Chapter 1: Units and Measurements

The side of the cube is given to four significant figures, so both derived quantities must be reported to four significant figures.

Total surface area of the cube (six identical square faces of side $$a = 7.203 \, \mathrm{m}$$):

$$S = 6a^2 = 6 \times (7.203)^2 \, \mathrm{m^2}$$

$$S = 6 \times 51.883209 \, \mathrm{m^2} = 311.299254 \, \mathrm{m^2}$$

Rounding to four significant figures, $$S = 311.3 \, \mathrm{m^2}$$.

Volume of the cube:

$$V = a^3 = (7.203)^3 \, \mathrm{m^3} = 373.714754 \, \mathrm{m^3}$$

Rounding to four significant figures, $$V = 373.7 \, \mathrm{m^3}$$.

The density is defined as mass per unit volume.

$$\rho = \dfrac{m}{V} = \dfrac{5.74 \, \mathrm{g}}{1.2 \, \mathrm{cm^3}} = 4.7833\ldots \, \mathrm{g \, cm^{-3}}$$

For multiplication or division, the result must be expressed to the same number of significant figures as the factor with the fewest. Mass $$5.74 \, \mathrm{g}$$ has $$3$$ significant figures and volume $$1.2 \, \mathrm{cm^3}$$ has $$2$$ significant figures. Hence the density is reported to $$2$$ significant figures.

$$\rho = 4.8 \, \mathrm{g \, cm^{-3}}$$

We compare the dimensions of the left-hand side (LHS) and the right-hand side (RHS) using $$[M]$$, $$[L]$$ and $$[T]$$ for mass, length and time.

Dimensions of LHS: $$\tfrac{1}{2}$$ is a pure number and is dimensionless.

$$[\tfrac{1}{2} m v^2] = [m][v]^2 = [M][L \, T^{-1}]^2 = [M \, L^2 \, T^{-2}]$$

Dimensions of RHS:

$$[m g h] = [M][L \, T^{-2}][L] = [M \, L^2 \, T^{-2}]$$

The dimensions on both sides are identical: $$[M^1 \, L^2 \, T^{-2}]$$. Hence the equation is dimensionally correct.

Note: Dimensional correctness does not by itself prove physical correctness; for instance, a missing dimensionless factor (such as $$\tfrac{1}{2}$$) cannot be detected by this method.

Assume the time period $$T$$ depends on the length $$l$$, mass $$m$$ and acceleration due to gravity $$g$$ as

$$T = k \, l^{a} \, m^{b} \, g^{c}$$

where $$k$$ is a dimensionless constant and $$a$$, $$b$$, $$c$$ are exponents to be found.

Writing the dimensions of each quantity:

$$[T] = [T^{1}], \quad [l] = [L], \quad [m] = [M], \quad [g] = [L \, T^{-2}]$$

Substituting,

$$[T^{1}] = [L]^{a} [M]^{b} [L \, T^{-2}]^{c} = [M^{b} \, L^{a+c} \, T^{-2c}]$$

Comparing exponents of $$M$$, $$L$$ and $$T$$ on both sides:

$$b = 0, \qquad a + c = 0, \qquad -2c = 1$$

Solving: $$c = -\tfrac{1}{2}$$ and hence $$a = +\tfrac{1}{2}$$, with $$b = 0$$.

Therefore

$$T = k \, l^{1/2} \, g^{-1/2} = k \sqrt{\dfrac{l}{g}}$$

Experimentally (or from a detailed analysis of the equation of motion) it is found that $$k = 2\pi$$, giving the well-known formula

$$T = 2\pi \sqrt{\dfrac{l}{g}}$$

Note that the time period is independent of the mass of the bob — a result predicted by dimensional analysis itself.

