📏 Chapter 2: Units and Measurements (Class 11 Physics) ⏱️
📌 Introduction: Need for Measurement
Physics is all about understanding the natural world through observation and experimentation. To describe and compare physical phenomena accurately, we need to measure physical quantities.
Physics natural world ko observation aur experiment se samajhne ke baare mein hai. Physical phenomena ko aAcche se describe aur compare karne ke liye, humein physical quantities ko measure karne ki zaroorat hai.- Measurement involves comparing a physical quantity with a standard, internationally accepted reference standard called a unit. Measurement mein ek physical quantity ko ek standard, internationally accepted reference standard jise **unit (ikayi)** kehte hain, se compare karna shamil hai.
- The result of a measurement is expressed as a number followed by a unit. (e.g., length = 10 metres). Measurement ka result ek number aur uske baad ek unit ke roop mein vyakt kiya jaata hai. (Jaise, lambai = 10 meter).
📐 Units of Measurement (Maapan Ki Ikaiyan)
Unit: A definite magnitude of a physical quantity, defined and adopted by convention or by law, that is used as a standard for measurement of the same kind of quantity.
Ikayi (Unit): Ek physical quantity ka nishchit parimaan, jo convention ya kanoon dwara paribhashit aur apnaya gaya hai, jiska upyog usi prakaar ki quantity ke maapan ke liye ek standard ke roop mein kiya jaata hai.Characteristics of a Good Unit:
- Well-defined: Should have a clear, unambiguous definition. (Spasht roop se paribhashit).
- Suitable Size: Neither too large nor too small compared to the quantity being measured. (Maapi jaane wali quantity ke comparison mein na bahut bada na bahut chhota).
- Reproducible: Should be easily reproducible everywhere. (Har jagah aasani se reproduce kiya ja sake).
- Invariant: Should not change with time or physical conditions (like temperature, pressure). (Samay ya bhautik sthitiyon ke saath badalna nahi chahiye).
- Internationally Accepted: Should be agreed upon globally for uniformity. 🌍🤝
🔢 Systems of Units (Ikaiyon Ki Pranaliyan)
Historically, different systems were used:
- FPS System: Foot (length), Pound (mass), Second (time). (British system).
- CGS System: Centimetre (length), Gram (mass), Second (time). (Gaussian system).
- MKS System: Metre (length), Kilogram (mass), Second (time).
- SI System (Système International d’Unités): The internationally accepted system, an improved and extended version of the MKS system. Highly Recommended. 🌍✅ SI System: Antarrashtriya star par svikrit pranali, MKS system ka behtar aur vistrit roop. Iska istemal karna chahiye.
📏 SI Units (SI Ikaiyan)
The SI system has:
- Seven Fundamental (Base) Units: For independent physical quantities. Saat Mool (Base) Ikaiyan: Svatantra physical quantities ke liye.
- Two Supplementary Units: For plane angle and solid angle.
- Derived Units: Units of all other physical quantities, obtained by combining fundamental units. Vyutpann (Derived) Ikaiyan: Anya sabhi physical quantities ki ikaiyan, jo mool ikaiyon ko milakar prapt hoti hain.
SI Base Units:
- Length (l): metre (m)
- Mass (m): kilogram (kg)
- Time (t): second (s)
- Electric Current (I): ampere (A)
- Thermodynamic Temperature (T): kelvin (K)
- Amount of Substance (n): mole (mol)
- Luminous Intensity (Iv): candela (cd)
SI Supplementary Units:
- Plane Angle: radian (rad)
- Solid Angle: steradian (sr)
Examples of Derived Units:
- Area (Length × Width): m × m = m²
- Volume (Length × Width × Height): m × m × m = m³
- Speed (Distance / Time): m / s = m s⁻¹
- Acceleration (Speed / Time): (m s⁻¹) / s = m s⁻²
- Force (Mass × Acceleration): kg × (m s⁻²) = kg m s⁻² (called Newton, N)
- Work/Energy (Force × Distance): (kg m s⁻²) × m = kg m² s⁻² (called Joule, J)
- Power (Work / Time): (kg m² s⁻²) / s = kg m² s⁻³ (called Watt, W)
- Pressure (Force / Area): (kg m s⁻²) / m² = kg m⁻¹ s⁻² (called Pascal, Pa)
💡 Solved Practice Problems: Units
Problem 1: Convert 72 km/h into m/s.
We know 1 km = 1000 m and 1 hour = 3600 seconds.
So, 1 kmh = 1000 m3600 s = 518 m/s.
