Current Affairs updated daily
KalviGavaniKalviGavani

Nature of Physical World and Measurement Exercise Solutions | KalviGavani

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement exercise solutions — SI units, dimensional analysis, significant figures, errors, vernier calipers, screw gauge, MCQs, worked numericals.

Updated 15 June 202622 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

View Solutions →

How to Use This Page

1. Solve first, check later. Attempt every exercise question on your own before reading the solution. 2. Follow the sequence. The sections mirror the textbook order: MCQs, fill-in-the-blanks, true-or-false, match-the-following, short answers, detailed answers. 3. Understand the method, not just the answer. Each solution explains the reasoning so you can tackle similar problems in the exam. 4. Practise numericals by re-working the worked examples with different numbers — the method stays the same.


Key Concepts (Quick Recap)

Physical Quantities and Units

TypeDefinitionExamples
Fundamental quantityCannot be expressed in terms of other quantitiesLength, Mass, Time, Temperature, Electric current, Luminous intensity, Amount of substance
Derived quantityExpressed using fundamental quantitiesForce, Energy, Power, Pressure, Frequency

The Seven SI Base Units

Fundamental quantitySI UnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric currentampereA
Thermodynamic temperaturekelvinK
Amount of substancemolemol
Luminous intensitycandelacd

Important Supplementary Units

QuantityUnitSymbol
Plane angleradianrad
Solid anglesteradiansr

Common Derived Units and Dimensional Formulae

Physical QuantityFormulaDimensional FormulaSI Unit
AreaL × L[M⁰L²T⁰]m²
VolumeL × L × L[M⁰L³T⁰]m³
Velocitydisplacement / time[M⁰LT⁻¹]m s⁻¹
Accelerationvelocity / time[M⁰LT⁻²]m s⁻²
Forcemass × acceleration[MLT⁻²]N (kg m s⁻²)
Pressureforce / area[ML⁻¹T⁻²]Pa (N m⁻²)
Work / Energyforce × distance[ML²T⁻²]J (N m)
Powerwork / time[ML²T⁻³]W (J s⁻¹)
Momentummass × velocity[MLT⁻¹]kg m s⁻¹
Impulseforce × time[MLT⁻¹]N s
Gravitational constant (G)F = Gm₁m₂/r²[M⁻¹L³T⁻²]N m² kg⁻²
Planck's constant (h)E = hν[ML²T⁻¹]J s
Coefficient of viscosity (η)F = ηA(dv/dx)[ML⁻¹T⁻¹]Pa s
Surface tension (T)force / length[MT⁻²]N m⁻¹
Specific heat capacity (c)Q = mcΔT[L²T⁻²K⁻¹]J kg⁻¹ K⁻¹
Universal gas constant (R)PV = nRT[ML²T⁻²K⁻¹mol⁻¹]J mol⁻¹ K⁻¹

Rules for Significant Figures

RuleExample
All non-zero digits are significant285 → 3 significant figures
Zeros between non-zero digits are significant2005 → 4 significant figures
Leading zeros are NOT significant0.025 → 2 significant figures
Trailing zeros after decimal are significant2.50 → 3 significant figures
Trailing zeros in a whole number without decimal are ambiguous1200 → use scientific notation
For multiplication/division, result has same number of sig figs as the least precise operand2.5 × 3.42 = 8.6 (2 sig figs)
For addition/subtraction, result has same number of decimal places as the least precise operand12.5 + 3.24 = 15.7 (1 decimal place)
Rounding: if digit ≥ 5, round up; if < 5, round down3.456 → 3.46 (3 sig figs)

Errors in Measurement

Error TypeFormulaNotes
Absolute errorΔa = \a_measured − a_true\Same units as the quantity
Mean absolute errorΔa_mean = (Σ\Δaᵢ\) / nAverage of absolute errors
Relative errorΔa_mean / a_meanDimensionless
Percentage error(Δa_mean / a_mean) × 100Expressed as %

Error propagation rules for combined measurements:

  • For x = a + b or x = a − b: Δx = Δa + Δb
  • For x = a × b or x = a / b: (Δx / x) = (Δa / a) + (Δb / b)
  • For x = aⁿ: (Δx / x) = n(Δa / a)

