Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
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Express 72 km h⁻¹ in m s⁻¹.
Given: v = 72 km h⁻¹
Use: 1 km = 1000 m and 1 h = 3600 s
v = 72 × 1000 m / 3600 s = 72 000 / 3600 = 20 m s⁻¹
Answer: 20 m s⁻¹
The density of a liquid is 1.25 g cm⁻³. Express it in kg m⁻³.
Given: ρ = 1.25 g cm⁻³
Use: 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³
1 g cm⁻³ = 10⁻³ kg / 10⁻⁶ m³ = 10³ kg m⁻³
ρ = 1.25 × 10³ kg m⁻³ = 1250 kg m⁻³
Answer: 1.25 × 10³ kg m⁻³
A light year is the distance light travels in one year. Taking c = 3.00 × 10⁸ m s⁻¹ and 1 y = 365.25 d, find 1 light year in metres. A star 4.24 light years away is how many metres away?
Given: c = 3.00 × 10⁸ m s⁻¹; 1 y = 365.25 × 24 × 3600 s = 3.1558 × 10⁷ s
Use: distance = speed × time
1 ly = 3.00 × 10⁸ × 3.1558 × 10⁷ = 9.467 × 10¹⁵ m ≈ 9.47 × 10¹⁵ m (three significant figures, as in c).
4.24 ly = 4.24 × 9.467 × 10¹⁵ = 4.01 × 10¹⁶ m (three significant figures).
Answer: 1 ly ≈ 9.47 × 10¹⁵ m; 4.24 ly ≈ 4.01 × 10¹⁶ m
The CGS unit of force is the dyne (g cm s⁻²). Express 1 newton in dyne.
Given: 1 N = 1 kg m s⁻²
Use: 1 kg = 10³ g and 1 m = 10² cm
1 N = 10³ g × 10² cm s⁻² = 10⁵ g cm s⁻² = 10⁵ dyne
Answer: 1 N = 10⁵ dyne
In a system the unit of mass is 10 kg, the unit of length is 100 m and the unit of time is 10 s. What is the magnitude of 1 J in the new units? (1 J = 1 kg m² s⁻².)
Given: Unit of mass 10 kg; of length 100 m; of time 10 s
Use: Numerical value × unit = same quantity, so n₂ = n₁ (M₁/M₂)(L₁/L₂)²(T₁/T₂)⁻²
n₂ = 1 × (1 kg / 10 kg) × (1 m / 100 m)² × (1 s / 10 s)⁻²
= 0.1 × 10⁻⁴ × 100 = 10⁻³
Answer: 1 J = 1 × 10⁻³ new units
A train moves at 54 km h⁻¹. Express its speed in km min⁻¹.
Given: v = 54 km h⁻¹
Use: 1 h = 60 min
v = 54 km / 60 min = 0.9 km min⁻¹
Answer: 0.9 km min⁻¹
State the number of significant figures in (a) 0.0450 (b) 5.00 × 10³ (c) 1200 (d) 1200. (with a decimal point) (e) 100.0 (f) 0.00302.
Given: Six numbers
Use: The counting rules of the chapter
(a) 0.0450: the leading zeros do not count; 4, 5 and the trailing 0 do → 3
(b) 5.00 × 10³: all digits of the base number count → 3
(c) 1200: trailing zeros without a decimal point do not count → 2
(d) 1200.: the decimal point makes the trailing zeros significant → 4
(e) 100.0: zeros between digits and a trailing zero after a decimal point → 4
(f) 0.00302: the zero between 3 and 2 counts, the leading zeros do not → 3
Answer: (a) 3, (b) 3, (c) 2, (d) 4, (e) 4, (f) 3
Round off to three significant figures: (a) 4.865 (b) 4.875 (c) 6.735 (d) 9.9951.
Given: Four numbers
Use: Rounding rule: a dropped 5 leaves an even preceding digit unchanged and raises an odd one
(a) 4.865: dropped digit 5, preceding digit 6 is even → 4.86
(b) 4.875: dropped 5, preceding 7 is odd → 4.88
(c) 6.735: dropped 5, preceding 3 is odd → 6.74
(d) 9.9951: the dropped part 51 is more than 5 → 9.99 + 0.01 = 10.0
Answer: (a) 4.86, (b) 4.88, (c) 6.74, (d) 10.0
A solid has mass 8.45 g and volume 3.2 cm³. Give its density to the correct number of significant figures.
