and torque have identical dimensions.

Reason: Same dimensions do not always mean same physical quantity.

Answer: (b)

Explanation: Both true, second is a caution.

Assertion Reason 17

Assertion: G has dimensions.

Reason: It appears in gravitational force law.

Answer: (a)

Explanation: Correct.

Assertion Reason 18

Assertion: A dimensionally wrong equation can be physically correct.

Reason: Physical equations must be homogeneous.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 19

Assertion: Trailing zeros after decimal are significant.

Reason: They indicate precision.

Answer: (a)

Explanation: Correct.

Assertion Reason 20

Assertion: In multiplication, percentage errors are added.

Reason: Relative uncertainties combine in products.

Answer: (a)

Explanation: Correct.

Assertion Reason 21

Assertion: SI base units are independent.

Reason: They define fundamental quantities.

Answer: (a)

Explanation: Both true and related.

Assertion Reason 22

Assertion: Dimensional analysis can find numerical constant 2π.

Reason: Numerical constants are dimensionless.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 23

Assertion: Leading zeros are significant.

Reason: Leading zeros only locate decimal point.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 24

Assertion: Relative error is dimensionless.

Reason: It is ratio of two quantities with same unit.

Answer: (a)

Explanation: Correct.

Assertion Reason 25

Assertion: sin θ requires θ dimensionless.

Reason: Trigonometric arguments are pure numbers.

Answer: (a)

Explanation: Correct.

Assertion Reason 26

Assertion: Work and torque have identical dimensions.

Reason: Same dimensions do not always mean same physical quantity.

Answer: (b)

Explanation: Both true, second is a caution.

Assertion Reason 27

Assertion: G has dimensions.

Reason: It appears in gravitational force law.

Answer: (a)

Explanation: Correct.

Assertion Reason 28

Assertion: A dimensionally wrong equation can be physically correct.

Reason: Physical equations must be homogeneous.

Answer: (d)

Explanation: Assertion false, reason true.

Assertion Reason 29

Assertion: Trailing zeros after decimal are significant.

Reason: They indicate precision.

Answer: (a)

Explanation: Correct.

Assertion Reason 30

Assertion: In multiplication, percentage errors are added.

Reason: Relative uncertainties combine in products.

Answer: (a)

Explanation: Correct.

17. Case Study Questions

Case Study: SI Units and Unit Conversion

A student converts speed, density and energy between SI and CGS units for a lab report.

Questions: Why use SI?; Convert 1 g cm-3; What is dimensional check?; Why powers matter?

Answers: SI is standard; 1000 kg m-3; compare dimensions; squared/cubed units change conversion factors.

Explanation: The passage connects formula revision with practical exam application.

Case Study: Significant Figures

A calculator gives 12.345678, but measured data have only three significant figures.

Questions: How many digits should be reported?; Why?; Which rule applies in multiplication?; What mistake is common?

Answers: 3 SF; precision is limited; least SF; writing too many calculator digits.

Explanation: The passage connects formula revision with practical exam application.

Case Study: Errors in Measurement

A length is 20.0 ± 0.2 cm and breadth is 10.0 ± 0.1 cm.

Questions: Find percentage errors; area error; why add?; final idea?

Answers: 1%, 1%; 2%; product rule; area uncertainty is 2%.

Explanation: The passage connects formula revision with practical exam application.

Case Study: Dimensional Analysis

A formula is proposed for pendulum time period.

Questions: What variables?; Why assume powers?; What can dimensions find?; What can it not find?

Answers: L and g; to compare M,L,T powers; dependence √(L/g); constant 2π.

Explanation: The passage connects formula revision with practical exam application.

Case Study: Principle of Homogeneity

A student writes s = ut + at.

Questions: Is it correct?; Dimension of ut?; Dimension of at?; Correct term?

Answers: No; [L]; [LT-1]; 1/2 at2.

Explanation: The passage connects formula revision with practical exam application.

Case Study: Dimensional Formulae in Mechanics and Electricity

A learner compares force, field, potential and resistance dimensions.

Questions: Dimension of force?; charge?; electric field?; resistance?

Answers: [MLT-2]; [IT]; [MLT-3I-1]; [ML2T-3I-2].

Explanation: The passage connects formula revision with practical exam application.

18. Quick Revision Notes

One-page Formula Sheet

  • Force: [MLT-2]
  • Energy: [ML2T-2]
  • Power: [ML2T-3]
  • Pressure: [ML-1T-2]

Most Important Tricks

  • All added terms must match dimensions
  • Constants cannot be found dimensionally
  • Arguments of sin, cos, ex, log x must be dimensionless
  • Products add percentage errors

Common Mistakes

  • Forgetting powers in unit conversion
  • Confusing units and dimensions
  • Writing too many significant figures
  • Subtracting errors in division
NEET/JEE focus: dimensions of constants, NCERT unit conversions, significant figures, error propagation and homogeneity checks.

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