Therefore, 72 km/h = 72 × 518 m/s
= 4 × 5 m/s = 20 m/s.
Answer: 20 m/s
Problem 2: The density of mercury is 13.6 g/cm³. Convert this to kg/m³.
Given density = 13.6 gcm³.
We know 1 g = 10⁻³ kg and 1 cm = 10⁻² m.
So, 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³.
Density = 13.6 × 10⁻³ kg10⁻⁶ m³
= 13.6 × 10⁻³⁺⁶ kg/m³
= 13.6 × 10³ kg/m³ = 13600 kg/m³.
Answer: 13600 kg/m³
Problem 3: Young’s modulus of steel is 1.9 × 10¹¹ N/m². Convert it to dyne/cm² (CGS unit). [Given 1 N = 10⁵ dyne, 1 m = 10² cm]
Given Y = 1.9 × 10¹¹ N/m².
Substitute units: 1 N = 10⁵ dyne, 1 m² = (10² cm)² = 10⁴ cm².
Y = 1.9 × 10¹¹ × (10⁵ dyne)(10⁴ cm²)
= 1.9 × 10¹¹ × 10⁵⁻⁴ dyne/cm²
= 1.9 × 10¹¹ × 10¹ dyne/cm²
= 1.9 × 10¹² dyne/cm².
Answer: 1.9 × 10¹² dyne/cm²
Problem 4: A derived unit for momentum is kg m/s. Express this in terms of base SI units.
The given unit is kg m/s.
kg (kilogram) is the base unit for Mass (M).
m (metre) is the base unit for Length (L).
s (second) is the base unit for Time (T).
Therefore, the unit kg m/s can be expressed in terms of base units directly as kg × m × s⁻¹ or M¹ L¹ T⁻¹ in dimensional terms.
Answer: kg m s⁻¹
Problem 5: Identify the fundamental quantities involved in the unit of Work (Joule), which is kg m² s⁻².
The unit of work is given as kg m² s⁻².
Comparing with base units:
- kg corresponds to Mass (M).
- m corresponds to Length (L).
- s corresponds to Time (T).
Therefore, the fundamental quantities involved are Mass, Length, and Time.
Answer: Mass, Length, Time
🎯 Significant Figures (Sarthak Ank)
Significant Figures: In a measured value, the significant figures are all the digits that are known reliably plus the first uncertain digit. They indicate the precision of a measurement.
Sarthak Ank: Ek maape gaye maan mein, sarthak ank ve sabhi ank hote hain jo vishvasniya roop se gyaat hote hain plus pehla anishchit ank. Ve maapan ki precision (yathaarthata) batate hain.Rules for Identifying Significant Figures:
- All non-zero digits are significant. (Ex: 123.45 has 5 sig figs).
- Zeros between two non-zero digits are significant. (Ex: 1002 has 4 sig figs; 5.03 has 3 sig figs).
- Leading zeros (zeros to the left of the first non-zero digit) in a number less than 1 are NOT significant. They just locate the decimal point. (Ex: 0.0025 has 2 sig figs (2, 5); 0.050 has 2 sig figs (5, 0)).
- Trailing zeros (zeros to the right of the last non-zero digit):
- They ARE significant if the number contains a decimal point. (Ex: 120.0 has 4 sig figs; 2.500 has 4 sig figs; 0.050 has 2 sig figs).
- They are NOT significant if the number does not contain a decimal point (ambiguous). (Ex: 1200 may have 2, 3, or 4 sig figs. It’s better written in scientific notation to avoid ambiguity). (Agar number mein decimal point hai, toh baad wale zero significant hain. Agar decimal nahi hai, toh baad wale zero significant NAHI maane jaate (jab tak scientific notation mein na likha ho)).
- Scientific Notation: All digits in the coefficient part are significant. (Ex: 1.20 × 10³ has 3 sig figs; 5.0 × 10⁻⁴ has 2 sig figs).
Rules for Calculations with Significant Figures:
- Addition/Subtraction: The final result should retain the same number of decimal places as the number with the fewest decimal places.
Antim result mein utne hi **decimal places** hone chahiye jitne uss number mein hain jisme sabse kam decimal places the.
Ex: 12.11 + 18.0 + 1.012 = 31.122. Least decimal places is 1 (in 18.0). Round result to 1 decimal place: 31.1.
- Multiplication/Division: The final result should retain the same number of significant figures as the number with the fewest significant figures.
Antim result mein utne hi **significant figures** hone chahiye jitne uss number mein hain jisme sabse kam significant figures the.
