Solved Examples

How to Use This Section

You have finished the theory. Sections 1 to 5 taught you units, significant figures, arithmetic with them, dimensional formulae and dimensional analysis, and each carried a dozen worked examples of its own. This section is different: it is one long problem set, 35 fully worked examples in an easy to medium to hard progression, and it deliberately mixes the topics up the way a real exam does.

Here is the thing about this chapter — nobody loses marks because the physics is hard. They lose marks because they converted $\mathrm{cm^3}$ to $\mathrm{m^3}$ with a factor of $10^{2}$ instead of $10^{6}$, or wrote 2.34 kg when the rule demanded 2.3 kg, or forgot that $[\eta]$ is not $[\mathrm{M\,L\,T^{-1}}]$. The only cure is volume. So work these.

The working method

1. Cover the solution. Genuinely attempt the example first, on paper, with a pen. Reading a solution feels like learning and is not.
2. Compare your steps, not just your answer. If you got the right number by a longer route, note the shorter one.
3. Read the Takeaway. Every example ends with one. That line is the transferable part; the numbers are disposable.

A last piece of advice before you start. Write the units at every line, not only at the end. Half the errors in this chapter announce themselves the moment you carry the units through, because you end up with an answer in $\mathrm{kg\ m\ s^{-1}}$ when you wanted a joule and you catch it instantly.

Solved Examples

Example 1: Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to ….. $\mathrm{m^3}$ (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ….. $\mathrm{(mm)^2}$ (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$^{-3}$.

Solution:

1. (a) Convert the side first, then cube it. $1\ \mathrm{cm} = 10^{-2}$ m, so $$V = (1\ \mathrm{cm})^3 = (10^{-2}\ \mathrm{m})^3 = 10^{-6}\ \mathrm{m^3}$$
2. (b) A solid cylinder has two flat circular ends plus the curved side, so the total surface area is $A = 2\pi r^2 + 2\pi r h = 2\pi r (r + h)$.
3. Substituting $r = 2.0$ cm and $h = 10.0$ cm: $$A = 2\pi (2.0)(2.0 + 10.0) = 2\pi \times 24.0 = 150.8\ \mathrm{cm^2}$$
4. Now convert. $1\ \mathrm{cm^2} = (10\ \mathrm{mm})^2 = 100\ \mathrm{mm^2}$, so $A = 150.8 \times 100 = 15080\ \mathrm{mm^2}$.
5. The data (2.0 and 10.0) carries 2 significant figures, so $$A = 1.5 \times 10^4\ \mathrm{mm^2}$$
6. (c) $18\ \mathrm{km\ h^{-1}} = \dfrac{18 \times 1000\ \mathrm{m}}{3600\ \mathrm{s}} = 5\ \mathrm{m\ s^{-1}}$. In 1 s it therefore covers 5 m.
7. (d) Relative density is a pure ratio: density of the substance divided by the density of water. Since water is $1\ \mathrm{g\ cm^{-3}}$, the density in $\mathrm{g\ cm^{-3}}$ is numerically the relative density, i.e. $11.3\ \mathrm{g\ cm^{-3}}$.
8. Convert: $1\ \mathrm{g\ cm^{-3}} = \dfrac{10^{-3}\ \mathrm{kg}}{10^{-6}\ \mathrm{m^3}} = 10^{3}\ \mathrm{kg\ m^{-3}}$, so $$\rho = 11.3 \times 10^3 = 1.13 \times 10^4\ \mathrm{kg\ m^{-3}}$$

Final Answer: (a) $10^{-6}\ \mathrm{m^3}$ (b) $1.5 \times 10^4\ \mathrm{mm^2}$ (c) 5 m (d) $11.3\ \mathrm{g\ cm^{-3}}$ or $1.13 \times 10^4\ \mathrm{kg\ m^{-3}}$.

Takeaway: The single most common error in this whole chapter lives in part (a). A prefix inside a squared or cubed bracket gets squared or cubed too: $\mathrm{cm^3} \to \mathrm{m^3}$ is $10^{-6}$, not $10^{-2}$. Part (d) is the other trap — relative density is dimensionless, and only becomes a density when you attach the density of water to it.

