Chapter 1: Units and Measurement (Physical World and Measurement)

(a) $1{\text{cm}}^3=(10^{-2}\text{m}{)}^3=10^{-6}{\text{m}}^3$.

(b) Total surface area $=2\pi r(r+h)=2\times 3.14\times 2.0\times (2.0+10.0)=150.7{\text{cm}}^2$. Since $1{\text{cm}}^2=100{\text{mm}}^2$, this is $1.5\times 10^4{\text{mm}}^2$ (2 significant figures, as r = 2.0 cm).

(c) $18{\text{km h}}^{-1}=18\times \frac{1000}{3600}=5{\text{m s}}^{-1}$, so 5 m in 1 s.

(d) Relative density 11.3 means density $=11.3{\text{g cm}}^{-3}=11.3\times \frac{10^{-3}\text{kg}}{10^{-6}{\text{m}}^3}=1.13\times 10^4{\text{kg m}}^{-3}$.

(a) 10⁻⁶ (b) 1.5 × 10⁴ (c) 5 (d) 11.3; 1.13 × 10⁴

(a) $1{\text{kg m}}^2{\text{s}}^{-2}=10^3\text{g}\times (10^2\text{cm}{)}^2{\text{s}}^{-2}=10^7{\text{g cm}}^2{\text{s}}^{-2}$.

(b) $1\text{ly}=9.46\times 10^{15}$ m, so $1\text{m}=\frac{1}{9.46\times 10^{15}}\text{ly}=1.06\times 10^{-16}\text{ly}\approx 10^{-16}$ ly.

(c) $3.0{\text{m s}}^{-2}=3.0\times \frac{10^{-3}\text{km}}{{\left(\frac{1}{3600}\text{h}\right)}^2}=3.0\times 10^{-3}\times 1.296\times 10^7=3.9\times 10^4{\text{km h}}^{-2}$.

(d) $1\text{N}=1{\text{kg m s}}^{-2}$, so $G=6.67\times 10^{-11}{\text{m}}^3{\text{s}}^{-2}{\text{kg}}^{-1}=6.67\times 10^{-11}\times \frac{10^6{\text{cm}}^3}{10^3\text{g}}{\text{s}}^{-2}=6.67\times 10^{-8}{\text{cm}}^3{\text{s}}^{-2}{\text{g}}^{-1}$.

(a) 10⁷ (b) 10⁻¹⁶ (c) 3.9 × 10⁴ (d) 6.67 × 10⁻⁸

Energy has dimensions $[M{L}^2{T}^{-2}]$. With $n_1=4.2$ in SI ($M_1=1$ kg, $L_1=1$ m, $T_1=1$ s) and new units $M_2=\alpha$ kg, $L_2=\beta$ m, $T_2=\gamma$ s:

$$n_2=n_1{\left(\frac{M_1}{M_2}\right)}^1{\left(\frac{L_1}{L_2}\right)}^2{\left(\frac{T_1}{T_2}\right)}^{-2}=4.2\left(\frac{1}{\alpha }\right){\left(\frac{1}{\beta }\right)}^2{\left(\frac{1}{\gamma }\right)}^{-2}=4.2{\alpha }^{-1}{\beta }^{-2}{\gamma }^2$$

1 calorie = 4.2 α⁻¹ β⁻² γ² new units. Shown.

A dimensional quantity is large or small only relative to some other quantity of the same kind. 1 mm is large for an atom but tiny for a city. So each statement needs a standard of comparison:

(a) Atoms are very small compared with a pin head (or a dust particle).

(b) A jet plane moves with great speed compared with a car (but slowly compared with light).

(c) The mass of Jupiter is very large compared with the mass of the Earth.

(d) The air in this room contains a number of molecules (about 10²⁷) far larger than, say, the number of people on the Earth (about 8 × 10⁹).

(e) Already has a standard: the proton is compared with the electron. No change needed.

(f) Already has a standard: sound is compared with light. No change needed.

A size is meaningful only relative to a standard; (e) and (f) already state one, the others are reframed above.

Speed of light = 1 new unit per second, so the new unit is the distance light travels in 1 s.

Time $=8\times 60+20=500$ s, so distance $=1\times 500=500$ new units.

500 new units (light-seconds)

The smaller the least count, the more precise the device.

(a) Vernier: $\text{LC}=\frac{1\text{mm}}{20}=0.05$ mm.

(b) Screw gauge: $\text{LC}=\frac{\text{pitch}}{N}=\frac{1\text{mm}}{100}=0.01$ mm.

(c) Optical instrument: about one wavelength of light $\approx 6000\text{A˚}=6\times 10^{-4}$ mm.

(c), the optical instrument (least count ~10⁻⁴ mm).

Real thickness $=\frac{\text{observed width}}{\text{magnification}}=\frac{3.5\text{mm}}{100}$.

