Chapter 2: Units and Measurement

Units and Systems

Fundamentals of Units

• Unit: Internationally accepted standard for measurement.
• Measurement = Number + Unit.
• Base units: Fundamental quantities (length, mass, time, etc.).
• Derived units: Combination of base units.
• System of units: Complete set of base and derived units.

Systems of Units

• CGS: Centimetre, gram, second (metric).
• FPS: Foot, pound, second (British system, non-metric).
• MKS: Metre, kilogram, second (metric).
• SI (Systeme Internationale): Internationally accepted, based on 7 base units and 2 supplementary units.

Seven SI Base Units

• Length: Metre (m) - distance light travels in (1/299,792,458) seconds.
• Mass: Kilogram (kg) - defined by Planck constant.
• Time: Second (s) - based on cesium atom vibrations.
• Electric current: Ampere (A) - flow of one coulomb per second.
• Temperature: Kelvin (K) - based on Boltzmann constant.
• Amount of substance: Mole (mol) - number of specified entities.
• Luminous intensity: Candela (cd) - light intensity in a given direction.

Supplementary Units

• Plane angle (radian, rad): dθ = ds/r
• Solid angle (steradian, sr): dΩ = dA/r²

• Radian: Angle subtended by arc equal to radius.
• Steradian: Solid angle subtended by area equal to square of radius.

Advantages of SI Units

• Coherent system: Units combine without extra factors.
• Rational system: Based on rational definitions.
• Absolute system: Based on universal constants.
• Metric system: Decimal-based units.

Solved Examples

Ex 1

• Find plane angle in radians if arc length ds = 5 m and radius r = 10 m.
• Solution: dθ = ds/r = 5/10 = 0.5 rad.

Practice Set

• Q1: Which is the base unit of electric current in SI?
  • A) Volt
  • B) Ampere
  • C) Ohm
  • D) Coulomb
• Q2: What is the SI unit of solid angle?
  • A) Radian
  • B) Steradian
  • C) Degree
  • D) Grad
• Q3: If arc length is 3 m and radius is 6 m, what is the plane angle in radians?
  • A) 0.5
  • B) 2
  • C) 18
  • D) 9

Answer Key

• Q1: B - Ampere is the SI base unit of electric current.
• Q2: B - Steradian is the SI unit of solid angle.
• Q3: A - Plane angle = 3/6 = 0.5 radians.

Memory Aids

• SI base units: MKS + A (Ampere), K (Kelvin), mol, cd.
• Radian = arc length / radius.
• Steradian = area / radius².
• SI advantages: Coherent, Rational, Absolute, Metric (CRAM).

Fundamentals of Units

• Unit: Internationally accepted standard for measurement.
• Measurement = Number + Unit.
• Base units: Fundamental quantities (length, mass, time, etc.).
• Derived units: Combination of base units.
• System of units: Complete set of base and derived units.

Systems of Units

• CGS: Centimetre, gram, second (metric).
• FPS: Foot, pound, second (British system, non-metric).
• MKS: Metre, kilogram, second (metric).
• SI (Systeme Internationale): Internationally accepted, based on 7 base units and 2 supplementary units.

Seven SI Base Units

• Length: Metre (m) - distance light travels in (1/299,792,458) seconds.
• Mass: Kilogram (kg) - defined by Planck constant.
• Time: Second (s) - based on cesium atom vibrations.
• Electric current: Ampere (A) - flow of one coulomb per second.
• Temperature: Kelvin (K) - based on Boltzmann constant.
• Amount of substance: Mole (mol) - number of specified entities.
• Luminous intensity: Candela (cd) - light intensity in a given direction.

Supplementary Units

• Plane angle (radian, rad): dθ = ds/r
• Solid angle (steradian, sr): dΩ = dA/r²

• Radian: Angle subtended by arc equal to radius.
• Steradian: Solid angle subtended by area equal to square of radius.

Advantages of SI Units

• Coherent system: Units combine without extra factors.
• Rational system: Based on rational definitions.
• Absolute system: Based on universal constants.
• Metric system: Decimal-based units.

Solved Examples

Ex 1

• Find plane angle in radians if arc length ds = 5 m and radius r = 10 m.
• Solution: dθ = ds/r = 5/10 = 0.5 rad.

Practice Set

• Q1: Which is the base unit of electric current in SI?
  • A) Volt
  • B) Ampere
  • C) Ohm
  • D) Coulomb
• Q2: What is the SI unit of solid angle?
  • A) Radian
  • B) Steradian
  • C) Degree
  • D) Grad
• Q3: If arc length is 3 m and radius is 6 m, what is the plane angle in radians?
  • A) 0.5
  • B) 2
  • C) 18
  • D) 9

Answer Key

• Q1: B - Ampere is the SI base unit of electric current.
• Q2: B - Steradian is the SI unit of solid angle.
• Q3: A - Plane angle = 3/6 = 0.5 radians.

Memory Aids

• SI base units: MKS + A (Ampere), K (Kelvin), mol, cd.
• Radian = arc length / radius.
• Steradian = area / radius².
• SI advantages: Coherent, Rational, Absolute, Metric (CRAM).

Fundamentals of Units

• Unit: Internationally accepted standard for measurement.
• Measurement = Number + Unit.
• Base units: Fundamental quantities (length, mass, time, etc.).
• Derived units: Combination of base units.
• System of units: Complete set of base and derived units.

Systems of Units

• CGS: Centimetre, gram, second (metric).
• FPS: Foot, pound, second (British system, non-metric).
• MKS: Metre, kilogram, second (metric).
• SI (Systeme Internationale): Internationally accepted, based on 7 base units and 2 supplementary units.

Seven SI Base Units

• Length: Metre (m) - distance light travels in (1/299,792,458) seconds.
• Mass: Kilogram (kg) - defined by Planck constant.
• Time: Second (s) - based on cesium atom vibrations.
• Electric current: Ampere (A) - flow of one coulomb per second.
• Temperature: Kelvin (K) - based on Boltzmann constant.
• Amount of substance: Mole (mol) - number of specified entities.
• Luminous intensity: Candela (cd) - light intensity in a given direction.

Supplementary Units

• Plane angle (radian, rad): dθ = ds/r
• Solid angle (steradian, sr): dΩ = dA/r²

• Radian: Angle subtended by arc equal to radius.
• Steradian: Solid angle subtended by area equal to square of radius.

Advantages of SI Units

• Coherent system: Units combine without extra factors.
• Rational system: Based on rational definitions.
• Absolute system: Based on universal constants.
• Metric system: Decimal-based units.

Solved Examples

Ex 1

• Find plane angle in radians if arc length ds = 5 m and radius r = 10 m.
• Solution: dθ = ds/r = 5/10 = 0.5 rad.

