Units and Measurements Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1(a) 1 cm = 10⁻² m, so (1 cm)³ = (10⁻²)³ m³ = 10⁻⁶ m³.
2. 2(b) Total surface area of a cylinder of radius r and height h is 2πr(h + r). With r = 2.0 cm and h = 10.0 cm: A = 2π(2.0)(12.0) = 48π ≈ 1.508 × 10² cm².
3. 31 cm² = (10 mm)² = 10² mm², so A = 1.508 × 10² × 10² = 1.508 × 10⁴ mm².
4. 4(c) 18 km h⁻¹ = 18 × (1000 m)/(3600 s) = 5 m s⁻¹, so the vehicle covers 5 m in 1 s.
5. 5(d) Relative density equals density in g cm⁻³, so density = 11.3 g cm⁻³. Since 1 g cm⁻³ = 10³ kg m⁻³ (multiply by 10⁻³/10⁻⁶ = 10³), density = 11.3 × 10³ kg m⁻³.

Step-by-step solution

1. 1(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.
2. 2(b) 1 ly = 9.46 × 10¹⁵ m, so 1 m = 1/9.46 × 10⁻¹⁵ ≈ 1.06 × 10⁻¹⁶ ly.
3. 3(c) 3.0 m s⁻² = 3.0 × 10⁻³ km s⁻². Since 1 h = 3600 s, 1 s⁻² = (3600)² h⁻² = 1.296 × 10⁷ h⁻².
4. 4So 3.0 × 10⁻³ × 1.296 × 10⁷ = 3.888 × 10⁴ ≈ 3.9 × 10⁴ km h⁻².
5. 5(d) G = 6.67 × 10⁻¹¹ N m² kg⁻² = 6.67 × 10⁻¹¹ (kg m s⁻²) m² kg⁻² = 6.67 × 10⁻¹¹ m³ s⁻² kg⁻¹.
6. 61 m³ = (10² cm)³ = 10⁶ cm³ and 1 kg = 10³ g: G = 6.67 × 10⁻¹¹ × 10⁶/10³ = 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹.

Step-by-step solution

1. 1Energy has dimensions [ML²T⁻²].
2. 2In SI, 1 calorie = 4.2 kg m² s⁻².
3. 3The new units are: mass α kg, length β m, time γ s. A numerical value transforms as n′ = n (u₁/u₁′)ᵃ (u₂/u₂′)ᵇ (u₃/u₃′)ᶜ, where a = 1, b = 2, c = −2 for [ML²T⁻²].
4. 4n′ = 4.2 × (1/α)¹ × (1/β)² × (1/γ)⁻² = 4.2 α⁻¹ β⁻² γ², which is exactly what had to be shown.

Step-by-step solution

1. 1A statement like ‘large’ or ‘small’ has meaning only relative to a chosen standard; the same quantity can be large compared with one standard and small compared with another.
2. 2(a) Reframe: atoms are very small compared with objects of everyday size (a few tenths of a nanometre).
3. 3(b) Reframe: a jet plane moves with great speed compared with a bicycle or car (about 250 m s⁻¹).
4. 4(c) Reframe: the mass of Jupiter is very large compared with the mass of the Earth (about 318 times).
5. 5(d) Reframe: the air inside this room contains a large number of molecules compared with, say, the number of people in the room (of order 10²⁷).
6. 6(e) Statement already relative: a proton is much more massive than an electron (about 1836 times).
7. 7(f) Statement already relative: the speed of sound is much smaller than the speed of light.

Step-by-step solution

1. 1Speed of light in the new unit is 1 (new unit of length) per (unit of time), by definition.
2. 2Time taken = 8 min 20 s = 8 × 60 + 20 = 500 s.
3. 3Distance = speed × time = 1 × 500 = 500 new units of length.

Step-by-step solution

1. 1Least count of the vernier callipers = 1 main-scale division / 20 = 1/20 mm = 0.05 mm = 5 × 10⁻⁵ m.
2. 2Least count of the screw gauge = pitch/divisions = 1 mm/100 = 0.01 mm = 1 × 10⁻⁵ m.
3. 3The optical instrument resolves down to one wavelength of light, λ ≈ 5 × 10⁻⁷ m.
4. 4Smallest least count ⇒ highest precision: 5 × 10⁻⁷ m < 1 × 10⁻⁵ m < 5 × 10⁻⁵ m.

Step-by-step solution

1. 1The observed width 3.5 mm is the magnified image width.
2. 2Actual thickness = observed width / magnification = 3.5 mm / 100 = 0.035 mm.
3. 30.035 mm = 3.5 × 10⁻² × 10⁻³ m = 3.5 × 10⁻⁵ m = 35 µm.

Step-by-step solution

1. 1(a) Wind the thread closely around a pencil or cylinder for N turns, measure the total length L of the N turns with the metre scale, then diameter = L/N. The more turns, the smaller the relative error in the single-turn width.
2. 2(b) No. Least count = pitch/divisions; increasing divisions reduces the least count, but accuracy is ultimately limited by the backlash of the screw, the zero error and the measurer’s ability to judge coincidence. Beyond a point, extra divisions only give false precision.
3. 3(c) Random errors average out over many readings; the random error of the mean falls roughly as 1/√N. With 100 readings the random scatter is strongly suppressed, so the mean is closer to the true value than with 5 readings. Systematic zero error, however, is not removed by either set.