The dimensional formula of energy is $$[E] = [M^1 \, L^2 \, T^{-2}]$$, so the numerical value of energy in any system is

$$n \propto \dfrac{1}{(\text{unit of mass})^1 \, (\text{unit of length})^2 \, (\text{unit of time})^{-2}}$$

If $$n_1$$ is the numerical value in one system (with units $$M_1, L_1, T_1$$) and $$n_2$$ in another (with units $$M_2, L_2, T_2$$), then

$$n_1 \, M_1^{1} L_1^{2} T_1^{-2} = n_2 \, M_2^{1} L_2^{2} T_2^{-2}$$

so

$$n_2 = n_1 \left(\dfrac{M_1}{M_2}\right)^{1} \left(\dfrac{L_1}{L_2}\right)^{2} \left(\dfrac{T_1}{T_2}\right)^{-2}$$

System 1 (SI): $$M_1 = 1 \, \mathrm{kg}$$, $$L_1 = 1 \, \mathrm{m}$$, $$T_1 = 1 \, \mathrm{s}$$, and $$n_1 = 4.2$$ (since $$1 \, \mathrm{cal} = 4.2 \, \mathrm{J} = 4.2 \, \mathrm{kg \, m^2 \, s^{-2}}$$).

System 2 (new): $$M_2 = \alpha \, \mathrm{kg}$$, $$L_2 = \beta \, \mathrm{m}$$, $$T_2 = \gamma \, \mathrm{s}$$.

Substituting,

$$n_2 = 4.2 \times \left(\dfrac{1 \, \mathrm{kg}}{\alpha \, \mathrm{kg}}\right)^{1} \left(\dfrac{1 \, \mathrm{m}}{\beta \, \mathrm{m}}\right)^{2} \left(\dfrac{1 \, \mathrm{s}}{\gamma \, \mathrm{s}}\right)^{-2}$$

$$n_2 = 4.2 \, \alpha^{-1} \, \beta^{-2} \, \gamma^{2}$$

Hence in the new units the magnitude of $$1 \, \mathrm{cal}$$ is $$4.2 \, \alpha^{-1} \beta^{-2} \gamma^{2}$$, as required.

In the new system the speed of light is unity, i.e. $$c = 1 \, (\text{new unit of length}) \, \mathrm{s^{-1}}$$. The new unit of length is therefore the distance light travels in $$1 \, \mathrm{s}$$ (a 'light-second').

Convert the travel time to seconds:

$$t = 8 \, \mathrm{min} + 20 \, \mathrm{s} = 8 \times 60 \, \mathrm{s} + 20 \, \mathrm{s} = 500 \, \mathrm{s}$$

The Sun–Earth distance is

$$d = c \, t = 1 \times 500 \, (\text{new unit}) = 500 \, \text{new units of length}$$

The microscope produces a magnified image whose linear dimension is

$$\text{(image size)} = M \times \text{(object size)}$$

where $$M = 100$$ is the linear magnification. Hence

$$\text{(thickness of hair)} = \dfrac{\text{image width}}{M} = \dfrac{3.5 \, \mathrm{mm}}{100}$$

$$= 3.5 \times 10^{-2} \, \mathrm{mm} = 0.035 \, \mathrm{mm} = 3.5 \times 10^{-5} \, \mathrm{m}$$

Area magnification is the square of the linear magnification:

$$M_{\text{area}} = M_{\text{linear}}^{2}$$

Convert both areas to the same unit. With $$1 \, \mathrm{m^2} = 10^{4} \, \mathrm{cm^2}$$,

$$A_{\text{screen}} = 1.55 \, \mathrm{m^2} = 1.55 \times 10^{4} \, \mathrm{cm^2}$$

$$M_{\text{area}} = \dfrac{A_{\text{screen}}}{A_{\text{slide}}} = \dfrac{1.55 \times 10^{4} \, \mathrm{cm^2}}{1.75 \, \mathrm{cm^2}} = 8.857 \times 10^{3}$$

Hence

$$M_{\text{linear}} = \sqrt{M_{\text{area}}} = \sqrt{8857.14} \approx 94.1$$

Convert the thickness to metres: $$t = 2.01 \, \mathrm{cm} = 0.0201 \, \mathrm{m}$$. The three measurements have $$4$$, $$4$$ and $$3$$ significant figures respectively; products must therefore be quoted to $$3$$ significant figures.