Given: m = 8.45 g (3 s.f.); V = 3.2 cm³ (2 s.f.)
Use: Division: keep the fewest significant figures
ρ = 8.45 / 3.2 = 2.640625 g cm⁻³
The volume has only two significant figures, so ρ = 2.6 g cm⁻³.
Answer: 2.6 g cm⁻³
Add 12.11 m, 18.0 m and 1.013 m and report the result correctly.
Given: 12.11 (2 d.p.); 18.0 (1 d.p.); 1.013 (3 d.p.)
Use: Addition: keep the fewest decimal places
Sum = 12.11 + 18.0 + 1.013 = 31.123 m
The least precise number, 18.0 m, has one decimal place, so the sum is 31.1 m.
Answer: 31.1 m
A rod has length 9.25 m and a piece of 6.8 m is cut off. Find the length left, correctly written.
Given: 9.25 m (2 d.p.); 6.8 m (1 d.p.)
Use: Subtraction: keep the fewest decimal places; round a dropped 5 to the even digit
9.25 − 6.8 = 2.45 m
One decimal place is allowed: 2.45 → the dropped digit is 5 and the preceding 4 is even, so 2.4 m.
Answer: 2.4 m
The radius of a circular disc is measured as 2.5 cm. Find its area in correct significant figures.
Given: r = 2.5 cm (2 s.f.)
Use: A = π r²; π is an exact constant, so the radius sets the precision
A = 3.14159 × (2.5)² = 3.14159 × 6.25 = 19.635 cm²
The radius has two significant figures, so A = 2.0 × 10¹ cm² (20 cm²).
Answer: 2.0 × 10¹ cm²
Each side of a cube is 4.12 cm. Find its total surface area and its volume to the correct number of significant figures.
Given: a = 4.12 cm (3 s.f.)
Use: A = 6a² and V = a³
A = 6 × (4.12)² = 6 × 16.9744 = 101.8464 cm² ≈ 102 cm²
V = (4.12)³ = 69.9345 cm³ ≈ 69.9 cm³
Both results keep three significant figures, the same as the side.
Answer: A ≈ 102 cm²; V ≈ 69.9 cm³
A length is measured as 25.4 ± 0.1 cm. Find the percentage error.
Given: l = 25.4 cm; Δl = 0.1 cm
Use: percentage error = (Δl / l) × 100
= (0.1 / 25.4) × 100 = 0.39 %
To one significant figure: 0.4 %.
Answer: ≈ 0.4 %
The sides of a rectangle are 12.5 ± 0.1 cm and 8.0 ± 0.1 cm. Find the area with its error.
Given: l = 12.5 ± 0.1 cm; b = 8.0 ± 0.1 cm
Use: For a product the percentage errors add
A = 12.5 × 8.0 = 100 cm²
Error in l: (0.1/12.5) × 100 = 0.8 %. Error in b: (0.1/8.0) × 100 = 1.25 %.
Error in A = 0.8 % + 1.25 % = 2.05 %, so ΔA = 100 × 0.0205 = 2.05 ≈ 2 cm².
Answer: A = 100 ± 2 cm²
Two masses, 2.5 g and 25.0 g, are each measured to ± 0.01 g. Compare the relative errors.
Given: Δm = 0.01 g in both
Use: relative error = Δm / m
For 2.5 g: (0.01/2.5) × 100 = 0.4 %.
For 25.0 g: (0.01/25.0) × 100 = 0.04 %.
The larger value has a ten times smaller relative error, though the absolute error is the same.
Answer: 0.4 % and 0.04 %
A body has mass 20.0 ± 0.1 g and volume 5.0 ± 0.1 cm³. Find its density with its error.
Given: m = 20.0 ± 0.1 g; V = 5.0 ± 0.1 cm³
Use: ρ = m/V; for a quotient the percentage errors add
ρ = 20.0 / 5.0 = 4.0 g cm⁻³
Percentage error = (0.1/20.0) × 100 + (0.1/5.0) × 100 = 0.5 % + 2 % = 2.5 %
Δρ = 4.0 × 0.025 = 0.1 g cm⁻³
Answer: ρ = 4.0 ± 0.1 g cm⁻³
Find the dimensional formula of pressure.