Step-by-step solution

1. 1Area magnification = screen area / slide area = 1.55 m² / 1.75 cm².
2. 2Convert: 1.55 m² = 1.55 × 10⁴ cm², so area magnification = 1.55 × 10⁴/1.75 ≈ 8857.
3. 3Linear magnification is the square root of the area magnification: m = √8857 ≈ 94.

Step-by-step solution

1. 1(a) Leading zeros are not significant: 0.007 has 1 significant figure.
2. 2(b) 2.64 × 10²⁴ has 3 significant figures.
3. 3(c) 0.2370 has 4 (the trailing zero after the decimal is significant).
4. 4(d) 6.320 has 4 (trailing zero is significant).
5. 5(e) 6.032 has 4 (the embedded zero counts).
6. 6(f) 0.0006032 has 4 (only leading zeros are not significant).

Step-by-step solution

1. 1Convert the thickness: 2.01 cm = 0.0201 m (3 significant figures).
2. 2Area = length × breadth = 4.234 × 1.005 = 4.25517 m².
3. 3The two factors have 4 and 4 significant figures, so the product is reported to 4: A = 4.255 m².
4. 4Volume = area × thickness = 4.25517 × 0.0201 = 0.0855289 m³.
5. 5The thickness has only 3 significant figures, so the volume is reported to 3: V = 0.0855 m³.

Step-by-step solution

1. 1Convert the gold masses: 20.15 g = 0.02015 kg and 20.17 g = 0.02017 kg.
2. 2(a) Total = 2.30 + 0.02015 + 0.02017 = 2.34032 kg.
3. 3In addition, the result has the same number of decimal places as the least precise term: 2.30 kg is precise only to the first decimal, so total = 2.3 kg.
4. 4(b) Difference = 20.17 − 20.15 = 0.02 g, keeping the two decimal places of the least precise piece.

Step-by-step solution

1. 1The argument of a square root must be dimensionless, since m₀ and m have the same dimensions [M].
2. 21 − v² is not dimensionless: v carries dimensions [LT⁻¹] and 1 is dimensionless, so the terms cannot be added.
3. 3For dimensional consistency the subtracted quantity must be (v²/c²), which is dimensionless because v and c both have speed dimensions.
4. 4Correct form: m₀ = m(1 − v²/c²)^(1/2), i.e. put c under v inside the bracket.

Step-by-step solution

1. 1Size of the atom ≈ 0.5 Å = 0.5 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m; taking this as the radius r.
2. 2Volume of one atom = (4/3)πr³ = (4/3)π(5 × 10⁻¹¹)³.
3. 3
   $$V_1=\frac{4}{3}\pi (1.25\times 10^{-31})\approx 5.24\times 10^{-31}{\text{m}}^3$$
4. 4A mole has NA = 6.023 × 10²³ atoms, so total volume = NA V₁.
5. 5
   $$V=6.023\times 10^{23}\times 5.24\times 10^{-31}\approx 3.15\times 10^{-7}{\text{m}}^3$$

Step-by-step solution

1. 1Molar volume of the gas = 22.4 L = 22.4 × 10⁻³ m³ = 2.24 × 10⁻² m³.
2. 2Size of a hydrogen molecule ≈ 1 Å, so take radius ≈ 0.5 × 10⁻¹⁰ m (as in Exercise 1.14).
3. 3Atomic volume of a mole of hydrogen ≈ 3.15 × 10⁻⁷ m³ (computed in Exercise 1.14).
4. 4
   $$\text{Ratio}=\frac{2.24\times 10^{-2}}{3.15\times 10^{-7}}\approx 7.1\times 10^4$$
5. 5The ratio is so large because gas molecules are widely separated — the gas itself is mostly empty space, so its molar volume is tens of thousands of times the volume actually filled by the molecules.

Step-by-step solution

1. 1When you look at nearby objects, the line of sight sweeps through a large angle in a short time as the train moves past, so the apparent angular speed is large — they seem to rush backwards.
2. 2For distant objects the line of sight swings through a tiny angle for the same displacement of the train, so their apparent angular motion is negligible — they seem fixed.
3. 3The same parallax idea applies to observing stars from two positions six months apart in the Earth’s orbit (Exercises 1.5 on light travel time illustrates the same large distance).
4. 4The apparent speed of an object therefore depends on its distance: apparent angular speed ∝ (train speed)/(distance).

Step-by-step solution

1. 1Guess: even though the Sun is a plasma, its density is expected to lie in the range of solids and liquids (roughly 10³ kg m⁻³), not of gases at ordinary pressure.
2. 2Volume of the Sun (sphere) = (4/3)πR³ = (4/3)π(7.0 × 10⁸)³.
3. 3
   $$V=\frac{4}{3}\pi (3.43\times 10^{26})\approx 1.437\times 10^{27}{\text{m}}^3$$
4. 4Density = mass/volume = 2.0 × 10³⁰/1.437 × 10²⁷ ≈ 1.4 × 10³ kg m⁻³.
5. 51.4 × 10³ kg m⁻³ is indeed in the range of densities of solids and liquids (about 10³ kg m⁻³), confirming the guess — the plasma has ionic densities far above ordinary gas.