Ex: 4.237 ÷ 2.51 = 1.688047… Least sig figs is 3 (in 2.51). Round result to 3 sig figs: 1.69.
- Rounding Off:
- If digit to be dropped is > 5, preceding digit increases by 1. (Ex: 1.688 round to 3 sig figs -> 1.69)
- If digit to be dropped is < 5, preceding digit stays same. (Ex: 1.683 round to 3 sig figs -> 1.68)
- If digit to be dropped is = 5, preceding digit increases by 1 if it’s odd, stays same if it’s even. (Ex: 1.675 -> 1.68; 1.685 -> 1.68) [This ‘even’ rule is common convention].
🎯 Solved Practice Problems: Significant Figures
Problem 1: State the number of significant figures in (a) 0.007 m² (b) 2.64 × 10²⁴ kg (c) 0.2370 g cm⁻³ (d) 6.320 J
(a) 0.007 m²: Leading zeros are not significant. Only 7 is significant. Ans: 1 sig fig.
(b) 2.64 × 10²⁴ kg: In scientific notation, only coefficient matters. 2, 6, 4 are non-zero. Ans: 3 sig figs.
(c) 0.2370 g cm⁻³: Leading zero not significant. 2, 3, 7 are significant. Trailing zero after decimal point is significant. Ans: 4 sig figs.
(d) 6.320 J: 6, 3, 2 are non-zero (significant). Trailing zero after decimal is significant. Ans: 4 sig figs.
Problem 2: Calculate the area of a rectangle with length 10.5 m and breadth 2.1 m, respecting significant figures.
Area = Length × Breadth
Area = 10.5 m × 2.1 m
Calculator result = 22.05 m²
Rule for Multiplication: Result should have fewest significant figures from original numbers.
10.5 has 3 significant figures.
2.1 has 2 significant figures.
Fewest sig figs = 2.
Round 22.05 to 2 significant figures. Digit to drop is 0 (< 5), preceding digit (2) remains same.
Answer: 22 m²
Problem 3: Add 2.31 cm, 0.5 cm, and 12.448 cm. Give the answer with correct significant figures.
Sum = 2.31 + 0.5 + 12.448
Aligning decimals:
2.31 0.5 +12.448 ------- 15.258
Rule for Addition: Result should have fewest decimal places from original numbers.
2.31 has 2 decimal places.
0.5 has 1 decimal place. (Fewest)
12.448 has 3 decimal places.
Result should be rounded to 1 decimal place.
Round 15.258 to 1 decimal place. Digit to drop is 5, preceding digit (2) is even, so it remains same [common rule]. (Some may round up based on subsequent non-zero digits).
Answer: 15.3 cm (Following simple ’round 5 up’ often taught)
OR
Answer: 15.2 cm (Following ’round 5 to nearest even’ rule)
(Note: Mention the rounding rule followed if ambiguity exists).
Problem 4: Round off 312.96 to three significant figures.
The first three significant figures are 3, 1, 2.
The digit to be dropped is 9 (the fourth digit).
Since 9 > 5, the preceding digit (2) is increased by 1.
Answer: 313
Problem 5: A substance weighs 4.84 g and has a volume of 2.2 cm³. Calculate its density in correct significant figures.
Density = Mass / Volume
Density = 4.84 g / 2.2 cm³
Calculator result = 2.2 g/cm³
Rule for Division: Result should have fewest significant figures.
Mass (4.84 g) has 3 significant figures.
Volume (2.2 cm³) has 2 significant figures. (Fewest)
Result should be rounded to 2 significant figures.
2.2 already has 2 significant figures.
Answer: 2.2 g/cm³
🔧 Dimensions and Dimensional Analysis (Vimayein Aur Vimiye Vishleshan)
Dimensions of a Physical Quantity: The powers (exponents) to which the fundamental (base) quantities must be raised to represent that quantity.
Bhautik Rashi Ki Vimayein: Mool (base) rashiyon ki ghaten (powers/exponents) jin par us rashi ko vyakt karne ke liye unhe uthana (raise karna) padta hai.We usually represent fundamental dimensions as:
- Length = [L]
- Mass = [M]
- Time = [T]
- Electric Current = [A] or [I]
- Temperature = [K] or [Θ]
- Amount of Substance = [mol] or [N]
- Luminous Intensity = [cd] or [J]
Dimensional Formula: An expression which shows how and which of the fundamental units are required to represent the unit of a physical quantity.