Given: Pressure = force / area
Use: [F] = [M L T⁻²] and [area] = [L²]
[P] = [M L T⁻²] / [L²] = [M L⁻¹ T⁻²]
Answer: [M L⁻¹ T⁻²]
Find the dimensional formulae of kinetic energy ½mv² and of power (energy / time).
Given: K = ½ m v²
Use: The factor ½ has no dimensions
[K] = [M][L T⁻¹]² = [M L² T⁻²]
[Power] = [M L² T⁻²] / [T] = [M L² T⁻³]
Answer: [M L² T⁻²] and [M L² T⁻³]
Check whether (a) s = ut + ½at² and (b) v = u + at² are dimensionally correct. (s: distance, u, v: speeds, a: acceleration, t: time)
Given: [s] = [L]; [u] = [v] = [L T⁻¹]; [a] = [L T⁻²]; [t] = [T]
Use: Every term must have the same dimensions
(a) [ut] = [L T⁻¹][T] = [L]; [½at²] = [L T⁻²][T²] = [L]; [s] = [L]. All equal → dimensionally correct.
(b) [v] = [L T⁻¹] but [at²] = [L T⁻²][T²] = [L], which is not [L T⁻¹] → dimensionally wrong (it should be v = u + at).
Answer: (a) correct; (b) incorrect
Newton's law of gravitation is F = G m₁m₂ / r². Find the dimensions of G.
Given: [F] = [M L T⁻²]; [m₁m₂] = [M²]; [r²] = [L²]
Use: G = F r² / (m₁ m₂)
[G] = [M L T⁻²][L²] / [M²] = [M⁻¹ L³ T⁻²]
Answer: [M⁻¹ L³ T⁻²]
The energy of a photon is E = hν, where ν is the frequency. Find the dimensions of h.
Given: [E] = [M L² T⁻²]; [ν] = [T⁻¹]
Use: h = E/ν
[h] = [M L² T⁻²] / [T⁻¹] = [M L² T⁻¹]
Answer: [M L² T⁻¹]
In the equation (P + a/V²)(V − b) = RT, P is pressure and V is volume. Find the dimensions of a and b.
Given: [P] = [M L⁻¹ T⁻²]; [V] = [L³]
Use: Terms added or subtracted must have the same dimensions
V − b requires [b] = [V] = [L³].
P + a/V² requires [a/V²] = [P], so [a] = [P][V²] = [M L⁻¹ T⁻²][L⁶] = [M L⁵ T⁻²].
Answer: [a] = [M L⁵ T⁻²]; [b] = [L³]
The speed of a wave on a stretched string is claimed to be v = √(F/μ), where F is the tension and μ the mass per unit length. Is this dimensionally consistent?
Given: [F] = [M L T⁻²]; [μ] = [M L⁻¹]
Use: Compare the dimensions of both sides
[F/μ] = [M L T⁻²] / [M L⁻¹] = [L² T⁻²]
√[F/μ] = [L T⁻¹], which is the dimension of speed → consistent.
Answer: Yes, dimensionally consistent
The speed v of a transverse wave on a string depends on the tension F, the length l and the mass m of the string. Use dimensions to find the form of v.
Given: v = k Fᵃ lᵇ mᶜ
Use: Equate the powers of [M], [L], [T]
[L T⁻¹] = [M L T⁻²]ᵃ [L]ᵇ [M]ᶜ = M^(a+c) L^(a+b) T^(−2a)
T: −2a = −1 → a = ½. M: a + c = 0 → c = −½. L: a + b = 1 → b = ½.
So v = k √(F l / m).
Answer: v = k √(F l / m)
The period T of a mass m on a spring depends on m and on the force constant k (force per unit extension). Find the form of T.
Given: T = C mᵃ kᵇ; [k] = [M T⁻²]
Use: Equate the powers
[T] = [M]ᵃ [M T⁻²]ᵇ = M^(a+b) T^(−2b)
M: a + b = 0. T: −2b = 1 → b = −½ and a = ½.
So T = C √(m/k), where C is a dimensionless constant that dimensions cannot give.
Answer: T = C √(m/k)
The force F needed to keep a mass m moving in a circle of radius r at speed v depends on m, v and r. Find the form of F.
Given: F = k mᵃ vᵇ rᶜ
Use: Equate the powers
[M L T⁻²] = [M]ᵃ [L T⁻¹]ᵇ [L]ᶜ = M^a L^(b+c) T^(−b)
M: a = 1. T: −b = −2 → b = 2. L: b + c = 1 → c = −1.
