Exercises — Units and Measurements

17 questions✓ Free · step-by-step

1.14 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.1
Fill in the blanks: (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h-1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm-3 or ....kg m-3.
Hint. For (b), a solid cylinder's total surface area includes both circular ends plus the curved side, so use 2πr(r + h).
Step 1 — (a) Volume of the cube. Side = 1 cm = 10⁻² m, so volume = (10⁻²)³ = 10⁻⁶ m³.

Step 2 — (b) Total surface area of the cylinder. Using 2πr(r + h) with r = 2.0 cm, h = 10.0 cm: 2π(2.0)(12.0) = 48π ≈ 150.8 cm². Since 1 cm² = 100 mm², this is 150.8 × 100 ≈ 1.5 × 10⁴ mm², kept to 2 significant figures because the radius (2.0 cm) has only two.

Step 3 — (c) Speed conversion. 18 km h⁻¹ = 18 × 1000 m / 3600 s = 5 m s⁻¹, so the vehicle covers 5 m in 1 s.

Step 4 — (d) Density of lead. Relative density (specific gravity) is density relative to water, so density = 11.3 × (density of water) = 11.3 g cm⁻³, which converts to 11.3 × 1000 kg m⁻³ = 1.13 × 10⁴ kg m⁻³, since 1 g cm⁻³ equals exactly 1000 kg m⁻³.

✦ Answer: (a) 10⁻⁶ m³ (b) ≈ 1.5 × 10⁴ mm² (c) 5 m (d) 11.3 g cm⁻³ = 1.13 × 10⁴ kg m³
Where students slip. Using only the curved surface area (2πrh) in (b) and forgetting the two circular end caps — a 'solid cylinder' has a closed top and bottom, so both end areas (2πr²) belong in the total.
1.24 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.2
Fill in the blanks by suitable conversion of units: (a) 1 kg m2 s-2 = ....g cm2 s-2 (b) 1 m = ..... ly (c) 3.0 m s-2 = .... km h-2 (d) G = 6.67 × 10-11 N m2 (kg)-2 = .... (cm)3 s-2 g-1.
Hint. For (d), first expand the newton into base units (kg m s⁻²) before converting each factor separately.
Step 1 — (a) 1 kg = 1000 g and 1 m² = 10⁴ cm², so 1 kg m² s⁻² = 1000 × 10⁴ g cm² s⁻² = 10⁷ g cm² s⁻².

Step 2 — (b) One light year is 9.46 × 10¹⁵ m, so 1 m equals 1/(9.46 × 10¹⁵) ly ≈ 1.06 × 10⁻¹⁶ ly.

Step 3 — (c) 1 m = 10⁻³ km and 1 s⁻² becomes (3600)² h⁻² = 1.296 × 10⁷ h⁻² (since there are 3600 s in an hour, converting per-second-squared to per-hour-squared multiplies by 3600²). So 3.0 m s⁻² = 3.0 × 10⁻³ × 1.296 × 10⁷ km h⁻² ≈ 3.9 × 10⁴ km h⁻².

Step 4 — (d) Expand N = kg m s⁻², so N m² kg⁻² = m³ s⁻² kg⁻¹. Converting m³ → cm³ multiplies by 10⁶, and kg⁻¹ → g⁻¹ multiplies by 10⁻³ (since 1 kg⁻¹ = 10⁻³ g⁻¹). So G = 6.67 × 10⁻¹¹ × 10⁶ × 10⁻³ = 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹.

✦ Answer: (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
Where students slip. In (d), converting kg⁻¹ to g⁻¹ by multiplying by 1000 instead of dividing — a per-kilogram quantity gets smaller in magnitude, not larger, once re-expressed per gram, because a gram is a smaller unit than a kilogram.
1.33 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.3
A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α-1 β-2 γ 2 in terms of the new units.
Hint. Use the general conversion rule n2 = n1 (M1/M2)^a (L1/L2)^b (T1/T2)^c, where a, b, c come straight from energy's dimensional formula [M L2 T-2].
Step 1 — Energy's dimensional formula is [M¹ L² T⁻²], so for a quantity with this formula, n₁M₁¹L₁²T₁⁻² = n₂M₂¹L₂²T₂⁻², giving n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)² (T₁/T₂)⁻².

