NCERT Solutions for Class 11 Physics Chapter 1: Units and Measurement

Concept. Each blank needs a unit conversion together with the correct number of significant figures. The golden rule is to treat units like algebraic symbols: replace each unit by its equivalent in the required unit, raise the conversion factor to the same power as the unit, and then multiply. Conversion never changes the number of significant figures.

(a) Volume of a cube of side 1 cm in m³

1. 1 cm = 10⁻² m, so 1 cm³ = (10⁻² m)³ = 10⁻⁶ m³ (the factor is cubed because volume has dimension [L³]).
2. Volume of cube = (side)³ = (1 cm)³ = 1 cm³.
3. Answer: 10⁻⁶ m³. (Check: 1 m³ = 10⁶ cm³ = 10³ litre.)

(b) Total surface area of a solid cylinder (r = 2.0 cm, h = 10.0 cm) in (mm)²

1. For a solid cylinder the surface consists of curved surface 2πrh plus two circular ends 2πr². Total = 2πr(r + h).
2. r + h = 2.0 cm + 10.0 cm = 12.0 cm.
3. Area = 2 × 3.14 × 2.0 cm × 12.0 cm = 150.8 cm² ≈ 1.5 × 10² cm² (two significant figures, because r = 2.0 cm has only two).
4. 1 cm = 10 mm ⇒ 1 cm² = 10² mm². Area = 1.5 × 10² × 10² mm².
5. Answer: 1.5 × 10⁴ (mm)².

(c) Distance covered in 1 s at 18 km h⁻¹

1. 18 km h⁻¹ = 18 × (1000 m) / (3600 s) = 18 × (5/18) m s⁻¹ = 5.0 m s⁻¹.
2. Distance = speed × time = 5.0 m s⁻¹ × 1 s.
3. Answer: 5.0 m (shortcut: to convert km h⁻¹ to m s⁻¹ multiply by 5/18).

(d) Density of lead (relative density 11.3)

1. Relative density = density of the substance / density of water at 4 °C. Density of water = 1 g cm⁻³ = 10³ kg m⁻³.
2. Density of lead = 11.3 × 1 g cm⁻³ = 11.3 g cm⁻³.
3. In SI: 11.3 × 10³ kg m⁻³ = 1.13 × 10⁴ kg m⁻³ (1 g cm⁻³ = 10⁻³ kg / 10⁻⁶ m³ = 10³ kg m⁻³).

Extra depth. Relative density (specific gravity) is a pure ratio and has no unit. The same numerical value gives density in g cm⁻³ directly, and in kg m⁻³ after multiplying by 1000. Always convert area by squaring and volume by cubing the length conversion factor: 1 cm² = 10² mm² = 10⁻⁴ m²; 1 cm³ = 10³ mm³ = 10⁻⁶ m³. For (b) the answer must contain two significant figures because the least precise datum, r = 2.0 cm, has two.

Conclusion: (a) 10⁻⁶; (b) 1.5 × 10⁴; (c) 5.0 m; (d) 11.3 g cm⁻³ and 1.13 × 10⁴ kg m⁻³.

Concept. Unit conversion by the factor-label method: write the quantity, multiply by conversion factors written so that the unwanted unit cancels, and carry the powers. Required constants: 1 kg = 10³ g, 1 m = 10² cm, 1 km = 10³ m, 1 h = 3600 s, and 1 light year (ly) = 9.46 × 10¹⁵ m.

Step-by-step for (c) (the part most students get wrong)

1. 3.0 m s⁻² = 3.0 × (10⁻³ km) / (1/3600 h)².
2. (1/3600 h)² = (1/3600)² h², so dividing by it multiplies by 3600² = 1.296 × 10⁷.
3. 3.0 × 10⁻³ × 1.296 × 10⁷ = 3.888 × 10⁴ ≈ 3.9 × 10⁴ km h⁻² (two significant figures).

Why (b) is a very small number: a light year is a unit of distance, the distance travelled by light in one year, so 1 ly = (3.00 × 10⁸ m s⁻¹)(3.156 × 10⁷ s) = 9.46 × 10¹⁵ m. One metre is therefore a tiny fraction (10⁻¹⁶) of it.