Example 2: Conversion of units

Fill in the blanks by suitable conversion of units: (a) $1\ \mathrm{kg\ m^2\ s^{-2}} = \ldots\ \mathrm{g\ cm^2\ s^{-2}}$ (b) $1\ \mathrm{m} = \ldots$ ly (c) $3.0\ \mathrm{m\ s^{-2}} = \ldots\ \mathrm{km\ h^{-2}}$ (d) $G = 6.67 \times 10^{-11}\ \mathrm{N\ m^2\ (kg)^{-2}} = \ldots\ \mathrm{(cm)^3\ s^{-2}\ g^{-1}}$

Solution:

1. (a) Replace each unit by its equivalent and multiply the numbers out: $$1\ \mathrm{kg\,m^2\,s^{-2}} = (10^3\ \mathrm{g})(10^2\ \mathrm{cm})^2 (\mathrm{s^{-2}}) = 10^3 \times 10^4 = 10^{7}\ \mathrm{g\,cm^2\,s^{-2}}$$
2. (b) One light year is the distance light travels in one year, $1\ \mathrm{ly} = 9.46 \times 10^{15}$ m. Inverting, $$1\ \mathrm{m} = \frac{1}{9.46 \times 10^{15}}\ \mathrm{ly} = 1.057 \times 10^{-16}\ \mathrm{ly}$$
3. (c) Here the time unit is squared, and this is where signs go wrong. Write each replacement separately: $1\ \mathrm{m} = 10^{-3}$ km, and $1\ \mathrm{s} = \dfrac{1}{3600}$ h, so $\mathrm{s^{-2}} = (3600)^{2}\ \mathrm{h^{-2}}$.
4. Therefore $$3.0\ \mathrm{m\,s^{-2}} = 3.0 \times 10^{-3} \times (3600)^2\ \mathrm{km\,h^{-2}} = 3.0 \times 10^{-3} \times 1.296 \times 10^{7}$$ $$= 3.888 \times 10^{4}\ \mathrm{km\,h^{-2}}$$
5. (d) First simplify the unit of $G$. Since $1\ \mathrm{N} = 1\ \mathrm{kg\,m\,s^{-2}}$, $$\mathrm{N\,m^2\,kg^{-2}} = (\mathrm{kg\,m\,s^{-2}})(\mathrm{m^2})(\mathrm{kg^{-2}}) = \mathrm{m^3\,s^{-2}\,kg^{-1}}$$
6. Now convert $\mathrm{m^3} \to \mathrm{cm^3}$ (a factor $10^6$) and $\mathrm{kg^{-1}} \to \mathrm{g^{-1}}$ (a factor $10^{-3}$, because a gram is smaller so there are fewer of the new unit in the denominator sense — safest is to write $\mathrm{kg^{-1}} = (10^3\ \mathrm{g})^{-1} = 10^{-3}\ \mathrm{g^{-1}}$): $$G = 6.67 \times 10^{-11} \times 10^{6} \times 10^{-3} = 6.67 \times 10^{-8}\ \mathrm{cm^3\,s^{-2}\,g^{-1}}$$

Final Answer: (a) $10^{7}$ (b) $1.057 \times 10^{-16}$ (c) $3.888 \times 10^{4}$ (d) $6.67 \times 10^{-8}$.

Takeaway: Never convert a compound unit in one mental leap. Break it into single units, write each replacement in brackets with its own power, and only then multiply the powers of ten. Part (d) also shows a habit worth keeping: simplify the unit before you convert it. Turning $\mathrm{N\,m^2\,kg^{-2}}$ into $\mathrm{m^3\,s^{-2}\,kg^{-1}}$ first makes the conversion a two-line job instead of a five-line one.