Measuring Instruments

Vernier Caliper:

  • Least Count (LC) = 1 MSD − 1 VSD
  • Standard: 1 MSD = 1 mm, 50 VSD → LC = 0.02 mm = 0.002 cm
  • Reading = MSR + (VSR × LC)

Screw Gauge (Micrometer):

  • Least Count (LC) = Pitch / Number of divisions on circular scale
  • Standard: Pitch = 0.5 mm, 50 divisions → LC = 0.01 mm = 0.001 cm
  • Reading = PSR + (CSR × LC)
  • Zero error: Positive zero error is subtracted; Negative zero error is added.

I. Choose the Best Answer (MCQs)

1. The symbol for the SI unit of thermodynamic temperature is

(a) °C &emsp; (b) °F &emsp; (c) K &emsp; (d) C

Answer(c) K
ExplanationThe SI base unit of thermodynamic temperature is the kelvin, symbol K. It does not use the degree symbol. Celsius (°C) and Fahrenheit (°F) are temperature scales, not SI units.

2. The dimensional formula for Planck's constant is

(a) [MLT⁻¹] &emsp; (b) [ML²T⁻¹] &emsp; (c) [ML²T⁻²] &emsp; (d) [ML²T⁻³]

Answer(b) [ML²T⁻¹]
ExplanationFrom E = hν, we get h = E / ν.
  • Energy E has dimensions [ML²T⁻²]
  • Frequency ν has dimensions [T⁻¹]
  • Therefore h = [ML²T⁻²] / [T⁻¹] = [ML²T⁻¹]

3. The number of significant figures in 0.00450 is

(a) 6 &emsp; (b) 5 &emsp; (c) 3 &emsp; (d) 2

Answer(c) 3
ExplanationThe leading zeros (0.00) are not significant — they only locate the decimal point. The digits 4, 5, and the trailing zero (after the decimal) are significant. So 0.00450 has 3 significant figures.

4. The least count of a vernier caliper with 50 divisions on the vernier scale is

(a) 0.01 mm &emsp; (b) 0.02 mm &emsp; (c) 0.1 mm &emsp; (d) 0.05 mm

Answer(b) 0.02 mm
Explanation
1 MSD = 1 mm 50 VSD = 49 MSD → 1 VSD = 49/50 mm = 0.98 mm LC = 1 MSD − 1 VSD = 1.00 mm − 0.98 mm = 0.02 mm

5. If the percentage errors in measuring mass and velocity are 2% and 3%, the percentage error in kinetic energy is

(a) 1% &emsp; (b) 5% &emsp; (c) 8% &emsp; (d) 6%

Answer(c) 8%
ExplanationKinetic energy KE = ½mv².
(ΔKE/KE) = (Δm/m) + 2(Δv/v) = 2% + 2 × 3% = 2% + 6% = 8%

6. Which of the following pairs has the same dimensional formula?

(a) Force and Impulse &emsp; (b) Work and Torque &emsp; (c) Momentum and Energy &emsp; (d) Pressure and Stress

Answer(d) Pressure and Stress
Explanation

Both have the same dimensional formula. (Work and Torque also share [ML²T⁻²], but the standard answer expected in TN Board exams is Pressure and Stress since they are physically analogous quantities.)

  • Pressure = Force / Area → [ML⁻¹T⁻²]
  • Stress = Force / Area → [ML⁻¹T⁻²]

7. A physical quantity is measured and its value is found to be 3.45 × 10⁴. The number of significant figures is

(a) 2 &emsp; (b) 3 &emsp; (c) 5 &emsp; (d) 4

Answer(b) 3
ExplanationIn scientific notation a × 10ⁿ, only the coefficient 'a' determines significant figures. Here 3.45 has 3 significant figures. The power of 10 does not add or remove significant figures.

II. Fill in the Blanks

1. The SI unit of electric current is ampere (A).

2. The dimensional formula for universal gravitational constant G is [M⁻¹L³T⁻²].

3. The number of significant figures in 0.00700 is 3 (7, 0, 0 — leading zeros are not significant, trailing zeros after decimal are significant).