Vimiye Sutra: Ek expression jo dikhata hai ki ek physical quantity ki unit ko represent karne ke liye kaun si aur kaise fundamental units ki zaroorat hai.Generally written as: [Ma Lb Tc Ad ...]
Dimensional Equation: An equation obtained by equating a physical quantity with its dimensional formula.
Vimiye Sameekaran: Ek physical quantity ko uske dimensional formula ke barabar rakh kar prapt equation.Examples of Dimensional Formulas:
- Volume = Length × Width × Height ⇒ [L]×[L]×[L] = [L³] or [M⁰L³T⁰]
- Speed = Distance / Time ⇒ [L] / [T] = [LT⁻¹] or [M⁰LT⁻¹]
- Acceleration = Speed / Time ⇒ [LT⁻¹] / [T] = [LT⁻²]
- Force = Mass × Acceleration ⇒ [M] × [LT⁻²] = [MLT⁻²]
- Work/Energy = Force × Distance ⇒ [MLT⁻²] × [L] = [ML²T⁻²]
- Pressure = Force / Area ⇒ [MLT⁻²] / [L²] = [ML⁻¹T⁻²]
Dimensionless Quantities: Have no dimensions [M⁰L⁰T⁰]. Ex: Angle (radian), Strain, Relative Density, Pure numbers.
⚙️ Dimensional Analysis and Its Applications
The method of studying physical phenomena based on dimensions is called dimensional analysis. It works on the **Principle of Homogeneity**:
Principle of Homogeneity of Dimensions:
- A physical equation will be dimensionally correct if and only if the dimensions of all the terms occurring on both sides of the equation are the same. Ek physical equation dimensionally tabhi sahi hogi jab equation ke dono taraf aane wale sabhi terms ki dimensions samaan hon.
- We can add or subtract only those physical quantities which have the same dimensions. Hum kewal unhi physical quantities ko jod ya ghata sakte hain jinki dimensions samaan hon.
Applications of Dimensional Analysis:
- 1. Checking the Dimensional Correctness of Equations: Verify if an equation is dimensionally consistent (LHS dimensions = RHS dimensions). Note: A dimensionally correct equation isn’t necessarily physically correct (may have wrong dimensionless constants). But a dimensionally incorrect equation *must* be wrong. ✅❓❌ Equations ki Dimensional Correctness Check Karna: Verify karna ki kya equation dimensionally sahi hai. (Dimensionally sahi equation zaroori nahi ki physically sahi ho, lekin dimensionally galat equation zaroori galat hogi).
- 2. Deriving Relationship between Physical Quantities: If we know the factors on which a quantity depends, we can sometimes derive a formula relating them (up to a dimensionless constant). 🔗 Physical Quantities ke Beech Sambandh Nikalna: Agar humein pata ho ki ek quantity kin factors par depend karti hai, toh hum unke beech formula derive kar sakte hain.
- 3. Converting Units from One System to Another: Use dimensional formula to find conversion factor between units in different systems (like SI to CGS). 🔄 Units ko Ek System se Doosre mein Convert Karna: Alag-alag systems mein units ke beech conversion factor nikalne ke liye dimensional formula use karna.
🔧 Solved Practice Problems: Dimensional Analysis
Problem 1: Check the dimensional correctness of the equation v = u + at (where v=final velocity, u=initial velocity, a=acceleration, t=time).
Dimensions of terms:
- [v] = [LT⁻¹]
- [u] = [LT⁻¹]
- [at] = [a] × [t] = [LT⁻²] × [T] = [LT⁻¹]
Dimensions of LHS = [v] = [LT⁻¹].
Dimensions of RHS terms: [u]=[LT⁻¹] and [at]=[LT⁻¹].
According to Principle of Homogeneity, we can add terms with same dimensions. The dimensions of all terms (v, u, at) are the same ([LT⁻¹]).
Answer: The equation is dimensionally correct.
Problem 2: Find the dimensional formula for Pressure (Pressure = Force/Area).
Dimension of Force [F] = [MLT⁻²].
Dimension of Area [Area] = [L²].
Dimension of Pressure [P] = [Force] / [Area]
[P] = [MLT⁻²][L²]
[P] = [ML¹⁻²T⁻²] = [ML⁻¹T⁻²].
Answer: [ML⁻¹T⁻²]
Problem 3: The frequency (f) of vibration of a string may depend upon length (l), tension (T – Force), and mass per unit length (m – Mass/Length). Derive the formula for frequency using dimensional analysis (assume proportionality constant k=1/2).