Example 3: The very small units — angstrom and fermi

(a) The wavelength of yellow sodium light is 5893 Å. Express it in m, in nm and in $\mu$m. (b) The radius of a gold nucleus is about 6.9 fermi. Express it in m and in Å. (c) Roughly how many times bigger is an atom than the nucleus inside it, taking the atom as 1 Å across and the nucleus as 1.2 fermi?

Solution:

1. Recall the two definitions. 1 Å $= 10^{-10}$ m (angstrom, the atomic-scale unit) and 1 fermi $= 10^{-15}$ m (also written fm, the nuclear-scale unit).
2. (a) In metres: $\lambda = 5893 \times 10^{-10} = 5.893 \times 10^{-7}$ m.
3. In nanometres: $1\ \mathrm{nm} = 10^{-9}$ m, so $\lambda = \dfrac{5.893 \times 10^{-7}}{10^{-9}} = 589.3$ nm.
4. In micrometres: $1\ \mu\mathrm{m} = 10^{-6}$ m, so $\lambda = \dfrac{5.893 \times 10^{-7}}{10^{-6}} = 0.5893\ \mu$m.
5. (b) $R = 6.9 \times 10^{-15}$ m. In angstrom: $R = \dfrac{6.9 \times 10^{-15}}{10^{-10}} = 6.9 \times 10^{-5}$ Å.
6. (c) Take the ratio directly, in metres: $$\frac{d_{\text{atom}}}{d_{\text{nucleus}}} = \frac{1 \times 10^{-10}}{1.2 \times 10^{-15}} = 8.3 \times 10^{4}$$

Final Answer: (a) $5.893 \times 10^{-7}$ m $=$ 589.3 nm $=$ 0.5893 $\mu$m. (b) $6.9 \times 10^{-15}$ m $= 6.9 \times 10^{-5}$ Å. (c) About $8.3 \times 10^4$ times, i.e. roughly $10^5$.

Takeaway: Learn the ladder 1 Å $= 10^{5}$ fermi and 1 nm $=$ 10 Å and you will never have to think about these again. Part (c) is worth remembering as a fact: the atom is about $10^5$ times the nucleus, which is why almost all of an atom is empty space.

Example 4: Pressure in four different units

(a) Show that $1\ \mathrm{N\ m^{-2}} = 10\ \mathrm{dyne\ cm^{-2}}$. (b) Standard atmospheric pressure is $1.013 \times 10^5$ Pa. Express it in dyne cm$^{-2}$ and in bar. (c) A normal systolic blood pressure reading is "120 mm of Hg". Convert it to pascal. Take $\rho_{\mathrm{Hg}} = 13600\ \mathrm{kg\ m^{-3}}$ and $g = 9.8\ \mathrm{m\ s^{-2}}$.

Solution:

1. (a) $1\ \mathrm{N} = 10^5$ dyne and $1\ \mathrm{m^2} = 10^4\ \mathrm{cm^2}$. Therefore $$1\ \mathrm{N\,m^{-2}} = \frac{10^{5}\ \mathrm{dyne}}{10^{4}\ \mathrm{cm^2}} = 10\ \mathrm{dyne\,cm^{-2}}$$
2. (b) Using the factor just proved, $$1\ \mathrm{atm} = 1.013 \times 10^{5} \times 10 = 1.013 \times 10^{6}\ \mathrm{dyne\,cm^{-2}}$$
3. The bar is defined as $1\ \mathrm{bar} = 10^{5}$ Pa exactly, so $$1\ \mathrm{atm} = \frac{1.013 \times 10^{5}}{10^{5}} = 1.013\ \mathrm{bar}$$
4. (c) "120 mm of Hg" is not a pressure unit in disguise — it means the pressure at the base of a mercury column 120 mm high, $P = \rho g h$.
5. Convert the height to SI first: $h = 120\ \mathrm{mm} = 0.120$ m. Then $$P = 13600 \times 9.8 \times 0.120 = 15993.6\ \mathrm{Pa}$$
6. The data has 2 significant figures (120), so $P = 1.6 \times 10^4$ Pa $= 16$ kPa.