Practice Set

• Q1: Which is the base unit of electric current in SI?
  • A) Volt
  • B) Ampere
  • C) Ohm
  • D) Coulomb
• Q2: What is the SI unit of solid angle?
  • A) Radian
  • B) Steradian
  • C) Degree
  • D) Grad
• Q3: If arc length is 3 m and radius is 6 m, what is the plane angle in radians?
  • A) 0.5
  • B) 2
  • C) 18
  • D) 9

Answer Key

• Q1: B - Ampere is the SI base unit of electric current.
• Q2: B - Steradian is the SI unit of solid angle.
• Q3: A - Plane angle = 3/6 = 0.5 radians.

Memory Aids

• SI base units: MKS + A (Ampere), K (Kelvin), mol, cd.
• Radian = arc length / radius.
• Steradian = area / radius².
• SI advantages: Coherent, Rational, Absolute, Metric (CRAM).

Fundamentals of Units

• Unit: Internationally accepted standard for measurement.
• Measurement = Number + Unit.
• Base units: Fundamental quantities (length, mass, time, etc.).
• Derived units: Combination of base units.
• System of units: Complete set of base and derived units.

Systems of Units

• CGS: Centimetre, gram, second (metric).
• FPS: Foot, pound, second (British system, non-metric).
• MKS: Metre, kilogram, second (metric).
• SI (Systeme Internationale): Internationally accepted, based on 7 base units and 2 supplementary units.

Seven SI Base Units

• Length: Metre (m) - distance light travels in (1/299,792,458) seconds.
• Mass: Kilogram (kg) - defined by Planck constant.
• Time: Second (s) - based on cesium atom vibrations.
• Electric current: Ampere (A) - flow of one coulomb per second.
• Temperature: Kelvin (K) - based on Boltzmann constant.
• Amount of substance: Mole (mol) - number of specified entities.
• Luminous intensity: Candela (cd) - light intensity in a given direction.

Supplementary Units

• Plane angle (radian, rad): dθ = ds/r
• Solid angle (steradian, sr): dΩ = dA/r²

• Radian: Angle subtended by arc equal to radius.
• Steradian: Solid angle subtended by area equal to square of radius.

Advantages of SI Units

• Coherent system: Units combine without extra factors.
• Rational system: Based on rational definitions.
• Absolute system: Based on universal constants.
• Metric system: Decimal-based units.

Solved Examples

Ex 1

• Find plane angle in radians if arc length ds = 5 m and radius r = 10 m.
• Solution: dθ = ds/r = 5/10 = 0.5 rad.

Practice Set

• Q1: Which is the base unit of electric current in SI?
  • A) Volt
  • B) Ampere
  • C) Ohm
  • D) Coulomb
• Q2: What is the SI unit of solid angle?
  • A) Radian
  • B) Steradian
  • C) Degree
  • D) Grad
• Q3: If arc length is 3 m and radius is 6 m, what is the plane angle in radians?
  • A) 0.5
  • B) 2
  • C) 18
  • D) 9

Answer Key

• Q1: B - Ampere is the SI base unit of electric current.
• Q2: B - Steradian is the SI unit of solid angle.
• Q3: A - Plane angle = 3/6 = 0.5 radians.

Memory Aids

• SI base units: MKS + A (Ampere), K (Kelvin), mol, cd.
• Radian = arc length / radius.
• Steradian = area / radius².
• SI advantages: Coherent, Rational, Absolute, Metric (CRAM).

Errors in Measurement

Types of Errors

• Systematic errors: Consistent bias in one direction.
• Random errors: Irregular fluctuations, unpredictable.
• Least count error: Due to instrument resolution limit.

Rules for SI Units

• Unit names not capitalized even if named after a person (e.g., newton, ampere).
• Symbols capitalized if named after a person (e.g., N for newton, A for ampere).
• No punctuation after unit symbols (e.g., kg m s^-2, not kg. m.s^-2).

Error Calculations

• Absolute error: |Δa_i| = |a_i - a_mean|
• Mean absolute error: Δa_mean = ∑ |Δa_i|/n
• Relative error: Δa_mean/a_mean
• Percentage error: δ_a = Δa_mean/a_mean × 100%

Combination of Errors

• Sum/Difference: ΔZ = ΔA + ΔB
• Product/Quotient: ΔZ/Z = ΔA/A + ΔB/B
• Power: ΔZ/Z = n ΔA/A if Z = Aⁿ
• General: Δy/y = l ΔA/A + m ΔB/B + n ΔC/C for y = K A^lB^m/Cⁿ

Solved Examples

Ex 1

• Given measurements: 10.2, 10.4, 10.3 cm. Find mean and mean absolute error.
• Mean: (10.2 + 10.4 + 10.3)/3 = 10.3 cm.
• Absolute errors: |10.2 - 10.3|=0.1, |10.4 - 10.3|=0.1, |10.3 - 10.3|=0.
• Mean absolute error: (0.1 + 0.1 + 0)/3 = 0.067 cm.

Practice Set

• Q1: What type of error is caused by a miscalibrated instrument?
  • A) Random error
  • B) Systematic error
  • C) Least count error
  • D) Human error
• Q2: If mean absolute error is 0.05 and mean value is 2.0, what is the relative error?
  • A) 0.025
  • B) 0.05
  • C) 0.1
  • D) 0.5
• Q3: For Z = A³, if relative error in A is 2%, what is relative error in Z?
  • A) 2%
  • B) 3%
  • C) 6%
  • D) 8%

Answer Key

• Q1: B - Systematic errors cause consistent bias like miscalibration.
• Q2: A - Relative error = 0.05 / 2.0 = 0.025.
• Q3: C - Relative error in Z = 3 × 2% = 6%.

Memory Aids

• Systematic = same direction error.
• Random = unpredictable fluctuations.
• Least count = instrument resolution limit.
• Errors add in sums, add relative errors in products.
• Power rule: multiply relative error by power.

Types of Errors

• Systematic errors: Consistent bias in one direction.
• Random errors: Irregular fluctuations, unpredictable.
• Least count error: Due to instrument resolution limit.

Rules for SI Units

• Unit names not capitalized even if named after a person (e.g., newton, ampere).
• Symbols capitalized if named after a person (e.g., N for newton, A for ampere).
• No punctuation after unit symbols (e.g., kg m s^-2, not kg. m.s^-2).

Error Calculations

• Absolute error: |Δa_i| = |a_i - a_mean|
• Mean absolute error: Δa_mean = ∑ |Δa_i|/n
• Relative error: Δa_mean/a_mean
• Percentage error: δ_a = Δa_mean/a_mean × 100%

Combination of Errors

• Sum/Difference: ΔZ = ΔA + ΔB
• Product/Quotient: ΔZ/Z = ΔA/A + ΔB/B
• Power: ΔZ/Z = n ΔA/A if Z = Aⁿ
• General: Δy/y = l ΔA/A + m ΔB/B + n ΔC/C for y = K A^lB^m/Cⁿ

Solved Examples

Ex 1

• Given measurements: 10.2, 10.4, 10.3 cm. Find mean and mean absolute error.
• Mean: (10.2 + 10.4 + 10.3)/3 = 10.3 cm.
• Absolute errors: |10.2 - 10.3|=0.1, |10.4 - 10.3|=0.1, |10.3 - 10.3|=0.
• Mean absolute error: (0.1 + 0.1 + 0)/3 = 0.067 cm.