Step 2 — Here n₁ = 4.2, with the old units being kg, m, s (M₁ = 1 kg, L₁ = 1 m, T₁ = 1 s), and the new units being M₂ = α kg, L₂ = β m, T₂ = γ s.

Step 3 — Substituting: n₂ = 4.2 × (1/α)¹ × (1/β)² × (1/γ)⁻² = 4.2 × α⁻¹ × β⁻² × γ², since raising (1/γ) to the power ⁻2 flips it to γ².

✦ Answer: n₂ = 4.2 α⁻¹ β⁻² γ², exactly as required to show.
Where students slip. Forgetting to invert the exponent on the time factor — T₁/T₂ raised to the power −2 becomes (T₂/T₁)² = γ², not γ⁻².
1.43 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.4
Explain this statement clearly: 'To call a dimensional quantity large or small is meaningless without specifying a standard for comparison'. In view of this, reframe the following statements wherever necessary: (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light.
Hint. Check each statement: does it already compare two named quantities of the same kind, or does it just call something 'big'/'small' with no explicit reference point?
Step 1 — Why the statement is true. Words like 'large' and 'small' are relative, meaning a value only counts as large or small once measured against some reference of the same kind — a mountain is small compared to a planet but enormous compared to an ant, so the word alone conveys nothing without stating what it is being compared to.

Step 2 — Reframing (a)–(d), which name no comparison object. (a) Atoms are very small objects compared to a tennis ball. (b) A jet plane moves with a speed greater than that of sound (or of a car on a highway). (c) The mass of Jupiter is very large compared to the mass of a cricket ball. (d) The air inside this room contains a far larger number of molecules than the air inside a football.

Step 3 — (e) and (f) need no reframing. Both already state a comparison between two named physical quantities of the same kind (proton's mass vs electron's mass; speed of sound vs speed of light), so they are already meaningful as written.

✦ Answer: (a)–(d) need an explicit comparison object added (examples above); (e) and (f) are already valid comparisons and require no change.
Where students slip. Reframing (e) or (f) unnecessarily — they already name two specific quantities being compared, which is exactly what the statement asks for; only the bare, comparison-free statements (a)–(d) need fixing.
1.52 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.5
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Hint. If c = 1 in the new unit, then distance = c × time collapses to distance = time (expressed in seconds), since multiplying by 1 changes nothing.
Step 1 — Convert the time to seconds. 8 min 20 s = 8 × 60 + 20 = 500 s.

Step 2 — Apply distance = speed × time with c = 1 new unit of length per second. Distance = c × t = 1 × 500 = 500 new units of length, since multiplying any number of seconds by 1 leaves that number unchanged.

✦ Answer: The Sun–Earth distance is 500 of the new units of length (this new unit is exactly what is normally called a 'light-second').
Where students slip. Leaving the time in minutes-and-seconds form instead of converting fully to seconds first — the new unit is defined per second, so the time must be a single number of seconds before multiplying.
1.62 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.6
Which of the following is the most precise device for measuring length: (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light?
Hint. Work out the least count of (a) and (b) in millimetres, then compare both against the wavelength of visible light, which is only a few hundred nanometres.
Step 1 — Least count of the vernier callipers. Assuming a 1 mm main-scale division, least count = 1 mm / 20 = 0.05 mm.

Step 2 — Least count of the screw gauge. Least count = pitch / number of circular divisions = 1 mm / 100 = 0.01 mm.

Step 3 — Compare with the optical instrument. Visible light has a wavelength of roughly 5 × 10⁻⁷ m = 5 × 10⁻⁴ mm, which is far finer than even the screw gauge's 0.01 mm, so the optical instrument resolves the smallest length differences of the three.

✦ Answer: (c), the optical instrument, is the most precise — its resolution (a wavelength of light, of order 10⁻⁴ mm) is far smaller than the screw gauge's 0.01 mm or the vernier callipers' 0.05 mm.
Where students slip. Assuming the screw gauge with more divisions (100 vs 20) is automatically the most precise overall — it does beat the vernier callipers, but both are still coarser than an optical method working at the scale of a light wavelength.
1.72 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.7
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?
Hint. The 3.5 mm width seen through the microscope is the magnified image, not the real hair — divide out the magnification to recover the true size.
Step 1 — Relate observed width to true thickness. Magnification is defined as (observed/apparent size) / (true size), so true thickness = observed width / magnification.