Final Answer: (a) $1\ \mathrm{N\,m^{-2}} = 10\ \mathrm{dyne\,cm^{-2}}$. (b) $1.013 \times 10^6\ \mathrm{dyne\,cm^{-2}} = 1.013$ bar. (c) About $1.6 \times 10^4$ Pa.

Takeaway: Two things to bank. First, the conversion factor $1\ \mathrm{Pa} = 10\ \mathrm{dyne\,cm^{-2}}$ (many students guess $10^{5}/10^{4}$ wrongly as $10^{9}$ by multiplying instead of dividing). Second, [NEET Important] any pressure quoted as a column height must go through $P = \rho g h$ — the mm is a length, not a pressure.

Example 5: The units of energy you actually meet

(a) Your electricity bill charges you per "unit", which is 1 kilowatt-hour. Express 1 kWh in joules. (b) A 2.0 kW geyser runs for 3.0 hours. How many units does it consume, and how much energy is that in joules? (c) Express 1 joule in erg and in electronvolt, given $1\ \mathrm{eV} = 1.6 \times 10^{-19}$ J.

Solution:

1. (a) A kilowatt-hour is a power multiplied by a time, so it is an energy. Convert both factors to SI: $$1\ \mathrm{kWh} = (1000\ \mathrm{W})(3600\ \mathrm{s}) = 3.6 \times 10^{6}\ \mathrm{J}$$
2. (b) Energy consumed $= P \times t = 2.0\ \mathrm{kW} \times 3.0\ \mathrm{h} = 6.0$ kWh, i.e. 6.0 units.
3. In joules: $6.0 \times 3.6 \times 10^{6} = 2.16 \times 10^{7}$ J.
4. (c) In CGS the unit of energy is the erg, and $1\ \mathrm{J} = 10^{7}$ erg (this is just $1\ \mathrm{kg\,m^2\,s^{-2}} = 10^3 \times 10^4\ \mathrm{g\,cm^2\,s^{-2}}$ from Example 2a).
5. In electronvolt: $$1\ \mathrm{J} = \frac{1}{1.6 \times 10^{-19}}\ \mathrm{eV} = 6.25 \times 10^{18}\ \mathrm{eV}$$

Final Answer: (a) $3.6 \times 10^6$ J (b) 6.0 units $= 2.16 \times 10^7$ J (c) $10^7$ erg $= 6.25 \times 10^{18}$ eV.

Takeaway: The joule is a hopeless unit for both ends of the scale, which is why physics keeps the eV for atoms and the kWh for kettles. All three are the same dimension $[\mathrm{M\,L^2\,T^{-2}}]$ — only the size differs. [JEE Tip] $1\ \mathrm{kWh} = 3.6 \times 10^6$ J and $1\ \mathrm{J} = 10^7$ erg are worth memorising outright.

Example 6: When the prefix gets squared or cubed

Express in SI base units: (a) $1\ \mathrm{mm^2}$ in $\mathrm{m^2}$ (b) $1\ \mathrm{\mu m^3}$ in $\mathrm{m^3}$ (c) $1\ \mathrm{nm^2}$ in $\mathrm{m^2}$ (d) a length of $4.5 \times 10^{-8}$ m in nm and in Å (e) a density of $2.7\ \mathrm{g\ cm^{-3}}$ in $\mathrm{kg\ m^{-3}}$.

Solution:

1. The one rule that governs (a) to (c): the prefix belongs to the unit before the power is applied, so the power hits the prefix too. Always rewrite the prefixed unit in brackets.
2. (a) $1\ \mathrm{mm^2} = (10^{-3}\ \mathrm{m})^2 = 10^{-6}\ \mathrm{m^2}$.
3. (b) $1\ \mathrm{\mu m^3} = (10^{-6}\ \mathrm{m})^3 = 10^{-18}\ \mathrm{m^3}$.
4. (c) $1\ \mathrm{nm^2} = (10^{-9}\ \mathrm{m})^2 = 10^{-18}\ \mathrm{m^2}$.
5. (d) $\dfrac{4.5 \times 10^{-8}}{10^{-9}} = 45$ nm, and $\dfrac{4.5 \times 10^{-8}}{10^{-10}} = 450$ Å.
6. (e) Numerator and denominator separately: $2.7\ \mathrm{g\,cm^{-3}} = \dfrac{2.7 \times 10^{-3}\ \mathrm{kg}}{(10^{-2}\ \mathrm{m})^{3}} = \dfrac{2.7 \times 10^{-3}}{10^{-6}} = 2.7 \times 10^{3}\ \mathrm{kg\,m^{-3}}$.