Practice Set

• Q1: What type of error is caused by a miscalibrated instrument?
  • A) Random error
  • B) Systematic error
  • C) Least count error
  • D) Human error
• Q2: If mean absolute error is 0.05 and mean value is 2.0, what is the relative error?
  • A) 0.025
  • B) 0.05
  • C) 0.1
  • D) 0.5
• Q3: For Z = A³, if relative error in A is 2%, what is relative error in Z?
  • A) 2%
  • B) 3%
  • C) 6%
  • D) 8%

Answer Key

• Q1: B - Systematic errors cause consistent bias like miscalibration.
• Q2: A - Relative error = 0.05 / 2.0 = 0.025.
• Q3: C - Relative error in Z = 3 × 2% = 6%.

Memory Aids

• Systematic = same direction error.
• Random = unpredictable fluctuations.
• Least count = instrument resolution limit.
• Errors add in sums, add relative errors in products.
• Power rule: multiply relative error by power.

Types of Errors

• Systematic errors: Consistent bias in one direction.
• Random errors: Irregular fluctuations, unpredictable.
• Least count error: Due to instrument resolution limit.

Rules for SI Units

• Unit names not capitalized even if named after a person (e.g., newton, ampere).
• Symbols capitalized if named after a person (e.g., N for newton, A for ampere).
• No punctuation after unit symbols (e.g., kg m s^-2, not kg. m.s^-2).

Error Calculations

• Absolute error: |Δa_i| = |a_i - a_mean|
• Mean absolute error: Δa_mean = ∑ |Δa_i|/n
• Relative error: Δa_mean/a_mean
• Percentage error: δ_a = Δa_mean/a_mean × 100%

Combination of Errors

• Sum/Difference: ΔZ = ΔA + ΔB
• Product/Quotient: ΔZ/Z = ΔA/A + ΔB/B
• Power: ΔZ/Z = n ΔA/A if Z = Aⁿ
• General: Δy/y = l ΔA/A + m ΔB/B + n ΔC/C for y = K A^lB^m/Cⁿ

Solved Examples

Ex 1

• Given measurements: 10.2, 10.4, 10.3 cm. Find mean and mean absolute error.
• Mean: (10.2 + 10.4 + 10.3)/3 = 10.3 cm.
• Absolute errors: |10.2 - 10.3|=0.1, |10.4 - 10.3|=0.1, |10.3 - 10.3|=0.
• Mean absolute error: (0.1 + 0.1 + 0)/3 = 0.067 cm.

Practice Set

• Q1: What type of error is caused by a miscalibrated instrument?
  • A) Random error
  • B) Systematic error
  • C) Least count error
  • D) Human error
• Q2: If mean absolute error is 0.05 and mean value is 2.0, what is the relative error?
  • A) 0.025
  • B) 0.05
  • C) 0.1
  • D) 0.5
• Q3: For Z = A³, if relative error in A is 2%, what is relative error in Z?
  • A) 2%
  • B) 3%
  • C) 6%
  • D) 8%

Answer Key

• Q1: B - Systematic errors cause consistent bias like miscalibration.
• Q2: A - Relative error = 0.05 / 2.0 = 0.025.
• Q3: C - Relative error in Z = 3 × 2% = 6%.

Memory Aids

• Systematic = same direction error.
• Random = unpredictable fluctuations.
• Least count = instrument resolution limit.
• Errors add in sums, add relative errors in products.
• Power rule: multiply relative error by power.

Significant Figures

Definition and Importance

• Digits accurately known in a measurement are significant figures.
• More significant figures mean higher measurement accuracy.

Rules for Counting Significant Figures

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros without decimal point are not significant.
• Powers of 10 are not significant figures.
• Trailing zeros in decimal numbers are significant.

Arithmetic Operations with Significant Figures

• Addition/Subtraction: Result has decimal places equal to the least precise measurement.
• Multiplication/Division: Result has significant figures equal to the number with least significant figures.

Rounding Off Rules

• If the digit to drop > 5, increase preceding digit by 1.
• If the digit to drop < 5, leave preceding digit unchanged.
• If digit to drop = 5 and preceding digit even, drop 5.
• If digit to drop = 5 and preceding digit odd, increase preceding digit by 1.

Solved Examples

• Ex 1: Add 734.52 g + 341.4 g + 0.402 g = 1076.3 g (rounded to 1 decimal place)
• Ex 2: Multiply 4.237 g / 2.51 cm³ = 1.69 g/cm³ (2 significant figures)

Practice Set

• Q1: How many significant figures are in 0.00450?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: What is the result of 12.11 + 0.023 + 1.1 rounded correctly?
  A. 13.233
  B. 13.2
  C. 14.2
  D. 13.23
• Q3: Multiply 3.00 × 10⁸ m/s by 3.156 × 10⁷ s. How many significant figures should the result have?
  A. 2
  B. 3
  C. 4
  D. 5

Answer Key

• Q1: B - Leading zeros not significant, 0.00450 has 3 significant figures.
• Q2: B - Least decimal places is 1, so round to 1 decimal place: 13.2.
• Q3: A - Least significant figures in factors is 2, so result has 2 significant figures.

Memory Aids

• Non-zero digits always count.
• Zeros between non-zero digits count.
• Leading zeros never count.
• Trailing zeros count only if decimal present.
• Add/Subtract: Match decimal places.
• Multiply/Divide: Match significant figures.

Definition and Importance

• Digits accurately known in a measurement are significant figures.
• More significant figures mean higher measurement accuracy.

Rules for Counting Significant Figures

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros without decimal point are not significant.
• Powers of 10 are not significant figures.
• Trailing zeros in decimal numbers are significant.

Arithmetic Operations with Significant Figures

• Addition/Subtraction: Result has decimal places equal to the least precise measurement.
• Multiplication/Division: Result has significant figures equal to the number with least significant figures.

Rounding Off Rules

• If the digit to drop > 5, increase preceding digit by 1.
• If the digit to drop < 5, leave preceding digit unchanged.
• If digit to drop = 5 and preceding digit even, drop 5.
• If digit to drop = 5 and preceding digit odd, increase preceding digit by 1.

Solved Examples

• Ex 1: Add 734.52 g + 341.4 g + 0.402 g = 1076.3 g (rounded to 1 decimal place)
• Ex 2: Multiply 4.237 g / 2.51 cm³ = 1.69 g/cm³ (2 significant figures)

Practice Set

• Q1: How many significant figures are in 0.00450?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: What is the result of 12.11 + 0.023 + 1.1 rounded correctly?
  A. 13.233
  B. 13.2
  C. 14.2
  D. 13.23
• Q3: Multiply 3.00 × 10⁸ m/s by 3.156 × 10⁷ s. How many significant figures should the result have?
  A. 2
  B. 3
  C. 4
  D. 5

Answer Key

• Q1: B - Leading zeros not significant, 0.00450 has 3 significant figures.
• Q2: B - Least decimal places is 1, so round to 1 decimal place: 13.2.
• Q3: A - Least significant figures in factors is 2, so result has 2 significant figures.