Extra depth. In (d) the conversion needs a prior step of writing the newton in base units, N = kg m s⁻². This is a standard idea: every derived unit must first be reduced to base units before converting. Check with dimensions: [G] = [M⁻¹ L³ T⁻²]; converting the three base units gives factors (10³)⁻¹ (10²)³ = 10³, so 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻² becomes 6.67 × 10⁻¹¹ × 10⁶/10³ = 6.67 × 10⁻⁸ in cgs. Generally, if a quantity has dimensions [M^a L^b T^c], its numerical value in cgs is the SI value multiplied by (10³)^a (10²)^b.

Conclusion: (a) 10⁷ g cm² s⁻²; (b) 1.06 × 10⁻¹⁶ ly; (c) 3.9 × 10⁴ km h⁻²; (d) 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹.

Concept. The numerical value of a physical quantity is inversely proportional to the size of the unit: nu = constant, i.e. n₁ u₁ = n₂ u₂. If the new unit is larger, the number measuring the same quantity is smaller. Use the dimensional formula of energy [M L² T⁻²] to relate the two systems.

1. Energy has dimensional formula [M¹ L² T⁻²]. In the old (SI) system 1 calorie = 4.2 J = 4.2 kg m² s⁻², so n₁ = 4.2 with units u₁ = kg m² s⁻².
2. In the new system the units are: mass M₂ = α kg, length L₂ = β m, time T₂ = γ s. The new unit of energy is u₂ = M₂ L₂² T₂⁻² = α β² γ⁻² (kg m² s⁻²).
3. Since the same energy is measured: n₁ u₁ = n₂ u₂, so n₂ = n₁ (u₁/u₂) = 4.2 × [1 / (α β² γ⁻²)].
4. Hence n₂ = 4.2 α⁻¹ β⁻² γ². Proved.

n₂ = n₁ [M₁/M₂]^a [L₁/L₂]^b [T₁/T₂]^c = 4.2 [1 kg/α kg]¹ [1 m/β m]² [1 s/γ s]⁻² = 4.2 α⁻¹ β⁻² γ²

Extra depth. Example 1: if the new unit of mass is a tonne (α = 10³) and β = γ = 1, the calorie becomes 4.2 × 10⁻³ new units, a smaller number because the new mass unit is larger. If the new system is CGS-like (α = 10⁻³, β = 10⁻², γ = 1), then 4.2 α⁻¹ β⁻² = 4.2 × 10³ × 10⁴ = 4.2 × 10⁷, which is the calorie in erg (1 J = 10⁷ erg). This general scaling is used whenever systems of units are changed.

Conclusion: Using n_1u1 = n_2u2 and the dimensional formula [M L² T⁻²], a calorie = 4.2 α⁻¹ β⁻² γ² in the new units.

Concept. A dimensional quantity is one that has a unit (length, speed, mass …). Its numerical value depends on the unit chosen: the same distance is 1 km, 1000 m or 10⁵ cm, and whether the number is ‘big’ is arbitrary. Words like ‘large’ and ‘small’ are therefore meaningful only when a standard (a quantity of the same kind) is specified and the ratio of the two is stated; a ratio of like quantities is a pure number that does not depend on units.

Example of the idea. Is 1 mm large or small? Compared with the size of an atom (10⁻¹⁰ m) it is 10⁷ times larger; compared with the size of the Earth (10⁷ m) it is 10⁻¹⁰ times smaller. The statement ‘1 mm is small’ is incomplete without the comparison.

Check with numbers. Number of molecules in 40 m³ of air at STP = (40 m³ / 22.4 × 10⁻³ m³ per mole) × 6.02 × 10²³ = 1.8 × 10³ mol × 6.02 × 10²³ ≈ 1 × 10²⁷, so the statement in (d) is valid with the standard ‘compared with the number of grains of sand in a bucket’.

Extra depth. In scientific statements, replace adjectives by ratios or orders of magnitude. A dimensionless number such as m_p/m_e = 1836, or the Mach number v/v_sound, is independent of the unit system and is therefore unambiguous. This is why physicists compare, for example, the strength of forces by dimensionless ratios (the gravitational force between two protons is about 10⁻³⁶ times the electrostatic force) rather than by saying a force is ‘weak’ or ‘strong’. In SI the ‘order of magnitude’ (Section 1.3) gives the quick comparison: the Earth’s diameter is 17 orders of magnitude larger than that of a hydrogen atom.

Conclusion: A dimensional quantity can be called large or small only relative to a specified standard of the same kind; statements (a) to (d) need a standard of comparison, whereas (e) and (f) already contain one.