4. The least count of a screw gauge with pitch 0.5 mm and 50 circular scale divisions is 0.01 mm.

5. The principle of homogeneity of dimensions states that every term in a physical equation must have the same dimensions.

6. Random errors can be reduced by taking multiple observations and averaging them.

7. If x = a + b, then the absolute error in x is Δx = Δa + Δb.

8. The SI unit of solid angle is steradian (sr).

9. 1 parsec = 3.086 × 10¹⁶ metres (or 3.26 light years).

10. The number of base quantities in the SI system is 7.

11. The dimensional formula for coefficient of viscosity is [ML⁻¹T⁻¹].

12. If x = aⁿ, then (Δx/x) = n(Δa/a).


III. True or False

1. All derived quantities have their own SI units with special names.

False. Not all derived quantities have special SI unit names. For example, velocity (m s⁻¹) and acceleration (m s⁻²) are derived quantities expressed in terms of base units without special names. Only some derived quantities like force (newton) and pressure (pascal) have special names.


2. The dimensions of impulse and momentum are the same.

True. Both impulse and momentum have the dimensional formula [MLT⁻¹]. Impulse = Force × time = [MLT⁻²] × [T] = [MLT⁻¹]. Momentum = mass × velocity = [M] × [LT⁻¹] = [MLT⁻¹].


3. Systematic errors can be eliminated by taking a large number of readings.

False. Systematic errors are consistent, repeatable errors caused by instrument calibration, experimental method, or environmental factors. Taking many readings reduces random errors, not systematic errors. Systematic errors must be eliminated by correcting the cause (calibrating the instrument, improving the method, etc.).


4. The trailing zeros in a number without a decimal point are significant.

False. Trailing zeros in a number without a decimal point are ambiguous. For example, 1200 could have 2, 3, or 4 significant figures. To avoid ambiguity, use scientific notation: 1.2 × 10³ (2 sig figs), 1.20 × 10³ (3 sig figs), or 1.200 × 10³ (4 sig figs).


5. The dimensional formula for surface tension is [ML⁰T⁻²].

True. Surface tension = Force / Length = [MLT⁻²] / [L] = [MT⁻²] = [ML⁰T⁻²].


IV. Match the Following

Column AColumn B
1.Force(a)[ML²T⁻²]
2.Surface tension(b)[MLT⁻²]
3.Energy(c)[MLT⁻¹]
4.Momentum(d)[MT⁻²]
5.Pressure(e)[ML⁻¹T⁻²]
6.Planck's constant(f)[ML²T⁻¹]
Answer

1 → (b) Force = [MLT⁻²] 2 → (d) Surface tension = [MT⁻²] 3 → (a) Energy = [ML²T⁻²] 4 → (c) Momentum = [MLT⁻¹] 5 → (e) Pressure = [ML⁻¹T⁻²] 6 → (f) Planck's constant = [ML²T⁻¹]


V. Answer Briefly (2-Mark Questions)

1. Distinguish between fundamental and derived quantities with two examples each.

Fundamental quantities are those that cannot be expressed in terms of other physical quantities. They are independent.

  • Examples: Mass (kg), Time (s)

Derived quantities are obtained by combining fundamental quantities using multiplication or division.

  • Examples: Velocity (m s⁻¹) = Length / Time, Force (N) = Mass × Acceleration

2. What are the limitations of dimensional analysis?

1. Dimensionless quantities cannot be determined — dimensional analysis gives no information about pure numbers or dimensionless constants (e.g., the ½ in KE = ½mv²). 2. Addition and subtraction — Dimensional analysis cannot handle equations involving addition/subtraction of quantities (e.g., s = ut + ½at² cannot be fully verified). 3. Trigonometric, logarithmic, and exponential functions — Dimensional analysis cannot derive relations involving sin θ, log x, eˣ etc. 4. Distinguishing quantities with same dimensions — Work and torque both have dimensions [ML²T⁻²] but are physically different quantities.