Step 2 — Substitute the values. True thickness = 3.5 mm / 100 = 0.035 mm = 3.5 × 10⁻⁵ m.

✦ Answer: The estimated thickness of the hair is 3.5 × 10⁻⁵ m (0.035 mm).
Where students slip. Reporting 3.5 mm itself as the hair's thickness — that figure is the size of the magnified image in the eyepiece, not the actual hair, so it must be divided by the magnification first.
1.83 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.8
Answer the following: (a) You are given a thread and a metre scale. How will you estimate the diameter of the thread? (b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Hint. For (a), think about how to turn a quantity too small to read directly on a metre scale into one large enough to read — by repeating it many times over.
Step 1 — (a) Measuring the thread's diameter. Wind the thread closely, turn after turn with no gaps or overlaps, around the metre scale (or a pencil) for a fixed number of turns, say N. Measure the total length L that these N turns occupy on the scale, and take diameter = L / N — this turns an unmeasurably thin single width into a measurable total length.

Step 2 — (b) Increasing circular-scale divisions on the screw gauge. No, accuracy cannot be increased without limit this way, since beyond a certain point the additional divisions become impossible to manufacture accurately and to read reliably, and effects such as backlash (looseness in the screw threads) and friction start to dominate the error — these do not shrink just because the scale is marked more finely.

Step 3 — (c) Why 100 measurements beat 5. Each individual reading carries some random error that fluctuates unpredictably in size and direction. Averaging over many more readings lets these random fluctuations increasingly cancel out, so the mean from 100 measurements sits closer to the true diameter than the mean from just 5, which means a small sample is much more likely to be skewed by a few unusually high or low readings.

✦ Answer: (a) diameter = (length of N close turns) / N (b) No — backlash, friction, and manufacturing limits cap the achievable accuracy regardless of division count (c) more readings average out random error better, giving a more reliable mean.
Where students slip. In (b), assuming more divisions always means more precision — the circular scale's fineness only helps up to the point where mechanical imperfections in the screw itself become the larger source of error.
1.92 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.9
The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement?
Hint. Area scales as the square of linear magnification, so take the square root of the area ratio to get the linear magnification.
Step 1 — Express both areas in the same unit. 1.55 m² = 1.55 × 10⁴ cm².

Step 2 — Find the area magnification. Area magnification = screen area / slide area = 1.55 × 10⁴ / 1.75 ≈ 8857.

Step 3 — Take the square root to get linear magnification. Linear magnification = √8857 ≈ 94.1, since area magnification is the square of linear magnification for a similar (proportionally scaled) shape.

✦ Answer: The linear magnification of the projector-screen arrangement is about 94.
Where students slip. Reporting the area ratio (≈ 8857) itself as the 'magnification' — linear magnification is the square root of that ratio, not the ratio itself.
1.103 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.10
State the number of significant figures in the following: (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm-3 (d) 6.320 J (e) 6.032 N m-2 (f) 0.0006032 m2
Hint. Leading zeros before the first non-zero digit are never significant; every digit from the first non-zero digit onward counts, including trailing zeros after a decimal point.
Step 1 — (a) 0.007 m²: only the digit 7 counts, since the zeros before it are just placeholders — 1 significant figure.

Step 2 — (b) 2.64 × 10²⁴ kg: the power of ten carries no significance; only 2, 6, 4 count — 3 significant figures.

Step 3 — (c) 0.2370 g cm⁻³: leading zero is not significant, but 2, 3, 7, 0 all are, including the trailing zero after the decimal — 4 significant figures.

Step 4 — (d) 6.320 J: 6, 3, 2, 0 are all significant, the trailing zero included because it follows a decimal point — 4 significant figures.

Step 5 — (e) 6.032 N m⁻²: 6, 0, 3, 2 all count, since a zero between two non-zero digits is always significant — 4 significant figures.

Step 6 — (f) 0.0006032 m²: the leading zeros are placeholders, so only 6, 0, 3, 2 count — 4 significant figures.

✦ Answer: (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4
Where students slip. Counting the leading zeros in (a), (c) and (f) as significant — zeros used only to place the decimal point never count, no matter how many of them there are.
1.113 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.11
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Hint. Convert every length to the same unit before multiplying, and remember that a multiplication's result can only keep as many significant figures as the least-precise input has.
Step 1 — Convert thickness to metres. 2.01 cm = 0.0201 m, which has 3 significant figures.