Concept. When a unit is defined so that a certain physical constant has the numerical value 1, all other quantities are measured in terms of it. Here the new unit of length is the distance light travels in vacuum in one unit of time (one second), because speed = 1 new length unit per second.

1. Given: speed of light c = 1 (new length unit) per second, i.e. 1 new unit = c × 1 s = 3 × 10⁸ m (the unit is the ‘light-second’).
2. Time taken by light from the Sun to reach the Earth: t = 8 min 20 s = 8 × 60 s + 20 s = 480 s + 20 s = 500 s.
3. Distance = speed × time = (1 new unit per second) × (500 s) = 500 new units.
4. Distance between Sun and Earth = 500 new units.

Cross-check in SI: d = c t = 3 × 10⁸ m s⁻¹ × 500 s = 1.5 × 10¹¹ m, the standard value of one astronomical unit. Dividing by the new unit (3 × 10⁸ m) gives 1.5 × 10¹¹/3 × 10⁸ = 500. The distance is thus 500 light-seconds.

Extra depth. This idea underlies natural units used in relativity and particle physics, where c = 1 and distances are quoted in light-seconds, light-minutes or light-years (1 ly = 9.46 × 10¹⁵ m). In the 2018 SI the metre itself is defined by fixing the numerical value of the speed of light c = 299 792 458 m s⁻¹ and defining the second through the caesium frequency; thus the accepted unit of length is already tied to c. Also note that this problem does not require the actual value of c: whatever the unit, if c = 1 then distance = time of light travel, so the answer is simply 500.

More to know — units based on constants. Setting c = 1 is routine in relativity: time and distance then have the same unit (one second of time corresponds to one light-second of distance, 3 × 10⁸ m), and the Sun–Earth distance of 1 astronomical unit is 500 light-seconds. Other familiar astronomical units: 1 light year = 9.46 × 10¹⁵ m; 1 parsec = 3.08 × 10¹⁶ m (about 3.26 ly); the Moon is only 1.3 light-seconds away. The delay of about 8 min 20 s is also the time lag with which we see the Sun: we always see the Sun as it was 500 s ago.

Conclusion: The Sun–Earth distance is 500 new units (500 light-seconds = 1.5 × 10¹¹ m).

Concept. The precision of a measuring device is set by its least count, the smallest length it can measure reliably. The device with the smallest least count is the most precise.

1. Least count of (a): 0.05 mm = 5 × 10⁻⁵ m.
2. Least count of (b): 0.01 mm = 10⁻⁵ m, five times finer than (a).
3. Least count of (c): about 5 × 10⁻⁷ m (a wavelength of light), twenty times finer than (b) and a hundred times finer than (a).
4. Hence (c) the optical instrument is the most precise.

Ranking: optical instrument (5 × 10⁻⁷ m) > screw gauge (10⁻⁵ m) > vernier callipers (5 × 10⁻⁵ m). A device cannot resolve distances smaller than the wavelength of the light used; that is why wavelength of light is the practical limit for optical measurement of length.

Extra depth. Observe that more precision is not the same as more accuracy: precision is the resolution of the instrument (how many digits can be read), accuracy is how close the reading is to the true value. A badly calibrated screw gauge with a zero error is precise but inaccurate. Instruments with still finer resolution (electron microscopes, scanning tunnelling microscopes) use radiation of shorter wavelength to measure atomic sizes of 10⁻¹⁰ m, which a visible-light optical instrument cannot do.

Conclusion: The most precise device is (c), the optical instrument, since its least count (about 5 × 10⁻⁷ m) is smaller than that of the screw gauge (10⁻⁵ m) and the vernier callipers (5 × 10⁻⁵ m).

Concept. A microscope of linear magnification m produces an image whose width is m times the real width: m = (size of image)/(size of object). So the real thickness = (width seen)/(magnification). The mean of 20 observations is used to reduce random error.

1. Magnification m = 100; mean observed (image) width = 3.5 mm.
2. Real thickness = observed width / magnification = 3.5 mm / 100.
3. Thickness = 0.035 mm = 3.5 × 10⁻² mm = 3.5 × 10⁻⁵ m.
4. The thickness of the hair is about 0.035 mm (3.5 × 10⁻⁵ m).

Meaning of the result. 35 μm is of the order of 10⁻⁵ m, typical for human hair (from about 20 to 100 μm); the hair is thinner than the least count of a vernier callipers (0.05 mm) and similar to that of a screw gauge (0.01 mm), so a microscope is the proper instrument.