Memory Aids

• Non-zero digits always count.
• Zeros between non-zero digits count.
• Leading zeros never count.
• Trailing zeros count only if decimal present.
• Add/Subtract: Match decimal places.
• Multiply/Divide: Match significant figures.

Definition and Importance

• Digits accurately known in a measurement are significant figures.
• More significant figures mean higher measurement accuracy.

Rules for Counting Significant Figures

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros without decimal point are not significant.
• Powers of 10 are not significant figures.
• Trailing zeros in decimal numbers are significant.

Arithmetic Operations with Significant Figures

• Addition/Subtraction: Result has decimal places equal to the least precise measurement.
• Multiplication/Division: Result has significant figures equal to the number with least significant figures.

Rounding Off Rules

• If the digit to drop > 5, increase preceding digit by 1.
• If the digit to drop < 5, leave preceding digit unchanged.
• If digit to drop = 5 and preceding digit even, drop 5.
• If digit to drop = 5 and preceding digit odd, increase preceding digit by 1.

Solved Examples

• Ex 1: Add 734.52 g + 341.4 g + 0.402 g = 1076.3 g (rounded to 1 decimal place)
• Ex 2: Multiply 4.237 g / 2.51 cm³ = 1.69 g/cm³ (2 significant figures)

Practice Set

• Q1: How many significant figures are in 0.00450?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: What is the result of 12.11 + 0.023 + 1.1 rounded correctly?
  A. 13.233
  B. 13.2
  C. 14.2
  D. 13.23
• Q3: Multiply 3.00 × 10⁸ m/s by 3.156 × 10⁷ s. How many significant figures should the result have?
  A. 2
  B. 3
  C. 4
  D. 5

Answer Key

• Q1: B - Leading zeros not significant, 0.00450 has 3 significant figures.
• Q2: B - Least decimal places is 1, so round to 1 decimal place: 13.2.
• Q3: A - Least significant figures in factors is 2, so result has 2 significant figures.

Memory Aids

• Non-zero digits always count.
• Zeros between non-zero digits count.
• Leading zeros never count.
• Trailing zeros count only if decimal present.
• Add/Subtract: Match decimal places.
• Multiply/Divide: Match significant figures.

Least Count

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = L.C = (b - a)/b × M
• Formula derivation: M - V = (b - a)/b × M

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: L.C = (10-9)/10 × 1 = 0.1 mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: L.C = 0.5/50 = 0.01 mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - L.C = (10-9)/10 × 1 = 0.1 mm
• Q2: A - L.C = 0.5/50 = 0.01 mm
• Q3: C - L.C = 1/100 = 0.01 mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = L.C = (b - a)/b × M
• Formula derivation: M - V = (b - a)/b × M

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: L.C = (10-9)/10 × 1 = 0.1 mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: L.C = 0.5/50 = 0.01 mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - L.C = (10-9)/10 × 1 = 0.1 mm
• Q2: A - L.C = 0.5/50 = 0.01 mm
• Q3: C - L.C = 1/100 = 0.01 mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = L.C = (b - a)/b × M
• Formula derivation: M - V = (b - a)/b × M

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: L.C = (10-9)/10 × 1 = 0.1 mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: L.C = 0.5/50 = 0.01 mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - L.C = (10-9)/10 × 1 = 0.1 mm
• Q2: A - L.C = 0.5/50 = 0.01 mm
• Q3: C - L.C = 1/100 = 0.01 mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Dimensions of Physical Quantities

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Dimensional Formulae and Dimensional Equations

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Dimensional Analysis

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

• Ex 1: Add 734.52 g + 341.4 g + 0.402 g = 1076.3 g (rounded to 1 decimal place)
• Ex 2: Multiply 4.237 g / 2.51 cm³ = 1.69 g/cm³ (2 significant figures)

Ex 1

• Given measurements: 10.2, 10.4, 10.3 cm. Find mean and mean absolute error.
• Mean: (10.2 + 10.4 + 10.3)/3 = 10.3 cm.
• Absolute errors: |10.2 - 10.3|=0.1, |10.4 - 10.3|=0.1, |10.3 - 10.3|=0.
• Mean absolute error: (0.1 + 0.1 + 0)/3 = 0.067 cm.

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Leading zeros not significant, 0.00450 has 3 significant figures.
• Q2: B - Least decimal places is 1, so round to 1 decimal place: 13.2.
• Q3: A - Least significant figures in factors is 2, so result has 2 significant figures.

Memory Aids

• Systematic = same direction error.
• Random = unpredictable fluctuations.
• Least count = instrument resolution limit.
• Errors add in sums, add relative errors in products.
• Power rule: multiply relative error by power.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Dimensional Consistency

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Unit Conversion

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Deducing Relations

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = K m^a l^b g^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K √(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = K m^a l^b g^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K √(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = K m^a l^b g^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K √(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = K m^a l^b g^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K √(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = K m^a l^b g^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K √(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = K m^a l^b g^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K √(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = K m^a l^b g^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K √(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = K m^a l^b g^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K √(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = K m^a l^b g^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K √(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = K m^a l^b g^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K √(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = K m^a l^b g^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K √(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = K m^a l^b g^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K √(l/g).
• Mass does not affect pendulum time period.

Limitations

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K √(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K √(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K √(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K √(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K √(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K √(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Fundamentals of Units

• Unit: Internationally accepted standard for measurement.
• Measurement = Number + Unit.
• Base units: Fundamental quantities (length, mass, time, etc.).
• Derived units: Combination of base units.
• System of units: Complete set of base and derived units.

Systems of Units

• CGS: Centimetre, gram, second (metric).
• FPS: Foot, pound, second (British system, non-metric).
• MKS: Metre, kilogram, second (metric).
• SI (Systeme Internationale): Internationally accepted, based on 7 base units and 2 supplementary units.

Seven SI Base Units

• Length: Metre (m) - distance light travels in 1/299,792,458 seconds.
• Mass: Kilogram (kg) - defined by Planck constant.
• Time: Second (s) - based on cesium atom vibrations.
• Electric current: Ampere (A) - flow of one coulomb per second.
• Temperature: Kelvin (K) - based on Boltzmann constant.
• Amount of substance: Mole (mol) - number of specified entities.
• Luminous intensity: Candela (cd) - light intensity in a given direction.

Supplementary Units

• Plane angle (radian, rad): dθ = ds/r
• Solid angle (steradian, sr): dΩ = dA/r²

• Radian: Angle subtended by arc equal to radius.
• Steradian: Solid angle subtended by area equal to square of radius.

Advantages of SI Units

• Coherent system: Units combine without extra factors.
• Rational system: Based on rational definitions.
• Absolute system: Based on universal constants.
• Metric system: Decimal-based units.

Solved Examples

Ex 1

• Find plane angle in radians if arc length \( ds = 5 \) m and radius \( r = 10 \) m.
• Solution: \( d\theta = \frac{ds}{r} = \frac{5}{10} = 0.5 \) rad.