Total surface area of a rectangular sheet (parallelepiped) of length $$L$$, breadth $$B$$, thickness $$t$$:

$$S = 2(LB + Bt + tL)$$

Computing each pair-wise product:

$$LB = 4.234 \times 1.005 \, \mathrm{m^2} = 4.25517 \, \mathrm{m^2}$$

$$Bt = 1.005 \times 0.0201 \, \mathrm{m^2} = 0.0202005 \, \mathrm{m^2}$$

$$tL = 0.0201 \times 4.234 \, \mathrm{m^2} = 0.0851034 \, \mathrm{m^2}$$

Sum $$= 4.25517 + 0.0202005 + 0.0851034 = 4.36047 \, \mathrm{m^2}$$

$$S = 2 \times 4.36047 = 8.72095 \, \mathrm{m^2}$$

To three significant figures, $$S \approx 8.72 \, \mathrm{m^2}$$.

Volume:

$$V = L \, B \, t = 4.234 \times 1.005 \times 0.0201 \, \mathrm{m^3}$$

$$V = 4.25517 \times 0.0201 = 0.0855289 \, \mathrm{m^3}$$

To three significant figures, $$V \approx 0.0855 \, \mathrm{m^3} = 8.55 \times 10^{-2} \, \mathrm{m^3}$$.

For addition and subtraction the result must be expressed to the same number of decimal places as the term with the fewest decimal places (i.e. with the largest uncertainty).

Express the gold-piece masses in kilograms: $$20.15 \, \mathrm{g} = 0.02015 \, \mathrm{kg}$$ and $$20.17 \, \mathrm{g} = 0.02017 \, \mathrm{kg}$$.

(a) Total mass of the box (box + two gold pieces):

$$m_{\text{total}} = 2.30 + 0.02015 + 0.02017 \, \mathrm{kg} = 2.34032 \, \mathrm{kg}$$

The least precise term is $$2.30 \, \mathrm{kg}$$, known to two decimal places. Rounding the sum to two decimal places:

$$m_{\text{total}} \approx 2.34 \, \mathrm{kg}$$

(b) Difference between the gold pieces (both known to two decimal places in grams):

$$\Delta m = 20.17 - 20.15 = 0.02 \, \mathrm{g}$$

The result $$0.02 \, \mathrm{g}$$ is correctly written to two decimal places (matching the precision of the inputs).

For the expression to make dimensional sense, the quantity inside the square root, $$1 - v^{2}$$, must be dimensionless — because we can only subtract a pure number from $$1$$.

$$v$$ has dimensions $$[L \, T^{-1}]$$, so $$v^2$$ has dimensions $$[L^{2} T^{-2}]$$ and is not dimensionless.

To make $$v^{2}$$ dimensionless, divide it by another quantity with the same dimensions. The only such constant available is the speed of light $$c$$ (with the same dimensions $$[L \, T^{-1}]$$). Hence the missing factor is $$c^{2}$$ in the denominator of $$v^{2}$$:

$$m = \dfrac{m_0}{\sqrt{1 - v^{2}/c^{2}}}$$

Now $$v/c$$ is dimensionless and the formula is dimensionally consistent. This is the standard relativistic mass formula.

Treat a hydrogen atom as a sphere of radius

$$r = 0.5 \, \mathrm{\AA} = 0.5 \times 10^{-10} \, \mathrm{m} = 5 \times 10^{-11} \, \mathrm{m}$$

Volume of one atom:

$$V_{1} = \dfrac{4}{3}\pi r^{3} = \dfrac{4}{3} \times 3.14159 \times (5 \times 10^{-11})^{3} \, \mathrm{m^3}$$

$$(5 \times 10^{-11})^{3} = 125 \times 10^{-33} = 1.25 \times 10^{-31}$$

$$V_{1} = \dfrac{4}{3} \times 3.14159 \times 1.25 \times 10^{-31} \, \mathrm{m^3} \approx 5.236 \times 10^{-31} \, \mathrm{m^3}$$

Total atomic volume in one mole (Avogadro's number $$N_A = 6.022 \times 10^{23}$$):

$$V_{\text{mole}} = N_A \times V_{1} = 6.022 \times 10^{23} \times 5.236 \times 10^{-31} \, \mathrm{m^3}$$

$$V_{\text{mole}} \approx 3.15 \times 10^{-7} \, \mathrm{m^3}$$

Take the 'size' of the hydrogen molecule to be its diameter $$\sim 1 \, \mathrm{\AA}$$, so its radius is $$r = 0.5 \, \mathrm{\AA} = 5 \times 10^{-11} \, \mathrm{m}$$.