Extra depth. Twenty observations are taken because the width varies from place to place along the hair and each reading carries random error: the mean of many readings is a more reliable estimate and its random error decreases as 1/√N. The result has only two significant figures because 3.5 mm has two. The microscope magnification (100) is an exact number and does not limit the significant figures.

More to know — magnification and measuring small sizes. The image seen in a microscope is larger than the object by the magnification: object size = image size/magnification, so a hair that appears 3.5 mm wide through a 100× microscope is really 0.035 mm = 35 μm wide. Typical scales for comparison: human hair 20–100 μm; red blood cell about 7 μm; bacterium about 1 μm; wavelength of visible light 0.4–0.7 μm; virus about 0.1 μm; hydrogen atom about 10⁻¹⁰ m = 10⁻⁴ μm. A vernier callipers (0.05 mm = 50 μm) cannot measure the hair directly because its least count is of the same order as the quantity measured, so the relative error would be about 100%.

Conclusion: The thickness of the hair is 3.5 mm / 100 = 0.035 mm (3.5 × 10⁻⁵ m).

Concept. These three parts test practical measurement ideas: magnifying a small quantity by repetition (a), the difference between least count and true accuracy (b) and the reduction of random error by averaging (c).

(a) Diameter of a thread with a metre scale

1. Wind the thread closely on a uniform pencil (or glass rod) so that the turns touch each other, with no overlap and no gap; count the number of turns N, say N = 20.
2. Using the metre scale measure the length L of the wound portion from the first to the last turn.
3. Diameter of the thread d = L / N. If L = 1.60 cm for N = 20 turns, d = 0.080 cm = 0.80 mm.
4. Repeat several times with different N and take the mean value to reduce error.

This method works because the metre scale cannot measure a very small length directly (its least count is 1 mm), but it can measure a large multiple N d accurately; the uncertainty 0.5 mm in L is divided by N, so the uncertainty in d becomes 0.5/20 = 0.025 mm.

(b) Can the accuracy of a screw gauge be increased arbitrarily by increasing the divisions?

No. The least count of the screw gauge is pitch / number of divisions = 1.0 mm / 200 = 0.005 mm. Increasing the divisions makes the least count numerically smaller but not the accuracy without limit, because accuracy is limited by other factors:

• Mechanical imperfections of the screw: the actual pitch of the thread is not uniform, and there is backlash (lost motion) in the thread.
• The faces of the screw and the stud are never perfectly flat or parallel, and the object may be slightly compressed, depending on the pressure applied (the ratchet reduces but does not remove this).
• As the divisions become closer it becomes physically impossible to see and read them separately (they cannot be closer than the width of a visible line), and human judgement of alignment introduces error.
• Temperature changes expand the screw and the object. So there is a practical limit beyond which more divisions give no gain in true accuracy.

(c) 100 measurements versus 5 measurements of the diameter

1. The reading of the vernier callipers scatters about the true value because of random errors (slightly different positions along the rod, slightly different pressure, judgement of the coinciding division). Random errors are as likely to be positive as negative.
2. When many readings are averaged, positive and negative errors tend to cancel, so the mean is closer to the true value.
3. The uncertainty of the mean decreases as 1/√N. For N = 100 the random error of the mean is smaller than for N = 5 by √(100/5) = √20 ≈ 4.5 times.
4. So a set of 100 measurements gives a more reliable estimate (it reduces random errors, not systematic errors such as a zero error, which must be corrected separately).

Extra depth. The same idea (averaging to beat random noise) is used in every experiment, from the 20 observations of Ex 1.7 to the thousands of events accumulated at a particle collider. Systematic errors (zero error, wrong calibration) are not reduced by taking more readings; they are removed only by correcting the instrument. A screw gauge of 200 divisions is therefore not necessarily better than one of 100 if its screw is of poor quality.

Conclusion: (a) Wind N close turns on a pencil and divide the length by N; (b) no, accuracy is limited by screw defects, backlash and reading difficulty; (c) averaging 100 readings cancels random errors much better than averaging 5.

Concept. Linear magnification m is the ratio of a length in the image to the corresponding length in the object: m = l_image/l_object. Areas scale as the square of lengths, so the ratio of areas equals m².