Practice Set

• Q1: Which is the base unit of electric current in SI?
  • A) Volt
  • B) Ampere
  • C) Ohm
  • D) Coulomb
• Q2: What is the SI unit of solid angle?
  • A) Radian
  • B) Steradian
  • C) Degree
  • D) Grad
• Q3: If arc length is 3 m and radius is 6 m, what is the plane angle in radians?
  • A) 0.5
  • B) 2
  • C) 18
  • D) 9

Answer Key

• Q1: B - Ampere is the SI base unit of electric current.
• Q2: B - Steradian is the SI unit of solid angle.
• Q3: A - Plane angle \( = \frac{3}{6} = 0.5 \) radians.

Memory Aids

• SI base units: MKS + A (Ampere), K (Kelvin), mol, cd.
• Radian = arc length / radius.
• Steradian = area / radius².
• SI advantages: Coherent, Rational, Absolute, Metric (CRAM).

Errors in Measurement

Types of Errors

• Systematic errors: Consistent bias in one direction.
• Random errors: Irregular fluctuations, unpredictable.
• Least count error: Due to instrument resolution limit.

Rules for SI Units

• Unit names not capitalized even if named after a person (e.g., newton, ampere).
• Symbols capitalized if named after a person (e.g., N for newton, A for ampere).
• No punctuation after unit symbols (e.g., kg m s^-2, not kg. m.s^-2).

Error Calculations

• Absolute error: |Δa_i| = |a_i - a_mean|
• Mean absolute error: Δa_mean = ∑|Δa_i|/n
• Relative error: Δa_mean/a_mean
• Percentage error: δ_a = Δa_mean/a_mean × 100%

Combination of Errors

• Sum/Difference: ΔZ = ΔA + ΔB
• Product/Quotient: ΔZ/Z = ΔA/A + ΔB/B
• Power: ΔZ/Z = n ΔA/A if Z = Aⁿ
• General: Δy/y = l ΔA/A + m ΔB/B + n ΔC/C for y = K A^lB^m/Cⁿ

Solved Examples

Ex 1

• Given measurements: 10.2, 10.4, 10.3 cm. Find mean and mean absolute error.
• Mean: (10.2 + 10.4 + 10.3)/3 = 10.3 cm.
• Absolute errors: |10.2 - 10.3|=0.1, |10.4 - 10.3|=0.1, |10.3 - 10.3|=0.
• Mean absolute error: (0.1 + 0.1 + 0)/3 = 0.067 cm.

Practice Set

• Q1: What type of error is caused by a miscalibrated instrument?
  • A) Random error
  • B) Systematic error
  • C) Least count error
  • D) Human error
• Q2: If mean absolute error is 0.05 and mean value is 2.0, what is the relative error?
  • A) 0.025
  • B) 0.05
  • C) 0.1
  • D) 0.5
• Q3: For Z = A³, if relative error in A is 2%, what is relative error in Z?
  • A) 2%
  • B) 3%
  • C) 6%
  • D) 8%

Answer Key

• Q1: B - Systematic errors cause consistent bias like miscalibration.
• Q2: A - Relative error = 0.05 / 2.0 = 0.025.
• Q3: C - Relative error in Z = 3 × 2% = 6%.

Memory Aids

• Systematic = same direction error.
• Random = unpredictable fluctuations.
• Least count = instrument resolution limit.
• Errors add in sums, add relative errors in products.
• Power rule: multiply relative error by power.

Types of Errors

• Systematic errors: Consistent bias in one direction.
• Random errors: Irregular fluctuations, unpredictable.
• Least count error: Due to instrument resolution limit.

Rules for SI Units

• Unit names not capitalized even if named after a person (e.g., newton, ampere).
• Symbols capitalized if named after a person (e.g., N for newton, A for ampere).
• No punctuation after unit symbols (e.g., kg m s^-2, not kg. m.s^-2).

Error Calculations

• Absolute error: \( |\Delta a_i| = |a_i - a_{mean}| \)
• Mean absolute error: \( \Delta a_{mean} = \frac{\sum |\Delta a_i|}{n} \)
• Relative error: \( \frac{\Delta a_{mean}}{a_{mean}} \)
• Percentage error: \( \delta_a = \frac{\Delta a_{mean}}{a_{mean}} \times 100\% \)

Combination of Errors

• Sum/Difference: \( \Delta Z = \Delta A + \Delta B \)
• Product/Quotient: \( \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \)
• Power: \( \frac{\Delta Z}{Z} = n \frac{\Delta A}{A} \) if \( Z = A^n \)
• General: \( \frac{\Delta y}{y} = l \frac{\Delta A}{A} + m \frac{\Delta B}{B} + n \frac{\Delta C}{C} \) for \( y = K \frac{A^l B^m}{C^n} \)

Solved Examples

Ex 1

• Given measurements: 10.2, 10.4, 10.3 cm. Find mean and mean absolute error.
• Mean: \( \frac{10.2 + 10.4 + 10.3}{3} = 10.3 \) cm.
• Absolute errors: |10.2 - 10.3|=0.1, |10.4 - 10.3|=0.1, |10.3 - 10.3|=0.
• Mean absolute error: \( \frac{0.1 + 0.1 + 0}{3} = 0.067 \) cm.

Practice Set

• Q1: What type of error is caused by a miscalibrated instrument?
  • A) Random error
  • B) Systematic error
  • C) Least count error
  • D) Human error
• Q2: If mean absolute error is 0.05 and mean value is 2.0, what is the relative error?
  • A) 0.025
  • B) 0.05
  • C) 0.1
  • D) 0.5
• Q3: For \( Z = A^3 \), if relative error in A is 2%, what is relative error in Z?
  • A) 2%
  • B) 3%
  • C) 6%
  • D) 8%

Answer Key

• Q1: B - Systematic errors cause consistent bias like miscalibration.
• Q2: A - Relative error = 0.05 / 2.0 = 0.025.
• Q3: C - Relative error in Z = 3 × 2% = 6%.

Memory Aids

• Systematic = same direction error.
• Random = unpredictable fluctuations.
• Least count = instrument resolution limit.
• Errors add in sums, add relative errors in products.
• Power rule: multiply relative error by power.

Significant Figures

Definition and Importance

• Digits accurately known in a measurement are significant figures.
• More significant figures mean higher measurement accuracy.

Rules for Counting Significant Figures

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros without decimal point are not significant.
• Powers of 10 are not significant figures.
• Trailing zeros in decimal numbers are significant.

Arithmetic Operations with Significant Figures

• Addition/Subtraction: Result has decimal places equal to the least precise measurement.
• Multiplication/Division: Result has significant figures equal to the number with least significant figures.

Rounding Off Rules

• If the digit to drop > 5, increase preceding digit by 1.
• If the digit to drop < 5, leave preceding digit unchanged.
• If digit to drop = 5 and preceding digit even, drop 5.
• If digit to drop = 5 and preceding digit odd, increase preceding digit by 1.