Volume of one molecule:

$$V_{1} = \dfrac{4}{3}\pi r^{3} = \dfrac{4}{3} \times 3.14159 \times (5 \times 10^{-11})^{3} \, \mathrm{m^3} \approx 5.236 \times 10^{-31} \, \mathrm{m^3}$$

Total molecular (atomic) volume in one mole:

$$V_{\text{atomic}} = N_A V_{1} = 6.022 \times 10^{23} \times 5.236 \times 10^{-31} \, \mathrm{m^3} \approx 3.15 \times 10^{-7} \, \mathrm{m^3}$$

Molar volume of the gas at STP:

$$V_{\text{molar}} = 22.4 \, \mathrm{L} = 22.4 \times 10^{-3} \, \mathrm{m^3} = 2.24 \times 10^{-2} \, \mathrm{m^3}$$

Required ratio:

$$\dfrac{V_{\text{molar}}}{V_{\text{atomic}}} = \dfrac{2.24 \times 10^{-2}}{3.15 \times 10^{-7}} \approx 7.1 \times 10^{4}$$

This is large because in a gas the molecules are separated by distances much greater than their own size. Most of the gas volume is empty space — the molecules occupy only about $$1/70{,}000$$ of the total volume at STP. (Equivalently, the mean intermolecular separation is roughly the cube-root, $$\sim (7 \times 10^{4})^{1/3} \approx 40$$ times the molecular diameter.)

The apparent motion of an external object as seen from a moving observer is governed by the rate at which the line of sight from observer to object turns — i.e. its angular velocity.

If $$v$$ is the train's speed and $$r$$ is the perpendicular distance of an object from the line of motion of the train, then in a short interval the line of sight to the object rotates through an angle

$$\Delta\theta \approx \dfrac{v \, \Delta t}{r}$$

so the apparent angular velocity is

$$\omega = \dfrac{\Delta\theta}{\Delta t} = \dfrac{v}{r}$$

For nearby objects (trees, houses, $$r$$ small) this $$\omega$$ is large; the objects sweep rapidly across the field of view, opposite to the train's motion.

For distant objects (hilltops, Moon, stars, $$r$$ very large) $$\omega \to 0$$; their angular position barely changes over the time of observation, so they appear practically stationary and seem to 'travel along with' the observer.

(The Moon, for example, is about $$3.8 \times 10^{8} \, \mathrm{m}$$ away; even at a train speed of $$30 \, \mathrm{m \, s^{-1}}$$ its line of sight rotates by only about $$8 \times 10^{-8} \, \mathrm{rad \, s^{-1}}$$, far too slow to perceive.)

Treat the Sun as a uniform sphere of radius $$R = 7.0 \times 10^{8} \, \mathrm{m}$$ and mass $$M = 2.0 \times 10^{30} \, \mathrm{kg}$$.

Volume:

$$V = \dfrac{4}{3}\pi R^{3} = \dfrac{4}{3} \times 3.14159 \times (7.0 \times 10^{8})^{3} \, \mathrm{m^3}$$

$$(7.0 \times 10^{8})^{3} = 343 \times 10^{24} = 3.43 \times 10^{26}$$

$$V = \dfrac{4}{3} \times 3.14159 \times 3.43 \times 10^{26} \, \mathrm{m^3} \approx 1.437 \times 10^{27} \, \mathrm{m^3}$$

Mean mass density:

$$\rho = \dfrac{M}{V} = \dfrac{2.0 \times 10^{30}}{1.437 \times 10^{27}} \, \mathrm{kg \, m^{-3}}$$

$$\rho \approx 1.39 \times 10^{3} \, \mathrm{kg \, m^{-3}} \approx 1.4 \, \mathrm{g \, cm^{-3}}$$

This is comparable to the density of water ($$10^{3} \, \mathrm{kg \, m^{-3}}$$) and lies in the range characteristic of solids and liquids, not gases (which at ordinary pressures have densities $$\sim 1 \, \mathrm{kg \, m^{-3}}$$).

The reason: although the Sun's material is a hot plasma (cannot be solid or liquid), the enormous self-gravitational compression squeezes the matter to a density typical of condensed matter. So the simple thermodynamic phase classification (solid/liquid/gas) does not directly predict density at astrophysical scales.