3. Define least count. State the least count of a standard vernier caliper and a standard screw gauge.

Least count is the smallest measurement that can be taken accurately with an instrument. It equals the difference between one main scale division and one vernier/circular scale division.

  • Standard vernier caliper: LC = 1 MSD − 1 VSD = 1 mm − 0.98 mm = 0.02 mm = 0.002 cm
  • Standard screw gauge: LC = Pitch / No. of circular divisions = 0.5 mm / 50 = 0.01 mm = 0.001 cm

4. Write the rules for determining significant figures in addition and multiplication.

Addition/Subtraction: The result should have the same number of decimal places as the quantity with the fewest decimal places.

Example: 12.5 + 3.24 = 15.74 → rounded to 15.7 (one decimal place, matching 12.5)

Multiplication/Division: The result should have the same number of significant figures as the quantity with the fewest significant figures.

Example: 2.5 (2 sig figs) × 3.42 (3 sig figs) = 8.550 → rounded to 8.6 (2 sig figs)

VI. Answer in Detail (5-Mark Questions)

1. Describe the principle, construction, and working of a vernier caliper with a worked example.

Principle: A vernier caliper uses the difference between one main scale division and one vernier scale division to measure lengths smaller than the smallest main scale division.

Construction:

  • Main scale: Graduated in centimetres and millimetres on a fixed steel strip.
  • Vernier scale: A sliding scale with 50 divisions, matching 49 mm on the main scale.
  • Fixed jaw (J₁): Attached to the main scale; holds the object on one side.
  • Movable jaw (J₂): Attached to the vernier scale; holds the object on the other side.
  • Depth probe: Extends from the vernier scale to measure depth of hollow objects.
Working1. Place the object between the two jaws and close them gently. 2. Note the main scale reading (MSR): the last main scale division before the zero of the vernier scale. 3. Note the vernier scale reading (VSR): the vernier division that coincides with any main scale division. 4. Total reading = MSR + (VSR × LC)

Worked example — Reading a vernier caliper:

A sphere is measured with a vernier caliper. The main scale reading is 2.4 cm and the 6th vernier division coincides with a main scale division. Find the diameter of the sphere.

Given: Main Scale Reading (MSR) = 2.4 cm Vernier Scale Reading (VSR) = 6 Least Count (LC) = 0.002 cm (standard vernier caliper) Total Reading = MSR + (VSR × LC) = 2.4 + (6 × 0.002) = 2.4 + 0.012 = 2.412 cm Diameter of the sphere = 2.412 cm

Zero error correction: If the zero of the vernier scale does not coincide with the zero of the main scale when the jaws are closed, there is a zero error.

  • Positive zero error: Subtract the correction from the reading.
  • Negative zero error: Add the correction to the reading.
Example with zero error: Measured reading = 2.412 cm Positive zero error = 0.004 cm (4th VSD coincides at closed position) Corrected reading = 2.412 − 0.004 = 2.408 cm

2. Explain errors in measurement — types, propagation of errors, and worked numerical examples.

Types of Errors:

TypeCauseRemedy
Systematic errorInstrument calibration, method, environmental factorsCorrect the source — calibrate instruments, improve method
Random errorUnpredictable fluctuationsTake multiple readings and average
Gross errorCarelessness — misreading, wrong recordingCareful observation and recording

Propagation of Errors:

For addition/subtraction (x = a ± b):

Δx = Δa + Δb (absolute errors add)

For multiplication/division (x = a × b or x = a / b):

(Δx / x) = (Δa / a) + (Δb / b) (relative errors add)

For power (x = aⁿ):

(Δx / x) = n × (Δa / a) (relative error is multiplied by the power)

Worked Example 1 — Error in derived quantity (multiplication):

The length and breadth of a rectangle are measured as (12.0 ± 0.1) cm and (8.0 ± 0.2) cm. Find the area with its error.