Solved Examples

• Ex 1: Add 734.52 g + 341.4 g + 0.402 g = 1076.3 g (rounded to 1 decimal place)
• Ex 2: Multiply 4.237 g / 2.51 cm³ = 1.69 g/cm³ (2 significant figures)

Practice Set

• Q1: How many significant figures are in 0.00450?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: What is the result of 12.11 + 0.023 + 1.1 rounded correctly?
  A. 13.233
  B. 13.2
  C. 14.2
  D. 13.23
• Q3: Multiply 3.00 × 10⁸ m/s by 3.156 × 10⁷ s. How many significant figures should the result have?
  A. 2
  B. 3
  C. 4
  D. 5

Answer Key

• Q1: B - Leading zeros not significant, 0.00450 has 3 significant figures.
• Q2: B - Least decimal places is 1, so round to 1 decimal place: 13.2.
• Q3: A - Least significant figures in factors is 2, so result has 2 significant figures.

Memory Aids

• Non-zero digits always count.
• Zeros between non-zero digits count.
• Leading zeros never count.
• Trailing zeros count only if decimal present.
• Add/Subtract: Match decimal places.
• Multiply/Divide: Match significant figures.

Definition and Importance

• Digits accurately known in a measurement are significant figures.
• More significant figures mean higher measurement accuracy.

Rules for Counting Significant Figures

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros without decimal point are not significant.
• Powers of 10 are not significant figures.
• Trailing zeros in decimal numbers are significant.

Arithmetic Operations with Significant Figures

• Addition/Subtraction: Result has decimal places equal to the least precise measurement.
• Multiplication/Division: Result has significant figures equal to the number with least significant figures.

Rounding Off Rules

• If the digit to drop > 5, increase preceding digit by 1.
• If the digit to drop < 5, leave preceding digit unchanged.
• If digit to drop = 5 and preceding digit even, drop 5.
• If digit to drop = 5 and preceding digit odd, increase preceding digit by 1.

Solved Examples

• Ex 1: Add 734.52 g + 341.4 g + 0.402 g = 1076.3 g (rounded to 1 decimal place)
• Ex 2: Multiply 4.237 g / 2.51 cm³ = 1.69 g/cm³ (2 significant figures)

Practice Set

• Q1: How many significant figures are in 0.00450?
  A. 2
  B. 3
  C. 4
  D. 5
• Q2: What is the result of 12.11 + 0.023 + 1.1 rounded correctly?
  A. 13.233
  B. 13.2
  C. 14.2
  D. 13.23
• Q3: Multiply 3.00 × 10⁸ m/s by 3.156 × 10⁷ s. How many significant figures should the result have?
  A. 2
  B. 3
  C. 4
  D. 5

Answer Key

• Q1: B - Leading zeros not significant, 0.00450 has 3 significant figures.
• Q2: B - Least decimal places is 1, so round to 1 decimal place: 13.2.
• Q3: A - Least significant figures in factors is 2, so result has 2 significant figures.

Memory Aids

• Non-zero digits always count.
• Zeros between non-zero digits count.
• Leading zeros never count.
• Trailing zeros count only if decimal present.
• Add/Subtract: Match decimal places.
• Multiply/Divide: Match significant figures.

Least Count

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = L.C = (b - a)/b × M
• Formula derivation: M - V = (b - a)/b × M

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: L.C = (10-9)/10 × 1 = 0.1 mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: L.C = 0.5/50 = 0.01 mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - L.C = (10-9)/10 × 1 = 0.1 mm
• Q2: A - L.C = 0.5/50 = 0.01 mm
• Q3: C - L.C = 1/100 = 0.01 mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = L.C = (b - a)/b × M
• Formula derivation: M - V = (b - a)/b × M

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: L.C = (10-9)/10 × 1 = 0.1 mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: L.C = 0.5/50 = 0.01 mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - L.C = (10-9)/10 × 1 = 0.1 mm
• Q2: A - L.C = 0.5/50 = 0.01 mm
• Q3: C - L.C = 1/100 = 0.01 mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = L.C = (b - a)/b × M
• Formula derivation: M - V = (b - a)/b × M

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: L.C = (10-9)/10 × 1 = 0.1 mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: L.C = 0.5/50 = 0.01 mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - L.C = (10-9)/10 × 1 = 0.1 mm
• Q2: A - L.C = 0.5/50 = 0.01 mm
• Q3: C - L.C = 1/100 = 0.01 mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Definition and Concept

• Smallest value measurable accurately by an instrument.
• Important for precision in measurements.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Vernier Constant (Least Count) = \( L.C = \frac{(b - a)}{b} \times M \)
• Formula derivation: \( M - V = (b - a)/b \times M \)

Least Count of Screw Gauge

• Least Count = Pitch / Number of divisions on circular scale.
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

• Ex 1: Vernier Callipers with M = 1 mm, a = 9, b = 10: \( L.C = \frac{(10-9)}{10} \times 1 = 0.1 \) mm
• Ex 2: Screw Gauge with pitch = 0.5 mm, 50 divisions: \( L.C = \frac{0.5}{50} = 0.01 \) mm

Practice Set

• Q1: If 10 main scale divisions equal 9 vernier divisions and M = 1 mm, what is the least count?
  A. 0.1 mm
  B. 0.09 mm
  C. 0.01 mm
  D. 1 mm
• Q2: A screw gauge has pitch 0.5 mm and 50 divisions on circular scale. What is its least count?
  A. 0.01 mm
  B. 0.1 mm
  C. 0.005 mm
  D. 0.05 mm
• Q3: If pitch is 1 mm and circular scale has 100 divisions, find least count.
  A. 0.01 mm
  B. 0.1 mm
  C. 0.001 mm
  D. 1 mm

Answer Key

• Q1: A - \( L.C = \frac{(10-9)}{10} \times 1 = 0.1 \) mm
• Q2: A - \( L.C = \frac{0.5}{50} = 0.01 \) mm
• Q3: C - \( L.C = \frac{1}{100} = 0.01 \) mm

Memory Aids

• Vernier L.C = (b - a)/b × M
• Screw Gauge L.C = Pitch / Number of divisions
• Pitch = distance per full rotation

Dimensions of Physical Quantities

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Concept and Definition

• Physical quantities are derived from base quantities: mass (M), length (L), time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = mass × acceleration = M × L × T^-2.
• Dimensions are expressed in square brackets, e.g., Force = [MLT^-2].

Dimension Examples

• Force: [MLT^-2] (1 in mass, 1 in length, -2 in time)
• Energy: [ML²T^-2] (1 in mass, 2 in length, -2 in time)

Solved Examples

• Ex 1: Find dimensions of force.
  • Force = mass × acceleration
  • Acceleration = length/time² = L T^-2
  • Force = M × L × T^-2 = [MLT^-2]
• Ex 2: Find dimensions of energy.
  • Energy = force × distance
  • Force = [MLT^-2], distance = L
  • Energy = [MLT^-2] × L = [ML²T^-2]

Practice Set

• Q1: What are the dimensions of velocity?
  • A. [LT^-1]
  • B. [MLT^-2]
  • C. [L²T^-1]
  • D. [MT^-1]
• Q2: The dimensions of power are:
  • A. [ML²T^-3]
  • B. [MLT^-2]
  • C. [M²L²T^-2]
  • D. [ML²T^-2]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  • A. [MLT^-2]
  • B. [ML²T^-2]
  • C. [M²LT^-2]
  • D. [ML²T^-1]

Answer Key

• Q1: A. [LT^-1] — Velocity = length/time.
• Q2: A. [ML²T^-3] — Power = energy/time = [ML²T^-2]/T.
• Q3: B. [ML²T^-2] — Work = force × displacement.