1. Area of the house on the slide A₁ = 1.75 cm² = 1.75 × 10⁻⁴ m² (1 cm² = 10⁻⁴ m²).
2. Area of the house on the screen A₂ = 1.55 m².
3. Area magnification = A₂/A₁ = 1.55 m² / (1.75 × 10⁻⁴ m²) = 8.857 × 10³.
4. Since A ∝ l²: m² = 8.857 × 10³, so m = √(8.857 × 10³) = 94.1.
5. Linear magnification m ≈ 94. (Three significant figures in data, so 94.1; the NCERT answer is quoted as 94.)

Why the square root? If every length is magnified by m, then both the width and the height of the picture become m times larger, so the area becomes m × m = m² times larger. The 35 mm in the question is the width of the slide film and is not needed for the calculation (it is a distractor).

Extra depth. Common error: quoting 8.86 × 10³ as the ‘magnification’. That is the area magnification. Linear magnification is the square root: lengths on the screen are 94 times the lengths on the slide, e.g. a house 1 cm tall on the slide appears 94 cm tall on the screen. This is typical of Class XI conversions of ratio of areas, volumes (m³) and lengths (m); in similar problems on volumes use the cube root.

More to know — scaling laws. If lengths scale by m, then areas scale by m² and volumes by m³. For the projector: 1 cm on the slide becomes 94 cm on the screen, area 1.75 cm² becomes 94² × 1.75 cm² = 1.55 × 10⁴ cm² = 1.55 m² ✓. These ideas return in optics (lateral magnification of lenses and mirrors, Class XII) and in similar-figure problems of biology and engineering: a model one-tenth the size has one-hundredth the area and one-thousandth the volume.

Conclusion: The linear magnification of the projector-screen arrangement is about 94.

Concept. Count digits from the first non-zero digit to the last digit, including trailing zeros if there is a decimal point. Leading zeros (before the first non-zero digit) are never significant; the power of ten in scientific notation never counts.

Rules used (Section 1.3)

• Non-zero digits are always significant.
• Zeros between non-zero digits are significant.
• Leading zeros (to the left of the first non-zero digit) are not significant.
• Trailing zeros after the decimal point are significant.
• The power of ten in scientific notation is not counted.

Scientific-notation check: (a) 7 × 10⁻³ (1 digit); (c) 2.370 × 10⁻¹ (4 digits); (f) 6.032 × 10⁻⁴ (4 digits). Writing numbers in scientific notation shows at once how many significant figures there are.

Extra depth. The number of significant figures does not change if the unit is changed: 0.2370 g cm⁻³ = 237.0 kg m⁻³ (still four). Be careful with a number like 2300 (no decimal point): the trailing zeros are ambiguous and are taken as not significant (two figures); writing 2.300 × 10³ removes the ambiguity (four figures). The number of significant figures indicates the precision: 0.007 m² is uncertain by 1 part in 7 (about 14%), while 0.0006032 m² is uncertain by 1 part in 6000.

More to know — why significant figures matter. A reading of 0.007 m² (one significant figure) implies a possible error of 0.0005 m² (about 7%), while 0.0006032 m² (four significant figures) implies an error of about 0.00000005 m² (about 0.01%). Reporting more figures than the instrument justifies would claim a precision that does not exist. Typical exam trap: 6.320 J must keep its trailing zero because it records that the hundredths–thousandths place was measured, whereas 6.32 J would claim less precision.

Conclusion: (a) 1; (b) 3; (c) 4; (d) 4; (e) 4; (f) 4.

Concept. Convert all dimensions to the same unit (metre), then compute and round to the least number of significant figures among the factors. The thickness 2.01 cm has only three significant figures, so the volume must be reported to three; the length and breadth have four.

1. l = 4.234 m (4 s.f.), b = 1.005 m (4 s.f.), t = 2.01 cm = 0.0201 m = 2.01 × 10⁻² m (3 s.f.).
2. Area of the sheet. A sheet has two faces of area l b and four edges: total surface area = 2(lb + bt + tl). lb = 4.234 × 1.005 = 4.2552; bt = 1.005 × 0.0201 = 0.0202; tl = 0.0201 × 4.234 = 0.0851.
3. Sum = 4.2552 + 0.0202 + 0.0851 = 4.3605 m²; area = 2 × 4.3605 = 8.721 m². Since the thickness has three significant figures the area is 8.72 m².
4. Volume. V = l b t = 4.234 m × 1.005 m × 2.01 × 10⁻² m = 8.5529 × 10⁻² m³. Rounding to three significant figures gives 8.55 × 10⁻² m³.