Given: l = 12.0 cm, Δl = 0.1 cm b = 8.0 cm, Δb = 0.2 cm Step 1: Calculate area A = l × b = 12.0 × 8.0 = 96.0 cm² Step 2: Calculate relative error in area (ΔA / A) = (Δl / l) + (Δb / b) = (0.1 / 12.0) + (0.2 / 8.0) = 0.00833 + 0.025 = 0.03333 Step 3: Calculate absolute error ΔA = A × (ΔA / A) = 96.0 × 0.03333 = 3.2 cm² Rounding: ΔA = 3 cm² (1 significant figure, matching the error) Step 4: Final answer A = (96 ± 3) cm² Percentage error = (3/96) × 100 = 3.1%

Worked Example 2 — Error in a power relation:

The resistance R = V / I is measured as R = (100 ± 5) Ω. If current I = (2.0 ± 0.1) A, find the power P = I²R with its error.

Given: R = 100 Ω, ΔR = 5 Ω I = 2.0 A, ΔI = 0.1 A Step 1: Calculate power P = I²R = (2.0)² × 100 = 4.0 × 100 = 400 W Step 2: Calculate relative error (ΔP / P) = 2(ΔI / I) + (ΔR / R) = 2 × (0.1 / 2.0) + (5 / 100) = 2 × 0.05 + 0.05 = 0.10 + 0.05 = 0.15 Step 3: Calculate absolute error ΔP = 400 × 0.15 = 60 W Step 4: Final answer P = (400 ± 60) W = (4.0 ± 0.6) × 10² W Percentage error = 15%

Worked Example 3 — Error in addition:

Two resistances R₁ = (100 ± 3) Ω and R₂ = (200 ± 4) Ω are connected in series. Find the total resistance with error.

In series: R_total = R₁ + R₂ Step 1: Total resistance R_total = 100 + 200 = 300 Ω Step 2: Absolute error (addition rule) ΔR_total = ΔR₁ + ΔR₂ = 3 + 4 = 7 Ω Step 3: Final answer R_total = (300 ± 7) Ω

3. Explain dimensional analysis with applications — checking dimensional consistency, deriving relations, and unit conversion, with worked examples.

Dimensional analysis is a method of studying physical quantities by examining their dimensions. It is based on the principle of homogeneity of dimensions: every term in a valid physical equation must have the same dimensions.

Applications:

Application 1 — Checking dimensional consistency:

Check the dimensional consistency of the equation: v² = u² + 2as

Left-hand side: v² Dimensions of v = [LT⁻¹] Dimensions of v² = [L²T⁻²] Right-hand side, term 1: u² Dimensions of u = [LT⁻¹] Dimensions of u² = [L²T⁻²] Right-hand side, term 2: 2as Dimensions of a = [LT⁻²] Dimensions of s = [L] Dimensions of 2as = [LT⁻²] × [L] = [L²T⁻²] All three terms have dimensions [L²T⁻²]. The equation is dimensionally consistent. ✓

Application 2 — Deriving a relation:

Derive an expression for the time period T of a simple pendulum using dimensional analysis, assuming T depends on length l and acceleration due to gravity g.

Step 1: Assume T = k × lᵃ × gᵇ where k is a dimensionless constant Step 2: Write dimensions of each quantity [T] = [L]ᵃ × [LT⁻²]ᵇ Step 3: Equate dimensions [M⁰L⁰T¹] = [L^(a+b) × T^(-2b)] Step 4: Equate powers For L: 0 = a + b ... (i) For T: 1 = −2b ... (ii) Step 5: Solve From (ii): b = −1/2 From (i): a = −b = 1/2 Step 6: Write the expression T = k × l^(1/2) × g^(−1/2) T = k × √(l / g) The dimensionless constant k = 2π (found experimentally). Therefore T = 2π√(l / g) ✓

Application 3 — Unit conversion:

Convert 1 newton (SI unit of force) to dyne (CGS unit of force).

1 newton = 1 kg m s⁻² We need to convert: kg → g: 1 kg = 10³ g m → cm: 1 m = 10² cm 1 N = 1 kg m s⁻² = 10³ g × 10² cm × s⁻² = 10⁵ g cm s⁻² = 10⁵ dyne Therefore, 1 N = 10⁵ dyne ✓

Worked Example 4 — Significant figures calculation:

Multiply 2.52 × 4.3 and express the result with correct significant figures.