Assume frequency f ∝ la Tb mc.
Dimensions:
- [f] = Frequency = 1/Time = [T⁻¹] (or [M⁰L⁰T⁻¹])
- [l] = Length = [L]
- [T] = Tension (Force) = [MLT⁻²]
- [m] = Mass per length = Mass/Length = [ML⁻¹]
Write dimensional equation:
[M⁰L⁰T⁻¹] = k [L]a [MLT⁻²]b [ML⁻¹]c
[M⁰L⁰T⁻¹] = k [La] [MbLbT⁻²ᵇ] [McL⁻ᶜ]
Combine powers of M, L, T on RHS:
[M⁰L⁰T⁻¹] = k [Mb+c La+b-c T⁻²ᵇ]
Equate powers of M, L, T on both sides (Principle of Homogeneity):
- For M: 0 = b + c ⇒ c = -b
- For L: 0 = a + b – c
- For T: -1 = -2b ⇒ b = 1/2
Solve for exponents:
- From T: b = 1/2
- From M: c = -b = -1/2
- From L: 0 = a + (1/2) – (-1/2) ⇒ 0 = a + 1/2 + 1/2 ⇒ 0 = a + 1 ⇒ a = -1
So, f ∝ l⁻¹ T1/2 m⁻¹ᐟ², or f ∝ 1l T1/2m1/2 = 1l √(T/m).
Using k = 1/2 (given/known constant): f = 12l √(T/m).
Answer: f = (1/2l)√(T/m)
Problem 4: Convert 1 Joule (SI unit of energy) into ergs (CGS unit of energy).
Dimensional formula for Energy = [ML²T⁻²].
SI unit: 1 Joule = 1 kg¹ m² s⁻²
CGS unit: 1 erg = 1 g¹ cm² s⁻²
Convert SI base units to CGS base units:
- 1 kg = 10³ g
- 1 m = 10² cm ⇒ 1 m² = (10² cm)² = 10⁴ cm²
- 1 s = 1 s
Substitute into Joule unit:
1 Joule = (10³ g)¹ × (10⁴ cm²)¹ × (1 s)⁻²
= 10³ × 10⁴ g¹ cm² s⁻²
= 10⁷ g cm² s⁻²
= 10⁷ ergs.
Answer: 1 Joule = 10⁷ ergs
Problem 5: Check dimensional correctness: E = mc², where E=Energy, m=mass, c=speed of light.
LHS Dimension [E] = Energy = [ML²T⁻²].
RHS Dimension [mc²] = [m] × [c]²
[m] = Mass = [M]
[c] = Speed = [LT⁻¹]
[c]² = [LT⁻¹]² = [L²T⁻²]
So, [mc²] = [M] × [L²T⁻²] = [ML²T⁻²].
Since dimensions of LHS ([ML²T⁻²]) = Dimensions of RHS ([ML²T⁻²]), the equation is dimensionally correct.
Answer: The equation is dimensionally correct.
❓ Sawal Jawab (Questions & Answers)
🤏 Very Short Answer Questions
1. Why is measurement needed in Physics?
To describe and compare physical quantities accurately.
2. What is a unit?
An accepted reference standard for measuring a physical quantity.
3. Name the international system of units.
SI System (Système International d’Unités).
4. How many fundamental units are there in SI?
Seven.
5. What is the SI unit of Mass?
kilogram (kg).
6. What is the SI unit of Electric Current?
ampere (A).
7. Name a derived physical quantity.
Speed, Force, Area, Volume etc.
8. What are significant figures?
Reliably known digits plus the first uncertain digit in a measurement.
9. How many significant figures are in 0.05020?
Four (5, 0, 2, 0).
10. How many significant figures are in 1500?
Ambiguous. Usually taken as 2, but better written in scientific notation (e.g., 1.5 × 10³ has 2).
11. What is the dimensional formula for Speed?
[LT⁻¹] or [M⁰LT⁻¹].
12. What is the dimensional formula for Force?
[MLT⁻²].
13. State the Principle of Homogeneity.
Dimensions of all terms on both sides of a correct physical equation must be the same.
14. Give one application of dimensional analysis.
Checking equation correctness / Deriving relations / Unit conversion.
15. Is the equation s = ut + 12at dimensionally correct (s=distance)?
No. [s]=L, [ut]=LT⁻¹T=L, [at]=LT⁻²T=LT⁻¹. Terms cannot be added.
📝 Short Answer Questions
1. Distinguish between fundamental and derived units.
- Fundamental Units: Units of base quantities (like metre, kg, second) that are independent.