So F = k m v² / r (in fact k = 1).
Answer: F = k m v² / r
The viscous force F on a small sphere moving through a liquid depends on the viscosity η (dimensions [M L⁻¹ T⁻¹]), the radius r and the speed v. Find the form of F.
Given: F = k ηᵃ rᵇ vᶜ
Use: Equate the powers
[M L T⁻²] = [M L⁻¹ T⁻¹]ᵃ [L]ᵇ [L T⁻¹]ᶜ = M^a L^(−a+b+c) T^(−a−c)
M: a = 1. T: −a − c = −2 → c = 1. L: −a + b + c = 1 → b = 1.
So F = k η r v (Stokes found k = 6π, which dimensions cannot give).
Answer: F = k η r v
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(a) 1 cm = 10⁻² m, so 1 cm³ = (10⁻²)³ m³ = 10⁻⁶ m³.
(b) Surface area = 2πr(r + h) = 2π × 2.0 × (2.0 + 10.0) = 150.8 cm². Since 1 cm² = 100 mm², this is 15 080 mm². With two significant figures (r = 2.0 cm): 1.5 × 10⁴ mm².
(c) 18 km h⁻¹ = 18 × 1000 m / 3600 s = 5 m s⁻¹, so the vehicle covers 5 m in 1 s.
(d) Relative density is the density as a multiple of that of water (1 g cm⁻³ = 10³ kg m⁻³). So the density of lead is 11.3 g cm⁻³ or 1.13 × 10⁴ kg m⁻³.
(a) 1 kg = 10³ g and 1 m² = 10⁴ cm², so 1 kg m² s⁻² = 10³ × 10⁴ g cm² s⁻² = 10⁷ g cm² s⁻².
(b) 1 light year = 9.46 × 10¹⁵ m, so 1 m = 1 / (9.46 × 10¹⁵) ly = 1.06 × 10⁻¹⁶ ly.
(c) 1 m = 10⁻³ km and 1 s = (1/3600) h, so 1 s⁻² = 3600² h⁻². Then 3.0 m s⁻² = 3.0 × 10⁻³ × 3600² km h⁻² = 3.0 × 10⁻³ × 1.296 × 10⁷ = 3.9 × 10⁴ km h⁻² (38 880, to two significant figures). Answer: 3.9 × 10⁴ km h⁻².
(d) 1 N m² kg⁻² = 1 m³ kg⁻¹ s⁻². Now 1 m³ = 10⁶ cm³ and 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.
1 cal = 4.2 J = 4.2 kg m² s⁻². The dimensions are [M¹ L² T⁻²]. If a quantity has numerical value n₁ in units (M₁, L₁, T₁) and n₂ in (M₂, L₂, T₂), then n₁ [M₁¹ L₁² T₁⁻²] = n₂ [M₂¹ L₂² T₂⁻²].
Here n₁ = 4.2, M₁ = 1 kg, L₁ = 1 m, T₁ = 1 s, and M₂ = α kg, L₂ = β m, T₂ = γ s.
n₂ = 4.2 (1 kg / α kg)¹ (1 m / β m)² (1 s / γ s)⁻² = 4.2 α⁻¹ β⁻² γ². Hence the calorie is 4.2 α⁻¹ β⁻² γ² in the new units.
A dimensional quantity has a numerical value that depends on the unit chosen (for example 1 km = 1000 m), so a number alone is neither large nor small. It is large or small only relative to another quantity of the same kind, which serves as the standard.
Reframed statements (the standards chosen here are examples):
Time taken by light = 8 min 20 s = 8 × 60 + 20 = 500 s.
The new unit of length is the distance light travels in one second, so that the speed of light is 1 unit per second. Distance = speed × time = 1 × 500 = 500 new units.
The precision of an instrument is set by its least count. Taking the main scale of the callipers to be in millimetres:
The smallest least count belongs to (c). The optical instrument is the most precise.
Magnification = 100, so the real thickness = apparent width / 100 = 3.5 mm / 100 = 0.035 mm.
Thickness ≈ 0.035 mm = 3.5 × 10⁻² mm (3.5 × 10⁻⁵ m).