Final Answer: (a) $10^{-6}\ \mathrm{m^2}$ (b) $10^{-18}\ \mathrm{m^3}$ (c) $10^{-18}\ \mathrm{m^2}$ (d) 45 nm $=$ 450 Å (e) $2.7 \times 10^3\ \mathrm{kg\,m^{-3}}$ (this is aluminium).

Takeaway: Notice that $\mathrm{\mu m^3}$ and $\mathrm{nm^2}$ both come out as $10^{-18}$ — pure coincidence, and exactly the kind of thing you will get wrong if you memorise answers instead of the method. Bracket the prefixed unit, apply the power, then multiply. [Board Important] The universal shortcut: to go from $\mathrm{g\ cm^{-3}}$ to $\mathrm{kg\ m^{-3}}$, multiply by $10^{3}$. Every time.

Example 7: Counting significant figures — a fresh drill

State the number of significant figures in each: (i) 0.05070 kg (ii) $5.0 \times 10^{-3}$ m (iii) 1200 s (iv) 1200.0 s (v) 30.00 mL (vi) 0.0001 m (vii) $3.0800 \times 10^{5}$ N.

Solution:

1. (i) 0.05070 kg. The leading zeros (before the 5) are not significant — they only place the decimal point. The sandwiched zero is significant, and the trailing zero after a decimal point is significant. Digits counted: 5, 0, 7, 0 $\to$ 4.
2. (ii) $5.0 \times 10^{-3}$ m. In scientific notation only the mantissa is counted; the power of ten never contributes. Digits: 5, 0 $\to$ 2.
3. (iii) 1200 s. Trailing zeros in a number without a decimal point are not significant. Digits: 1, 2 $\to$ 2.
4. (iv) 1200.0 s. The decimal point is now present, so every trailing zero counts. Digits: 1, 2, 0, 0, 0 $\to$ 5.
5. (v) 30.00 mL. Sandwiched zero plus two trailing zeros after a decimal point, all significant $\to$ 4.
6. (vi) 0.0001 m. All four zeros are leading zeros. Only the 1 survives $\to$ 1.
7. (vii) $3.0800 \times 10^{5}$ N. Mantissa only: 3, 0, 8, 0, 0 $\to$ 5.

Final Answer: (i) 4 (ii) 2 (iii) 2 (iv) 5 (v) 4 (vi) 1 (vii) 5.

Takeaway: Compare (iii) and (iv). Adding a single decimal point took the count from 2 to 5 without changing the value at all — that is how much information the notation carries. [JEE Tip] If a number in an exam is written as 1200 with no decimal point and no scientific notation, treat it as 2 significant figures unless the question tells you otherwise.

Example 8: Into and out of scientific notation

(a) Write the astronomical unit, 149600000 km, in scientific notation, and then round it to 3 significant figures. (b) The charge on an electron is 0.00000000000000000016 C. Write it in scientific notation. (c) The speed of light is 299792458 m s$^{-1}$. Write it to 3 significant figures. (d) Avogadro's number is 602200000000000000000000 per mole. Write it in scientific notation.