Note on the meaning of ‘area’. For a sheet, ‘area’ means the total surface area (both faces and the edges), the standard NCERT answer being 8.72 m². If only the area of one face were wanted, it would be 4.255 m². The two edge strips together contribute only 0.2% of the area, but they are included when the sheet is taken as a cuboid.

Extra depth. The rule applied is that in a product the result has as many significant figures as the factor with the fewest significant figures: here 2.01 cm (three). In the addition inside the bracket the rule is by decimal places, but the smaller terms (0.0202, 0.0851) have more decimal places than needed, so the final area is governed by the least precise factor, the thickness, giving 8.72. Always keep one extra digit during the calculation (4.2552, not 4.26) and round only at the end.

Conclusion: The sheet has a surface area of 8.72 m² and a volume of 8.55 × 10⁻² m³.

Concept. For addition and subtraction the rule is in terms of decimal places, not significant figures: the result must have as many decimal places as the number with the fewest decimal places, after all the quantities are expressed in the same unit.

(a) Total mass of the box

1. Convert the gold masses to kilograms: 20.15 g = 0.02015 kg and 20.17 g = 0.02017 kg.
2. Total = 2.30 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg.
3. The balance reads 2.30 kg, with only two decimal places (hundredths of a kilogram), so the answer must have two decimal places.
4. Rounding 2.34032 to two decimal places: 2.34 kg.

(b) Difference in the masses of the gold pieces

1. Difference = 20.17 g − 20.15 g = 0.02 g.
2. Both numbers have two decimal places (the measurement precision is 0.01 g), so the answer keeps two decimal places: 0.02 g (one significant figure).

Meaning. The grocer’s balance cannot detect the gold pieces separately: adding 40.32 g to a box of 2.30 kg changes the reading from 2.30 kg to only 2.34 kg. The difference of 0.02 g between the two pieces is much smaller than the balance’s precision (10 g) and so is only meaningful if the gold pieces were weighed on a more sensitive balance, as the 0.01 g precision of their masses shows.

Extra depth. In subtraction, significant figures can be lost: 20.17 g and 20.15 g each have four significant figures, but their difference 0.02 g has only one. This is why subtracting two nearly equal numbers is bad practice experimentally: the relative error of the difference becomes large (here ±0.01 g in 0.02 g is 50%). The grocer’s-balance total, on the other hand, is limited by the least precise item, the 2.30 kg reading.

More to know — absolute and relative error. The grocer’s balance has an absolute uncertainty of ±0.01 kg (10 g) in 2.30 kg (relative error 0.4%), so extra 40 g changes the reading by about four units in the last place only. The gold pieces, with ±0.01 g in 20.15 g, have a relative error of 0.05%. Their difference, 0.02 g, has an absolute uncertainty of up to ±0.02 g (errors add in subtraction), i.e. up to 100% relative error: a clear illustration of why differences of nearly equal numbers lose significance.

Conclusion: (a) The total mass is 2.34 kg; (b) the difference in the masses of the gold pieces is 0.02 g.

Concept. Use the principle of homogeneity of dimensions: only quantities with the same dimensions can be added or subtracted, and the argument of a power such as (1 − x)^1/2 must be dimensionless if ‘1’ (a pure number) appears in it.

1. In the expression (1 − v²)^1/2 the number 1 is dimensionless, so v² must also be dimensionless for the subtraction 1 − v² to be meaningful.
2. But v² has dimensions [L² T⁻²] (it is a speed squared). To make it dimensionless divide by another quantity with the same dimensions: the only speed in the problem is c, with [c²] = [L² T⁻²].
3. Hence the correct term is v²/c², which is dimensionless.
4. The relation is therefore m = m_o / (1 − v²/c²)^1/2.

Physical check. For v ≪ c, v²/c² ≈ 0 and m ≈ m_o: the mass is the rest mass, as in everyday motion. As v → c the denominator tends to zero and m grows without limit: no massive particle can reach the speed of light. At v = 0.6c, m = m_o/√(1 − 0.36) = m_o/0.8 = 1.25 m_o.

Extra depth. Dimensional analysis shows where c must occur but cannot fix dimensionless numbers or functions: it cannot tell (1 − v²/c²)^1/2 from, say, (1 − v²/c²)^1/2 multiplied by a pure number, or from a function of v/c with the same dimensions. The factor v² and the exponent 1/2 are taken from the boy’s memory of Einstein’s relation; dimensional analysis only certifies that the corrected formula is consistent. This is exactly the limitation stated in Section 1.6: a dimensionally correct equation need not be correct.