- Derived Units: Units of other quantities obtained by combining fundamental units (like m/s, kg/m³, Newton).
2. List the 7 fundamental quantities and their SI units.
- Length (metre, m), Mass (kilogram, kg), Time (second, s), Electric Current (ampere, A), Temperature (kelvin, K), Amount of Substance (mole, mol), Luminous Intensity (candela, cd).
3. Explain the rules for counting significant figures regarding zeros.
- Zeros between non-zeros ARE significant (102).
- Leading zeros (before first non-zero) are NOT significant (0.012).
- Trailing zeros AFTER decimal ARE significant (1.20).
- Trailing zeros WITHOUT decimal are NOT significant / ambiguous (1200).
4. Calculate (2.5 × 1.25) + 0.51 with proper significant figures.
- Multiplication: 2.5 (2 sig figs) × 1.25 (3 sig figs) = 3.125. Result should have 2 sig figs → 3.1.
- Addition: 3.1 (1 decimal place) + 0.51 (2 decimal places) = 3.61. Result should have 1 decimal place → 3.6.
- Answer: 3.6
5. Find the dimensions of Density (Mass/Volume).
- [Mass] = [M]
- [Volume] = [L³]
- [Density] = [Mass]/[Volume] = [M]/[L³] = [ML⁻³T⁰].
6. Find the dimensions of Power (Work/Time).
- [Work/Energy] = [ML²T⁻²]
- [Time] = [T]
- [Power] = [Work]/[Time] = [ML²T⁻²] / [T] = [ML²T⁻³].
7. Can a dimensionally correct equation be physically incorrect? Explain.
- Yes. Dimensional correctness only checks the consistency of dimensions on both sides.
- It doesn’t say anything about dimensionless constants or the exact physical relationship.
- Example: We might derive time period T ∝ √(l/g). Dimensionally correct. But correct formula is T = 2π√(l/g). Dimensional analysis cannot find the ‘2π’.
8. Write the dimensional formula of gravitational constant G from F = G m₁m₂/r².
- Rearrange: G = F r² / (m₁m₂).
- Dimensions: [F]=MLT⁻², [r]=L, [m₁]=[m₂]=M.
- [G] = [MLT⁻²] [L]² / ([M][M]) = [ML³T⁻²] / [M²].
- [G] = [M¹⁻² L³ T⁻²] = [M⁻¹L³T⁻²].
9. Convert force of 100 dynes (CGS) into Newtons (SI).
- We know 1 N = 10⁵ dynes, so 1 dyne = 10⁻⁵ N.
- 100 dynes = 100 × 10⁻⁵ N = 10² × 10⁻⁵ N = 10⁻³ N.
- Answer: 10⁻³ N (or 0.001 N).
10. What are supplementary units in SI?
- They are units for two dimensionless quantities:
- Plane Angle: radian (rad).
- Solid Angle: steradian (sr).
11. Round off 0.04597 to three significant figures.
- First three sig figs are 4, 5, 9.
- Digit to drop is 7 (> 5).
- Preceding digit (9) increases by 1, making it 10. Carry over 1.
- Result: 0.0460 (Trailing zero after decimal is significant).
📜 Long Answer Questions
1. Explain the need for a standard system of units like SI. Discuss characteristics of a good unit.
Need for Standard System:
- Early measurements used local/body parts (handspan, cubit), which varied person-to-person and place-to-place.
- Led to confusion and inconsistency in trade and science.
- A standard system ensures uniformity, allowing scientists/engineers worldwide to communicate results unambiguously.
- SI system provides a coherent, rationalised system based on seven base units.
Characteristics of Good Unit:
- Well-defined, Unambiguous.
- Suitable size.
- Reproducible easily.
- Invariant (doesn’t change with time/conditions).
- Internationally accepted.
2. Define Fundamental and Derived quantities & units with examples.
- Fundamental Quantities: Physical quantities that are independent of each other and cannot be expressed in terms of other physical quantities. SI has 7: Length, Mass, Time, Current, Temperature, Amount of Substance, Luminous Intensity.
- Fundamental Units: The SI units chosen for measuring fundamental quantities. Ex: metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), candela (cd).
- Derived Quantities: Physical quantities whose definition is based on (derived from) other physical quantities. Ex: Speed (Distance/Time), Force (Mass × Acceleration).
- Derived Units: Units obtained by combining fundamental units according to the definition of the derived quantity. Ex: Unit of Speed = m/s (or ms⁻¹). Unit of Force = kg m s⁻² (or Newton, N).