0.035 mm

(a) Wind the thread on a uniform rod (or pencil) so that the turns touch each other, say n turns. Measure the length L of the winding with the metre scale. Diameter $=\frac{L}{n}$. Taking many turns makes the error small.

(b) In principle, $\text{LC}=\frac{\text{pitch}}{N}$ falls as N rises. In practice, no: very fine divisions cannot be read by the eye, and errors such as backlash and uneven pitch remain. So accuracy cannot be increased without limit.

(c) Random errors are as likely to be positive as negative. The more readings we average, the more these errors cancel, so the mean of 100 readings is closer to the true value than the mean of 5.

(a) Wind n close turns, measure their length L: d = L/n. (b) No, not arbitrarily. (c) Averaging more readings reduces random error.

Areal magnification $=\frac{1.55{\text{m}}^2}{1.75{\text{cm}}^2}=\frac{1.55\times 10^4}{1.75}=8857$.

Linear magnification $=\sqrt{\text{areal magnification}}=\sqrt{8857}$.

94.1

(a) Leading zeros do not count: only 7. (b) The power of 10 does not count: 2, 6, 4. (c) Leading zero no; trailing zero after the decimal point counts: 2, 3, 7, 0. (d) 6, 3, 2, 0. (e) The zero between non-zero digits counts: 6, 0, 3, 2. (f) 6, 0, 3, 2.

(a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4

$l=4.234$ m, $b=1.005$ m, $t=2.01$ cm $=0.0201$ m (3 significant figures, so the answers keep 3).

Area of the sheet (all faces) $=2(lb+bt+tl)=2(4.2552+0.0202+0.0851)=2(4.3605)=8.721{\text{m}}^2\approx 8.72{\text{m}}^2$.

Volume $=l\times b\times t=4.234\times 1.005\times 0.0201=0.08553{\text{m}}^3\approx 0.0855{\text{m}}^3$.

Area 8.72 m²; volume 0.0855 m³

(a) Total $=2.30+0.02015+0.02017=2.34032$ kg. The least precise term, 2.30 kg, has two decimal places, so the total is 2.34 kg. (The NCERT answer key rounds this further to 2.3 kg.)

(b) Difference $=20.17-20.15=0.02$ g (two decimal places, as both masses have two).

(a) 2.34 kg (NCERT key: 2.3 kg) (b) 0.02 g

m and m₀ have the same dimensions, so $(1-v^2{)}^{1/2}$ must be dimensionless. Since 1 is a pure number, $v^2$ must also be dimensionless, which needs $\frac{v^2}{c^2}$.

$m=\frac{m_0}{{\left(1-\frac{v^2}{c^2}\right)}^{1/2}}$

Radius $r=0.5\text{A˚}=0.5\times 10^{-10}$ m. Volume of one atom $=\frac{4}{3}\pi r^3=\frac{4}{3}\times 3.14\times (0.5\times 10^{-10}{)}^3=5.23\times 10^{-31}{\text{m}}^3$.

For one mole: $5.23\times 10^{-31}\times 6.023\times 10^{23}=3.15\times 10^{-7}{\text{m}}^3$.

About 3 × 10⁻⁷ m³

Taking size 1 Å (radius 0.5 Å), the atomic volume of a mole is $3.15\times 10^{-7}{\text{m}}^3$ (as in 1.14). Molar volume $=22.4\text{L}=22.4\times 10^{-3}{\text{m}}^3$.

Ratio $=\frac{22.4\times 10^{-3}}{3.15\times 10^{-7}}=7.1\times 10^4\approx 10^4$ (order of magnitude).

The ratio is large because the separation between gas molecules is much larger than the size of a molecule; most of a gas is empty space.

About 10⁴ (≈ 7 × 10⁴); intermolecular gaps in a gas are much larger than molecular size.

The apparent motion depends on how fast the line of sight to an object turns (parallax). For a nearby object, a small movement of the train changes the angle it makes at the eye by a large amount, so it seems to rush backwards. For a very distant object, the same movement changes the angle very little, so it seems almost stationary. Since we know we are moving, a body whose direction does not change seems to move along with us.

Near objects change their angle at the eye quickly (large parallax), distant ones hardly at all.

$$\rho =\frac{M}{\frac{4}{3}\pi R^3}=\frac{2.0\times 10^{30}}{\frac{4}{3}\times 3.14\times (7.0\times 10^8{)}^3}=\frac{2.0\times 10^{30}}{1.44\times 10^{27}}\approx 1.4\times 10^3{\text{kg m}}^{-3}$$

This is the density of liquids/solids (water is $10^3$ kg m⁻³), not of gases (about 1 kg m⁻³). It is high because the outer layers are pulled inwards strongly by the gravity of the inner layers.

ρ ≈ 1.4 × 10³ kg m⁻³: in the range of liquids/solids, not gases.