Step 2 — Compute the face area. Area = length × breadth = 4.234 × 1.005 = 4.25517 m². Both inputs have 4 significant figures, so the area rounds to 4 significant figures: 4.255 m².

Step 3 — Compute the volume. Volume = area × thickness = 4.25517 × 0.0201 ≈ 0.085529 m³. The thickness has only 3 significant figures, so this is the limiting factor and the volume rounds to 3 significant figures: 8.55 × 10⁻² m³.

✦ Answer: Area ≈ 4.255 m²; Volume ≈ 8.55 × 10⁻² m³.
Where students slip. Rounding the area to 3 significant figures because the thickness (used later, for volume) has only 3 — each multiplication step is rounded according to its own inputs, not a figure that only becomes relevant in a later step.
1.122 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.12
The mass of a box measured by a grocer's balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?
Hint. For addition and subtraction, the rule is about decimal places, not significant-figure counts — round the result to match the least number of decimal places among the inputs.
Step 1 — (a) Add the masses in the same unit. 2.30 kg + 20.15 g + 20.17 g = 2.30 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg. The grocer's balance value (2.30 kg) is reliable to only 2 decimal places, so the sum must be rounded to 2 decimal places: 2.34 kg.

Step 2 — (b) Subtract the gold-piece masses. 20.17 g − 20.15 g = 0.02 g. Both values are known to 2 decimal places, so the difference is correctly reported to 2 decimal places as well: 0.02 g.

✦ Answer: (a) Total mass ≈ 2.34 kg (b) Difference in mass = 0.02 g.
Where students slip. Reporting the raw sum 2.34032 kg unrounded — the grocer's balance limits the whole measurement to 2 decimal places in kg, so the extra digits from the precise gold-piece masses are false precision.
1.132 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.13
A famous relation in physics relates 'moving mass' m to the 'rest mass' m0 of a particle in terms of its speed v and the speed of light, c. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: m = m0 / (1 - v2)1/2. Guess where to put the missing c.
Hint. Whatever sits inside the square root must be dimensionless, since you can never take the square root of '1 minus a squared speed'.
Step 1 — Check dimensions of the boy's version. Inside the root he has 1 − v², but 1 is a pure number while v² has dimensions of speed squared — these cannot be subtracted, since only quantities with matching dimensions can be added or subtracted.

Step 2 — Fix it with c. Dividing v by c gives a dimensionless ratio (v/c), so (v/c)² can validly be subtracted from the pure number 1.

✦ Answer: m = m₀ / √(1 − v²/c²) — the missing c² belongs under v² inside the root, giving the dimensionless ratio v²/c².
Where students slip. Placing c outside the square root instead of dividing v by it inside — that would leave 1 − v² (a dimensionally invalid subtraction) untouched inside the root.
1.143 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.14
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10-10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?
Hint. Find the volume of one atom by treating it as a tiny sphere of radius 0.5 Å, then scale up by Avogadro's number for one mole.
Step 1 — Volume of a single hydrogen atom. Treating the atom as a sphere of radius r = 0.5 Å = 5 × 10⁻¹¹ m, V = (4/3)πr³ = (4/3)π × (5 × 10⁻¹¹)³ ≈ 5.24 × 10⁻³¹ m³.

Step 2 — Scale to one mole. One mole contains Avogadro's number of atoms, Nₐ ≈ 6.02 × 10²³, so total atomic volume = 5.24 × 10⁻³¹ × 6.02 × 10²³ ≈ 3.15 × 10⁻⁷ m³.

✦ Answer: The total atomic volume of a mole of hydrogen atoms is about 3.15 × 10⁻⁷ m³.
Where students slip. Using the angstrom value (0.5) directly as the radius in metres instead of converting it to 5 × 10⁻¹¹ m first — skipping the unit conversion throws the whole answer off by a factor of 10¹⁰.
1.153 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.15
One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?
Hint. Reuse the same sphere-volume method as the previous question, this time with the molecule's own radius (1 Å, not the atom's 0.5 Å), then compare against the 22.4 L gas volume.
Step 1 — Molecular volume for one mole. With radius r = 1 Å = 10⁻¹⁰ m, volume of one molecule = (4/3)πr³ ≈ 4.19 × 10⁻³⁰ m³. For one mole (Nₐ ≈ 6.02 × 10²³ molecules), total molecular volume ≈ 4.19 × 10⁻³⁰ × 6.02 × 10²³ ≈ 2.52 × 10⁻⁶ m³.