Step 1: Perform the multiplication 2.52 × 4.3 = 10.836 Step 2: Determine the number of significant figures 2.52 has 3 significant figures 4.3 has 2 significant figures Step 3: Result must have the same number of sig figs as the least precise operand (2 sig figs) Step 4: Round 10.836 to 2 significant figures 10.836 → 11 Answer: 2.52 × 4.3 = 11 (2 significant figures)

Worked Example 5 — Screw gauge reading with zero error:

A wire is measured using a screw gauge with a pitch of 0.5 mm and 50 divisions on the circular scale. The main scale reading is 2 mm, the circular scale reading is 35, and the instrument has a positive zero error of +3 divisions. Find the correct diameter.

Given: Pitch = 0.5 mm Circular scale divisions = 50 Main Scale Reading (PSR) = 2 mm Circular Scale Reading (CSR) = 35 Positive zero error = +3 divisions Step 1: Calculate least count LC = Pitch / No. of circular divisions = 0.5 / 50 = 0.01 mm Step 2: Calculate observed reading Observed reading = PSR + (CSR × LC) = 2 + (35 × 0.01) = 2 + 0.35 = 2.35 mm Step 3: Calculate zero correction Positive zero error = +3 divisions Zero correction = −(3 × 0.01) = −0.03 mm Step 4: Calculate correct reading Correct reading = Observed reading + Zero correction = 2.35 + (−0.03) = 2.32 mm Diameter of the wire = 2.32 mm = 0.232 cm

Common Mistakes to Avoid

1. Mixing units in a single calculation. Always convert all quantities to the same unit system (preferably SI) before computing. Example: Do not add a length in cm to one in m without converting.

2. Confusing absolute error with relative error. Absolute error has the same unit as the quantity; relative error is dimensionless. Percentage error = relative error × 100.

3. Applying multiplication error rule to addition. For x = a + b, use Δx = Δa + Δb (not Δx/x = Δa/a + Δb/b). The relative error rule is for products and quotients only.

4. Forgetting the power in error propagation. For x = aⁿbᵐ, the relative error is (Δx/x) = n(Δa/a) + m(Δb/b). The powers n and m multiply the individual relative errors.

5. Counting leading zeros as significant figures. In 0.0045, the zeros before 4 are not significant. Only the 4 and 5 count — 2 significant figures.

6. Ignoring zero error in instruments. Always check for zero error before taking measurements with a vernier caliper or screw gauge, and apply the appropriate correction.

7. Confusing dimensional consistency with correctness. A dimensionally consistent equation may still be wrong (e.g., s = ut + at² is dimensionally correct but physically wrong — the correct equation is s = ut + ½at²). Dimensional analysis is necessary but not sufficient.

8. Using the wrong number of significant figures after rounding. After multiplication, match the least number of significant figures. After addition, match the least number of decimal places. Do not mix up these two rules.


Exam Checklist

  • [ ] I can list all 7 SI base quantities with their units and symbols
  • [ ] I can write the dimensional formula for at least 15 common physical quantities
  • [ ] I can derive dimensional formulae from the defining equation of a quantity
  • [ ] I can check dimensional consistency of equations term by term
  • [ ] I can use dimensional analysis to derive relations (e.g., time period of pendulum)
  • [ ] I can convert units between SI and CGS using the dimensional method
  • [ ] I can identify the number of significant figures in any number (including scientific notation)
  • [ ] I can apply significant figure rules for addition/subtraction and multiplication/division
  • [ ] I can calculate least count for vernier calipers and screw gauges
  • [ ] I can take a reading from a vernier caliper or screw gauge, including zero error correction
  • [ ] I can calculate absolute error, relative error, and percentage error from a set of measurements
  • [ ] I can propagate errors through addition, subtraction, multiplication, division, and power operations
  • [ ] I can distinguish between systematic, random, and gross errors and state remedies for each
  • [ ] I can state the limitations of dimensional analysis (at least 4 points)

Frequently Asked Questions

Only physical quantities with the same dimensional formula can be added or subtracted. Every term in a physically meaningful equation must have identical dimensions.