Conclusion: The missing c belongs under v²: m = m_o / (1 − v²/c²)^1/2.

Concept. The atom is treated as a sphere whose volume is (4/3)πr³, and one mole contains Avogadro’s number N_A = 6.022 × 10²³ atoms. The total atomic volume of a mole is N_A times the volume of one atom. The ‘size’ 0.5 Å is taken as the radius of the hydrogen atom, as in the NCERT solution.

1. Radius r = 0.5 Å = 0.5 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m.
2. Volume of one hydrogen atom V₁ = (4/3)πr³ = (4/3) × 3.14 × (5 × 10⁻¹¹ m)³ = (4/3) × 3.14 × 1.25 × 10⁻³¹ m³ = 5.2 × 10⁻³¹ m³.
3. Total atomic volume of one mole = N_A V₁ = 6.022 × 10²³ × 5.2 × 10⁻³¹ m³ = 3.2 × 10⁻⁷ m³.
4. The size is given to one significant figure (0.5 Å), so the answer is about 3 × 10⁻⁷ m³.

Sense of scale. 3 × 10⁻⁷ m³ is only 0.3 cm³ (the size of a pea) for a whole mole (6.0 × 10²³ atoms, mass about 1 g). The space occupied by the matter of atoms is tiny; the volume of a mole of hydrogen gas under normal conditions is 22.4 litres (Ex 1.15), much larger. A solid or liquid, in which atoms are packed closely, has a molar volume of the order of 10 cm³ — still about a factor of 30 more than the sum of the atomic volumes because atoms are not tightly packed spheres.

Extra depth. If the 0.5 Å had been the diameter, the radius would be 0.25 Å and the volume eight times smaller. The question states ‘size’ and NCERT takes it as the radius (using standard Bohr radius of hydrogen 0.53 Å, which supports the radius interpretation). The calculation relies on the order of magnitude: 10⁻³¹ m³ for one atom, 10⁻⁷ m³ for a mole.

Conclusion: The total atomic volume of a mole of hydrogen atoms is about 3 × 10⁻⁷ m³.

Concept. The molar volume of an ideal gas at STP is the volume occupied by the container space for 6.022 × 10²³ molecules; the atomic volume is the volume of the molecules themselves. Their ratio tells us how much of the gas is actually matter and how much is empty space.

1. Molar volume V_m = 22.4 L = 22.4 × 10⁻³ m³ = 2.24 × 10⁻² m³.
2. Atomic volume of a mole of hydrogen (Ex 1.14, size 1 Å taken as for the atom, radius 0.5 Å) V_a = 3 × 10⁻⁷ m³.
3. Ratio V_m / V_a = 2.24 × 10⁻² m³ / 3 × 10⁻⁷ m³ = 7.5 × 10⁴ (with the more precise V_a = 3.15 × 10⁻⁷ m³, 7.1 × 10⁴).
4. The ratio is about 7 × 10⁴, i.e. of the order of 10⁵.

Why is the ratio so large?

• In a gas the molecules are very far apart compared with their own size: the average separation at STP is about 3.3 nm = 33 Å, roughly 30 times the molecular diameter (1 Å).
• The molecules actually occupy only about 1 part in 10⁵ of the volume available; the rest is empty space. Since volume goes as (length)³, a separation of 30 diameters gives a volume ratio of about 30³ ≈ 3 × 10⁴, of the order observed.
• The intermolecular forces are weak at such distances, so molecules move almost independently; this is why gases are compressible and why an ideal gas can treat molecules as point particles.
• In solids and liquids the molecules are in contact (separation about one diameter), so the corresponding ratio is only of order 1–10.

Check using the volume per molecule. V_m/N_A = 2.24 × 10⁻² m³ / 6.02 × 10²³ = 3.7 × 10⁻²⁶ m³ per molecule, a cube of side (3.7 × 10⁻²⁶)^1/3 = 3.3 × 10⁻⁹ m. Dividing by the molecular diameter 10⁻¹⁰ m gives about 33. Cubing gives 3.6 × 10⁴, consistent in order of magnitude with the ratio found above (the extra factor 2 arises from the sphere factor 4π/3 × (1/2)³ = 0.52 for the volume of the molecule).