Step 2 — Molar (gas) volume in the same unit. 22.4 L = 22.4 × 10⁻³ m³ = 2.24 × 10⁻² m³.

Step 3 — Take the ratio. Ratio = 2.24 × 10⁻² / 2.52 × 10⁻⁶ ≈ 8.9 × 10³, of order 10⁴.

Step 4 — Why the ratio is so large. The molecules themselves take up only a tiny fraction of the total gas volume, which means almost all of the 22.4 L is empty space between molecules — unlike a liquid or solid, where molecules sit packed against one another with almost no gap.

✦ Answer: The ratio is of order 10⁴ (≈ 8.9 × 10³), because gas molecules are spread far apart with mostly empty space between them, whereas in a liquid or solid the same molecules would be packed close together.
Where students slip. Reusing the atomic radius (0.5 Å) from the previous question instead of the molecular radius (1 Å) this question specifies — the two questions deliberately use different sizes.
1.162 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.16
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train's motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary.
Hint. This is the same parallax effect the chapter uses to measure the distance to stars — think about how the apparent angular shift of an object depends on how far away it is.
Step 1 — What parallax says about angular shift. When an observer moves sideways by a distance s, an object at distance d appears to shift through an angle roughly θ ≈ s/d, so for the same sideways displacement s, a nearby object (small d) sweeps through a much larger angle than a distant one (large d).

Step 2 — Applying this to the train. As the train moves, the observer's position (and line of sight) keeps changing rapidly. Nearby trees and houses sit at a small d, so this changing viewpoint produces a large, fast-changing angular shift, making them appear to whip past. Distant hills, the Moon, and stars sit at an enormous d in comparison, so the very same motion of the observer produces a practically zero angular shift for them, which is why they appear to stay still.

✦ Answer: Nearby objects have a small parallax distance, so the same motion of the observer produces a large apparent angular shift and they seem to fly past; distant objects have a parallax angle too small to notice, so they appear stationary.
Where students slip. Explaining this only as 'the train is moving fast' without mentioning distance — the train's speed is the same for near and far objects alike, so speed alone cannot explain why only the nearby ones appear to move; the deciding factor is how distance changes the size of the apparent angular shift.
1.173 marksNCERT Cl-11 Physics Part I, Ch1 Exercises, Q1.17
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10^7 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0 ×10^30 kg, radius of the Sun = 7.0 × 10^8 m.
Hint. Treat the Sun as a uniform sphere, divide mass by volume, and then compare the number against everyday densities of solids/liquids (~10³ kg m⁻³) versus gases (~1 kg m⁻³).
Step 1 — Volume of the Sun as a sphere. V = (4/3)πR³ = (4/3)π × (7.0 × 10⁸)³ ≈ 1.44 × 10²⁷ m³.

Step 2 — Density = mass / volume. ρ = 2.0 × 10³⁰ / 1.44 × 10²⁷ ≈ 1.4 × 10³ kg m⁻³.

Step 3 — Compare with familiar densities. This value (≈ 1400 kg m⁻³) sits in the same range as ordinary liquids and solids (order 10³ kg m⁻³), not gases (order 1 kg m⁻³ at everyday pressure) — a surprising result, since matter this hot cannot chemically be a solid or liquid at all.

✦ Answer: The Sun's mean density works out to about 1.4 × 10³ kg m⁻³, in the range typical of liquids and solids rather than gases — even though the Sun is entirely plasma, its own immense gravity compresses that plasma to a density comparable to water.
Where students slip. Assuming that because the Sun is described as a 'hot plasma/ionized gas', its density must fall in the low, everyday gas range — gravitational self-compression at the Sun's enormous mass pushes the density far above ordinary gas values.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph101.pdf) — one end-of-chapter Exercises set (17 questions, 1.1–1.17); the chapter has no separate in-text question sets, only worked Examples that are not student exercises. Questions are referenced from the NCERT textbook for identification.

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