(a) Wind the thread closely, turn by turn without gaps, around a pencil to make a coil of N turns (say 20–50). Measure the length l of the coil with the metre scale. The diameter of the thread is d = l / N.
(b) The least count is pitch / divisions = 1.0 mm / 200 = 0.005 mm. No, accuracy cannot be raised without limit: the screw itself has mechanical imperfections (back-lash, a pitch that is not exactly uniform), the divisions become too fine to read, and the pressure applied on the object varies, so other errors dominate beyond some point.
(c) Random errors are as likely to be positive as negative, so they tend to cancel in the average. The more measurements are taken, the smaller the effect of random errors on the mean (the uncertainty falls roughly as 1/√N), so 100 readings give a more reliable mean than 5.
Areas scale as the square of the linear magnification m: area on screen / area on slide = m².
Area on slide = 1.75 cm² = 1.75 × 10⁻⁴ m². So m² = 1.55 / (1.75 × 10⁻⁴) = 8.86 × 10³, and m = √(8.86 × 10³) = 94.1.
The linear magnification is about 94.
Convert the thickness: 2.01 cm = 0.0201 m. The least number of significant figures in the data is three (2.01 cm), so the results have three.
Area of the sheet (two faces, two long edges and two short edges) = 2(lb + bt + tl) = 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234) = 2(4.25517 + 0.02020 + 0.08510) = 8.7209 m² ≈ 8.72 m².
Volume = l b t = 4.234 × 1.005 × 0.0201 = 0.085529 m³ ≈ 8.55 × 10⁻² m³.
(a) Express all masses in kg: 2.30 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg. The least precise number, 2.30 kg, has two decimal places, so the total is 2.34 kg.
(b) 20.17 g − 20.15 g = 0.02 g. Both numbers have two decimal places, so the difference is 0.02 g (it has only one significant figure).
The quantity subtracted from 1 must be dimensionless, since 1 is a pure number. In (1 − v²) the term v² has the dimensions of speed squared, [L² T⁻²], so it must be divided by another speed squared. The only speed available is c.
m = m₀ / (1 − v²/c²)^(1/2). Now v²/c² is dimensionless, and m has the same dimensions as m₀.
Take the radius of the atom as r = 0.5 Å = 0.5 × 10⁻¹⁰ m. Volume of one atom = (4/3)πr³ = (4/3) × 3.14 × (0.5 × 10⁻¹⁰)³ = 5.2 × 10⁻³¹ m³.
Number of atoms in one mole = 6.02 × 10²³. Total atomic volume = 5.2 × 10⁻³¹ × 6.02 × 10²³ = 3.2 × 10⁻⁷ m³, that is about 3 × 10⁻⁷ m³.
A molecule of size (diameter) 1 Å has a radius 0.5 Å, so the volume of a mole of molecules is the same as in 1.14: about 3 × 10⁻⁷ m³ (3.15 × 10⁻⁷ m³).
Molar volume = 22.4 L = 22.4 × 10⁻³ m³. Ratio = 22.4 × 10⁻³ / 3.15 × 10⁻⁷ = 7.1 × 10⁴, that is about 7 × 10⁴.
Why so large: in a gas the molecules are very far apart. The average separation between molecules is about ten times the size of a molecule or more, so almost all of the volume of a gas is empty space between the molecules, and the molecules themselves occupy only a tiny fraction.
What we notice is the change in the direction (angle) in which an object is seen. When the train moves a certain distance, the line of sight to a nearby object turns through a large angle in a short time, so the object seems to rush backwards.
For a very distant object, the same movement of the train changes the line of sight by an almost negligible angle, because the angle changed is roughly (distance moved by train) / (distance of the object). It therefore appears to be stationary (or to travel along with you). This is the effect called parallax.
At such temperatures the matter is a gas (plasma), so one would guess a density in the range of gases, which is very low (about 1 kg m⁻³ or less).
Check with the data: volume of the Sun = (4/3)πR³ = (4/3) × 3.14 × (7.0 × 10⁸)³ = 1.4 × 10²⁷ m³. Density = mass / volume = 2.0 × 10³⁰ / 1.4 × 10²⁷ = 1.4 × 10³ kg m⁻³.
So the guess is wrong: the density is in the range of the densities of solids and liquids (water is 1.0 × 10³ kg m⁻³). The reason is the Sun's huge gravitational attraction, which compresses the matter to such a high density even though it is a hot gas.