Memory Aids

• Force dimension: "M L T to the minus two" (M L T^-2).
• Energy dimension: "M L squared T to the minus two" (M L² T^-2).
• Velocity dimension: length/time = L T^-1.
• Power dimension: energy/time = M L² T^-3.

Dimensional Formulae and Dimensional Equations

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Concepts

• Dimensional Formula: Expression showing base quantities representing a physical quantity's dimension.
  Example: Volume = [M⁰L³T⁰], Velocity = [M⁰LT^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
  Example: Force [F] = [MLT^-2], Mass density [ρ] = [ML^-3T⁰]
• Dimensionless Quantities: Quantities without dimensions.
  Examples include:
  • Pure numbers
  • Ratios (e.g., density of substance/density of water)
  • Exponential functions (e.g., e^x)
  • Logarithmic functions (e.g., log x)
  • Trigonometric functions (e.g., sin x)

Solved Examples

Ex 1: Find the dimensional formula of speed.
Solution:
Speed = Distance/Time = L/T = [M⁰LT^-1]

Ex 2: Write the dimensional equation of force.
Solution:
Force = Mass × Acceleration = M × LT^-2 = [MLT^-2]

Practice Set

• Q1: What is the dimensional formula of volume?
  A) [M⁰L³T⁰] B) [MLT^-2] C) [M¹L²T^-1] D) [M⁰L²T¹]
• Q2: Which of the following is a dimensionless quantity?
  A) Force B) Pure number C) Velocity D) Mass density
• Q3: The dimensional formula of mass density is:
  A) [ML^-3T⁰] B) [MLT^-2] C) [M⁰L³T⁰] D) [M¹L¹T^-1]

Answer Key

• Q1: A) [M⁰L³T⁰]
  Volume is length cubed, no mass or time dimension.
• Q2: B) Pure number
  Pure numbers have no dimensions, unlike force, velocity, or density.
• Q3: A) [ML^-3T⁰]
  Mass density = Mass/Volume = M/L³, no time dimension.

Memory Aids

• Volume = L³, no M or T.
• Force = M × L × T^-2.
• Dimensionless = pure numbers, ratios, trig/log/exponential functions.

Dimensional Analysis

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use n₂ = n₁ (Q₁/Q₂) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done W = F × d

• Force F: [MLT^-2]
• Displacement d: [L]
• Work done W: [ML²T^-2]
• Check: [MLT^-2] × [L] = [ML²T^-2] ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. 10³
  • B. 10⁴
  • C. 10⁵
  • D. 10⁶
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = 10⁵ dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use n₂ = n₁ (Q₁/Q₂) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Key Concepts

• Check dimensional correctness: Dimensions on both sides of an equation must match for physical validity.
• Convert units: Physical quantity remains same; use \( n_2 = n_1 \left( \frac{Q_1}{Q_2} \right) \) to convert between unit systems.
• Find relations: Use dimensional homogeneity to derive formulas relating physical quantities.
• Find dimensions of constants: Use known dimensions of other quantities in the equation.

Solved Examples

Ex 1: Check dimensional consistency of work done \( W = F \times d \)

• Force \( F \): \([MLT^{-2}]\)
• Displacement \( d \): \([L]\)
• Work done \( W \): \([ML^2T^{-2}]\)
• Check: \([MLT^{-2}] \times [L] = [ML^2T^{-2}]\) ✓ consistent

Practice Set

• Q1: Which principle ensures dimensions on both sides of an equation are equal?
  • A. Conservation of energy
  • B. Principle of homogeneity
  • C. Newton's second law
  • D. Law of conservation of mass
• Q2: Convert 1 Newton (MKS) to dynes (CGS). What is the factor?
  • A. \(10^3\)
  • B. \(10^4\)
  • C. \(10^5\)
  • D. \(10^6\)
• Q3: Which of the following is NOT checked by dimensional analysis?
  • A. Dimensional consistency
  • B. Value of constants
  • C. Relation among quantities
  • D. Unit conversion

Answer Key

• Q1: B - Principle of homogeneity states dimensions must be same on both sides.
• Q2: C - 1 N = \(10^5\) dynes by converting kg to g and m to cm.
• Q3: B - Dimensional analysis cannot determine constant values.

Memory Aids

• Check dimensions match on both sides for correctness.
• Use \( n_2 = n_1 \left( \frac{Q_1}{Q_2} \right) \) for unit conversions.
• Constants' dimensions found by balancing equation dimensions.

Dimensional Consistency

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done W = F × d is dimensionally consistent.
• Example 2: Equation S = ut + 1/2at² has all terms with dimension [L].

Solved Examples

Ex 1: Verify S = ut + 1/2at²

• [S] = [L]
• [ut] = [L T^-1] × [T] = [L]
• [at²] = [L T^-2] × [T²] = [L]
• All terms have dimension [L] ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. F = ma
  • B. S = ut + at
  • C. V = u + at²
  • D. P = 2πr
• Q3: Dimensions of acceleration a are?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - F = ma is dimensionally consistent.
• Q3: B - Acceleration has dimension [L T^-2].

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Key Concepts

• Principle of homogeneity: Each term in an equation must have the same dimensions.
• Example 1: Work done \( W = F \times d \) is dimensionally consistent.
• Example 2: Equation \( S = ut + \frac{1}{2}at^2 \) has all terms with dimension \([L]\).

Solved Examples

Ex 1: Verify \( S = ut + \frac{1}{2}at^2 \)

• \( [S] = [L] \)
• \( [ut] = [L T^{-1}] \times [T] = [L] \)
• \( [at^2] = [L T^{-2}] \times [T^2] = [L] \)
• All terms have dimension \([L]\) ✓ consistent

Practice Set

• Q1: What must be true for an equation to be dimensionally consistent?
  • A. Dimensions of terms differ
  • B. Dimensions of terms are same
  • C. Only first term has dimension
  • D. Dimensions are ignored
• Q2: Which equation is dimensionally consistent?
  • A. \( F = ma \)
  • B. \( S = ut + at \)
  • C. \( V = u + at^2 \)
  • D. \( P = 2\pi r \)
• Q3: Dimensions of acceleration \( a \) are?
  • A. \( [L T^{-1}] \)
  • B. \( [L T^{-2}] \)
  • C. \( [M L T^{-2}] \)
  • D. \( [M L^2 T^{-2}] \)

Answer Key

• Q1: B - All terms must have same dimensions for consistency.
• Q2: A - \( F = ma \) is dimensionally consistent.
• Q3: B - Acceleration has dimension \( [L T^{-2}] \).