Extra depth. The empty-space picture explains why gases can be compressed to the liquid state, why they diffuse and mix quickly, and why the ideal gas law (Chapter 12) neglects molecular volume. Compare the ratio for liquid water: 18 cm³ per mole against the atomic volume of about 10 cm³ (a factor of about 2): in liquids and solids the molecules are almost touching. The huge gas ratio of about 10⁵ is the order-of-magnitude result NCERT expects.

Conclusion: The ratio of molar volume to atomic volume is about 10⁵ (7 × 10⁴) because the gas molecules are very far apart compared with their size, so most of the volume is empty space.

Concept. The effect is called parallax: the apparent shift in the position of an object against a distant background when the observer changes position. What the eye notices is the angular velocity of the line of sight to the object, not its linear velocity, and this depends strongly on the distance of the object.

1. Let the train (observer) move with speed v. In a short time t it moves a distance s = v t along the track, from O₁ to O₂.
2. For an object at perpendicular distance d from the track, the line of sight turns through an angle θ (in radian) ≈ s/d = v t/d. The angular velocity of the line of sight is ω = θ/t = v/d.
3. Nearby trees and houses have small d, so ω = v/d is large: the line of sight turns quickly and the object sweeps backwards across your field of view. The sense is opposite to the train’s motion because the observer moves forward relative to the object.
4. Distant objects (hill tops, the Moon, stars) have very large d, so ω = v/d is almost zero; their direction in the sky hardly changes during the observation, so they appear stationary (or, since you know you are moving, seem to move along with you).

Application. The same principle is used to measure the distance of nearby stars (annual parallax) and in range-finding: from the angular shift θ for a known baseline b the distance is d = b/θ. Our two eyes use binocular parallax for depth perception.

Extra depth. Notice that the linear velocity relative to the train is the same for all objects (all appear to move with speed v relative to the train in the train frame); the difference lies in the angle turned through per second, which is what the eye senses. This is why in a flying aeroplane, ground objects directly below seem to rush past while distant clouds drift slowly. Parallax is also the reason that the stars show no apparent motion over a human lifetime even though the Earth moves at 30 km s⁻¹ around the Sun.

Conclusion: Because the angle turned through by the line of sight in a given time is large for near objects (θ ≈ vt/d) and nearly zero for distant objects, near objects seem to move rapidly backwards and distant objects seem stationary (parallax).

Concept. At such high temperatures one might guess that the Sun’s matter is a gas of very low density. The data allow us to test this guess by computing the mean density ρ = mass / volume, treating the Sun as a sphere of volume (4/3)πR³.

1. Volume of the Sun V = (4/3)πR³ = (4/3) × 3.14 × (7.0 × 10⁸ m)³ = (4/3) × 3.14 × 3.43 × 10²⁶ m³ = 1.44 × 10²⁷ m³.
2. Mean density ρ = M/V = 2.0 × 10³⁰ kg / 1.44 × 10²⁷ m³ = 1.4 × 10³ kg m⁻³.
3. Compare: density of water = 1.0 × 10³ kg m⁻³; air at STP = 1.3 kg m⁻³. The Sun’s mean density, 1.4 × 10³ kg m⁻³ = 1.4 g cm⁻³, is of the order of the density of liquids and solids, not of gases.
4. Guess (‘gas-like’) is wrong: the mean density is about 1.4 times that of water and about 1000 times that of air.

Why is the density so high although the matter is a gas (plasma)? The Sun’s own enormous gravity compresses the matter: the pressure at the centre is about 10¹⁶ Pa, which keeps the particles close together despite the very high temperature (the gas pressure balances gravity). The matter is strongly ionised, so electrons and nuclei are separate and can be packed much more densely than in ordinary atoms. The density is not uniform: it is highest at the centre (about 10⁵ kg m⁻³, much larger than the mean) and falls to about 10⁻⁴ kg m⁻³ near the surface, so 1.4 × 10³ kg m⁻³ is only the average value (supplementary data, not from the NCERT text).

Extra depth. The test illustrates a lesson of the chapter: use the order of magnitude to judge the reasonableness of a result (Section 1.3). The mean density of the Earth is 5.5 × 10³ kg m⁻³, so the Sun is about one quarter as dense as the Earth although it is 1.3 million times larger in volume. The mass of the Sun (2.0 × 10³⁰ kg) is about 3.3 × 10⁵ times the mass of the Earth (6.0 × 10²⁴ kg).

Conclusion: The mean density of the Sun is about 1.4 × 10³ kg m⁻³, in the range of densities of solids and liquids, not of gases, because of the huge gravitational compression.