3. Explain the rules for determining significant figures in a measurement with examples.
Rules for Significant Figures:
- Non-zero digits: Always significant. Ex: 285 (3 sig figs).
- Zeros between non-zeros: Always significant. Ex: 2005 (4 sig figs); 7.03 (3 sig figs).
- Leading Zeros (before 1st non-zero digit): NOT significant. Ex: 0.0032 (2 sig figs); 0.040 (2 sig figs).
- Trailing Zeros (after last non-zero digit):
- Significant *if* number has a decimal point. Ex: 3.500 (4 sig figs); 600.0 (4 sig figs).
- NOT significant (ambiguous) *if* number has no decimal point. Ex: 3500 (Ambiguous: 2, 3, or 4 sig figs). Better as 3.5×10³ (2 sig figs), 3.50×10³ (3 sig figs) etc.
- Scientific Notation (
A × 10B): All digits in the coefficient ‘A’ are significant. Ex: 6.022×10²³ (4 sig figs).
4. Explain the rules for arithmetic operations (+, -, ×, ÷) with significant figures.
Rules for Calculation:
- Addition and Subtraction:
- Perform operation normally.
- Final result should be rounded to have the same number of **decimal places** as the number with the *fewest* decimal places in the calculation.
- Ex: 2.1 + 10.33 + 0.005 = 12.435. Fewest decimal places is 1 (in 2.1). Round result to 12.4.
- Multiplication and Division:
- Perform operation normally.
- Final result should be rounded to have the same number of **significant figures** as the number with the *fewest* significant figures in the calculation.
- Ex: 3.12 (3 sig figs) × 1.5 (2 sig figs) = 4.68. Fewest sig figs is 2. Round result to 4.7.
- (Remember rounding rules for digit 5: Round to nearest even, or simply round up if taught).
5. What is Dimensional Analysis? State the Principle of Homogeneity and give two major applications of dimensional analysis.
- Dimensional Analysis: Method of studying physical phenomena using the dimensions of the quantities involved.
- Principle of Homogeneity: States that a physical equation is dimensionally valid only if the dimensions of all terms on both sides are identical. Also implies only quantities with same dimensions can be added/subtracted. (Equation tabhi valid hai jab dono taraf ke sabhi terms ki dimension same ho).
- Applications:
- **Checking Correctness:** Verify if equations like `v² = u² + 2as` are dimensionally consistent. LHS=[LT⁻¹]²=[L²T⁻²]. RHS: [u²]=[L²T⁻²], [2as]=[LT⁻²][L]=[L²T⁻²]. Since all terms have same dimension, equation is correct.
- **Deriving Relations:** Finding how one quantity depends on others (e.g., time period of pendulum T ∝ la gb).
- **Unit Conversion:** Converting values between systems (e.g., Newton to dyne).
6. Define the terms: Length, Mass, Time and their SI units. Why are they considered fundamental?
- Length: The measurement or extent of something from end to end. SI Unit: **metre (m)**. Definition based on speed of light.
- Mass: The amount of matter in a physical body or object. It is also a measure of inertia. SI Unit: **kilogram (kg)**. Definition based on Planck’s constant.
- Time: The indefinite continued progress of existence and events in the past, present, and future regarded as a whole. SI Unit: **second (s)**. Definition based on caesium atomic clock frequency.
- Why Fundamental?: These quantities (along with Ampere, Kelvin, Mole, Candela) are considered fundamental because they are **independent** of each other. They cannot be expressed or derived simply by combining other physical quantities using multiplication or division. All other physical quantities (like speed, force, energy) can be derived from these base quantities.
7. The radius of a sphere is measured to be (2.1 ± 0.1) cm. Calculate its surface area with error limits using significant figure rules where applicable (Treat errors simply). [NEET/JEE hint – Basic error idea]
(Note: Rigorous error propagation is more advanced, we’ll use a simplified approach relevant to sig figs concept).
- Radius r = 2.1 cm (2 significant figures). Uncertainty Δr = 0.1 cm.
- Surface Area formula A = 4πr². Use π ≈ 3.14 (3 sig figs).
- Calculate area using mean value: A = 4 × 3.14 × (2.1)² = 4 × 3.14 × 4.41 = 55.3896 cm².
- Sig Fig Rule for Multiplication: The numbers involved are 4 (exact), 3.14 (3 sig figs), 2.1 (2 sig figs), 2.1 (2 sig figs). The least number of significant figures is 2 (from radius).