Memory Aids

• All terms in an equation must have identical dimensions.
• Acceleration dimension: length/time².
• Use dimensional check to verify formula correctness.

Unit Conversion

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: Q₁ n₁ = Q₂ n₂ where Q is unit, n is value.
• Convert value: n₂ = n₁ (Q₁/Q₂).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• n₂ = 1 × (Kg/g)¹ × (m/cm)¹ × (s/s)^-2
• = 1 × 1000 × 100 × 1 = 10⁵
• 1 N = 10⁵ dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. 10^-3
  • B. 10^-4
  • C. 10^-5
  • D. 10^-6
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = 10^-5 Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = 10⁵ dynes.
• Multiply or divide by powers of 10 for unit changes.

Key Concepts

• Formula: \( Q_1 n_1 = Q_2 n_2 \) where \( Q \) is unit, \( n \) is value.
• Convert value: \( n_2 = n_1 \left( \frac{Q_1}{Q_2} \right) \).
• Example: Convert 1 Newton (MKS) to dynes (CGS).
• Use mass, length, time conversion factors in powers of dimensions.

Solved Examples

Ex 1: Convert 1 N to dynes

• \( n_2 = 1 \times \left( \frac{Kg}{g} \right)^1 \times \left( \frac{m}{cm} \right)^1 \times \left( \frac{s}{s} \right)^{-2} \)
• \( = 1 \times 1000 \times 100 \times 1 = 10^5 \)
• \( 1 N = 10^5 \) dynes

Practice Set

• Q1: To convert 1 m to cm, multiply by?
  • A. 0.01
  • B. 10
  • C. 100
  • D. 1000
• Q2: 1 dyne equals how many Newtons?
  • A. \(10^{-3}\)
  • B. \(10^{-4}\)
  • C. \(10^{-5}\)
  • D. \(10^{-6}\)
• Q3: Which factor is used to convert seconds to milliseconds?
  • A. 100
  • B. 1000
  • C. 0.001
  • D. 10

Answer Key

• Q1: B - 1 m = 100 cm, so multiply by 100.
• Q2: C - 1 dyne = \(10^{-5}\) Newtons.
• Q3: B - 1 s = 1000 ms.

Memory Aids

• Use ratio of units raised to power of dimensions for conversion.
• 1 N = \(10^5\) dynes.
• Multiply or divide by powers of 10 for unit changes.

Deducing Relations

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = Km^al^bg^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K√(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = Km^al^bg^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K√(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = Km^al^bg^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K√(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = Km^al^bg^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K√(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = Km^al^bg^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K√(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = Km^al^bg^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K√(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = Km^al^bg^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K√(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = K m^a l^b g^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1{2}, b=1{2}
• Relation: T = K √(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = Km^al^bg^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K√(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = Km^al^bg^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K√(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = Km^al^bg^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K√(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = Km^al^bg^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K√(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = Km^al^bg^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K√(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = Km^al^bg^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K√(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = Km^al^bg^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K√(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: T = Km^al^bg^c.
• Use dimensional homogeneity to find powers a, b, c.
• Example: Time period of pendulum T depends on mass m, length l, gravity g.
• Result: T = K√(l/g), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume T = Km^al^bg^c
• Dimensions: [T] = [M]^a [L]^b [L T^-2]^c = [M]^a [L]^b+c [T]^-2c
• Equate powers: a=0, b+c=0, -2c=1 ⇒ c=-1/2, b=1/2
• Relation: T = K√(l/g)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If T ∝ l^1/2 g^-1/2, what is the dimension of g?
  • A. [L T^-1]
  • B. [L T^-2]
  • C. [M L T^-2]
  • D. [M L² T^-2]
• Q3: If T = Km^al^bg^c and a=0, what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension [L T^-2].
• Q3: B - a=0 means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period T = K√(l/g).
• Mass does not affect pendulum time period.

Key Concepts

• Express physical quantity as product of powers of influencing quantities: \( T = K m^a l^b g^c \).
• Use dimensional homogeneity to find powers \( a, b, c \).
• Example: Time period of pendulum \( T \) depends on mass \( m \), length \( l \), gravity \( g \).
• Result: \( T = K \sqrt{\frac{l}{g}} \), mass does not affect time period.

Solved Examples

Ex 1: Find relation for pendulum time period

• Assume \( T = K m^a l^b g^c \)
• Dimensions: \( [T] = [M]^a [L]^b [L T^{-2}]^c = [M]^a [L]^{b+c} [T]^{-2c} \)
• Equate powers: \( a=0, b+c=0, -2c=1 \Rightarrow c=-\frac{1}{2}, b=\frac{1}{2} \)
• Relation: \( T = K \sqrt{\frac{l}{g}} \)

Practice Set

• Q1: Which quantity does NOT affect pendulum time period?
  • A. Mass
  • B. Length
  • C. Gravity
  • D. Amplitude
• Q2: If \( T \propto l^{1/2} g^{-1/2} \), what is the dimension of \( g \)?
  • A. \( [L T^{-1}] \)
  • B. \( [L T^{-2}] \)
  • C. \( [M L T^{-2}] \)
  • D. \( [M L^2 T^{-2}] \)
• Q3: If \( T = K m^a l^b g^c \) and \( a=0 \), what does it imply?
  • A. Time period depends on mass
  • B. Time period independent of mass
  • C. Mass affects length
  • D. Mass affects gravity

Answer Key

• Q1: A - Mass does not affect pendulum time period.
• Q2: B - Gravity has dimension \( [L T^{-2}] \).
• Q3: B - \( a=0 \) means no mass dependence.

Memory Aids

• Use dimensional powers to find relations.
• Pendulum time period \( T = K \sqrt{l/g} \).
• Mass does not affect pendulum time period.

Limitations

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant K in T = K√(l/g).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.

Key Concepts

• Cannot derive relations involving addition, subtraction, or trigonometric functions.
• Cannot find numerical values of constants.
• Cannot solve for more than three unknowns in mechanics using dimensional method.
• Physical quantities with same dimensions may differ in nature.

Solved Examples

Ex 1: Dimensional method cannot find constant \( K \) in \( T = K \sqrt{l/g} \).

Practice Set

• Q1: Which is NOT a limitation of dimensional analysis?
  • A. Cannot find constants
  • B. Cannot derive trigonometric relations
  • C. Can solve all equations exactly
  • D. Limited to multiplication/division relations
• Q2: Dimensional analysis fails when number of unknowns is?
  • A. Less than 3
  • B. Equal to 3
  • C. More than 3
  • D. Exactly 1
• Q3: Physical quantities with same dimensions are always?
  • A. Same nature
  • B. Different nature
  • C. Identical
  • D. None of these

Answer Key

• Q1: C - Dimensional analysis cannot solve all equations exactly.
• Q2: C - More than 3 unknowns cannot be solved dimensionally.
• Q3: B - Same dimensions may represent different physical quantities.

Memory Aids

• Dimensional analysis limited to multiplication/division.
• Cannot find constants or trigonometric relations.
• Check dimensional consistency only for complex cases.