- So, the calculated area should be rounded to 2 significant figures.
- Rounding 55.3896 to 2 sig figs: 55 cm².
Simplified Error Estimation: (Class 11 intro often focuses on % error, let’s calculate result based on sig figs of input)
- Result should be based on the precision of the least precise input used in multiplication/division. Radius has 2 sig figs.
Answer (Respecting input precision): The area should be reported to 2 significant figures, i.e., 55 cm².
(Advanced error calculation would involve relative errors: ΔA/A ≈ 2(Δr/r) leading to ΔA ≈ 5.3 cm², Area = (55 ± 5) cm² approximately, but basic sig fig rules limit result to 55 cm²).
8. Check the dimensional correctness of Stokes’ Law formula: F = 6πηrv (where F=Viscous Force, η=Coefficient of viscosity, r=radius, v=velocity).
We need the dimension of coefficient of viscosity (η).
- From formula: η = F / (6πrv). Since 6π is dimensionless, [η] = [F] / ([r][v]).
- [F] = Force = [MLT⁻²].
- [r] = Radius (length) = [L].
- [v] = Velocity = [LT⁻¹].
- Therefore, [η] = [MLT⁻²] / ([L][LT⁻¹]) = [MLT⁻²] / [L²T⁻¹] = [ML¹⁻²T⁻²⁻⁽⁻¹⁾] = **[ML⁻¹T⁻¹]**.
Now check dimensions of RHS of F = 6πηrv:
- [RHS] = [6π][η][r][v]
- [6π] is dimensionless.
- [RHS] = [ML⁻¹T⁻¹] × [L] × [LT⁻¹]
- Combine powers: [RHS] = [M¹ L⁻¹⁺¹⁺¹ T⁻¹⁻¹] = [ML¹T⁻²] = [MLT⁻²].
Dimension of LHS [F] = [MLT⁻²].
Since dimensions of LHS = Dimensions of RHS, the equation is dimensionally correct.
Answer: The formula F = 6πηrv is dimensionally correct.
9. Write the dimensions of: (a) Pressure Gradient (b) Angular Velocity (c) Specific Heat Capacity.
- (a) Pressure Gradient: Pressure change per unit distance = ΔP / Δx.
- [Pressure, P] = [ML⁻¹T⁻²] (from Problem 2).
- [Distance, x] = [L].
- [Pressure Gradient] = [P]/[x] = [ML⁻¹T⁻²] / [L] = [ML⁻²T⁻²].
- (b) Angular Velocity (ω): Angle turned per unit time = Angle / Time.
- [Angle] = Dimensionless = [M⁰L⁰T⁰] or [1].
- [Time] = [T].
- [Angular Velocity] = [1] / [T] = [T⁻¹] or [M⁰L⁰T⁻¹].
- (c) Specific Heat Capacity (s): From Q = msΔT, s = Q / (mΔT) where Q=Heat Energy, m=mass, ΔT=Temperature change.
- [Heat Energy, Q] = Work/Energy = [ML²T⁻²].
- [Mass, m] = [M].
- [Temperature Change, ΔT] = [K].
- [Specific Heat, s] = [ML²T⁻²] / ([M][K]) = [L²T⁻²K⁻¹] or [M⁰L²T⁻²K⁻¹].
10. If Velocity (V), Force (F) and Time (T) are chosen as fundamental quantities, find the dimensions of Mass.
We need to express Mass [M] in terms of [V], [F], [T].
- Write dimensions of chosen fundamentals in terms of M, L, T:
- [V] = [LT⁻¹]
- [F] = [MLT⁻²]
- [T] = [T]
- We want to find exponents a, b, c such that:
[M] = [V]a [F]b [T]c - Substitute dimensions:
[M¹L⁰T⁰] = [LT⁻¹]a [MLT⁻²]b [T]c [M¹L⁰T⁰] = [LaT⁻ᵃ] [MbLbT⁻²ᵇ] [Tc]- Combine powers on RHS:
[M¹L⁰T⁰] = [Mb La+b T-a-2b+c] - Equate powers of M, L, T on both sides:
- M: 1 = b ⇒ b = 1
- L: 0 = a + b ⇒ a = -b ⇒ a = -1
- T: 0 = -a – 2b + c ⇒ 0 = -(-1) – 2(1) + c ⇒ 0 = 1 – 2 + c ⇒ 0 = -1 + c ⇒ c = 1
So, the dimensions of Mass in terms of V, F, T are [V⁻¹F¹T¹].
Answer: [F V⁻¹ T]