Solution:

1. The recipe. Move the decimal point until exactly one non-zero digit sits to its left. Moving it $n$ places left gives $\times 10^{+n}$; moving it $n$ places right gives $\times 10^{-n}$.
2. (a) 149600000 km. The decimal is currently after the final zero; move it 8 places left: $$1.496 \times 10^{8}\ \mathrm{km}$$ That is 4 significant figures. To 3 significant figures the digit dropped is 6, which is more than 5, so the 9 rounds up to 10 and carries: $1.50 \times 10^{8}$ km.
3. (b) Move the decimal 19 places right: $1.6 \times 10^{-19}$ C — 2 significant figures.
4. (c) $299792458 = 2.99792458 \times 10^{8}$. To 3 figures, the digit dropped is 7, so 9 rounds up and carries again: $3.00 \times 10^{8}\ \mathrm{m\ s^{-1}}$.
5. (d) Move the decimal 23 places left: $6.022 \times 10^{23}\ \mathrm{mol^{-1}}$ — 4 significant figures.

Final Answer: (a) $1.496 \times 10^8$ km, i.e. $1.50 \times 10^8$ km to 3 s.f. (b) $1.6 \times 10^{-19}$ C (c) $3.00 \times 10^8\ \mathrm{m\ s^{-1}}$ (d) $6.022 \times 10^{23}\ \mathrm{mol^{-1}}$.

Takeaway: Parts (a) and (c) both hide a carry. Rounding $1.496 \to 1.50$ and $2.998 \to 3.00$ requires the trailing zeros to be written, because dropping them would falsely claim fewer significant figures. [Board Important] Writing $3 \times 10^8$ when the question asked for 3 significant figures is a mark lost — it says 1 significant figure.

Example 9: The volume of a mole of atoms

The unit of length convenient on the atomic scale is known as an angstrom, 1 Å $= 10^{-10}$ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in $\mathrm{m^3}$ of one mole of hydrogen atoms?

Solution:

1. Read "size" carefully. The size of the atom is given as 0.5 Å, and the standard reading is that this is the radius. So $r =$ 0.5 Å $= 0.5 \times 10^{-10}$ m $= 5 \times 10^{-11}$ m.
2. Volume of one atom, treating it as a sphere: $$V_1 = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.5 \times 10^{-10})^3$$
3. Cube the bracket first: $(0.5 \times 10^{-10})^3 = 0.125 \times 10^{-30} = 1.25 \times 10^{-31}\ \mathrm{m^3}$.
4. Multiply by $\frac{4}{3}\pi = 4.19$: $$V_1 = 4.19 \times 1.25 \times 10^{-31} = 5.24 \times 10^{-31}\ \mathrm{m^3}$$
5. Scale up to a mole. One mole contains $N_A = 6.022 \times 10^{23}$ atoms, and Avogadro's number is a defined (exact) quantity here: $$V = N_A V_1 = 6.022 \times 10^{23} \times 5.24 \times 10^{-31}$$
6. Multiply mantissas and add exponents: $6.022 \times 5.24 = 31.5$, and $10^{23} \times 10^{-31} = 10^{-8}$, giving $31.5 \times 10^{-8} = 3.15 \times 10^{-7}\ \mathrm{m^3}$.
7. The data ("about 0.5 Å") carries 1 significant figure, so honestly this is $3 \times 10^{-7}\ \mathrm{m^3}$.

Final Answer: $V \approx 3.15 \times 10^{-7}\ \mathrm{m^3}$, i.e. about $3 \times 10^{-7}\ \mathrm{m^3}$ (roughly 0.3 mL).

Takeaway: A mole of hydrogen atoms, packed solid, would fit in a third of a millilitre. Hold on to that number — the next example uses it to make a point about gases. [NEET Important] The cubing step is where marks vanish: $(0.5)^3 = 0.125$, not 0.5, and $(10^{-10})^3 = 10^{-30}$, not $10^{-13}$.

Example 10: Why a gas is mostly nothing

One mole of an ideal gas at STP occupies 22.4 L (the molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? Take the size of a hydrogen molecule to be about 1 Å. Why is this ratio so large?

Solution:

1. Molar volume in SI. $1\ \mathrm{L} = 10^{-3}\ \mathrm{m^3}$, so $$V_{\text{molar}} = 22.4\ \mathrm{L} = 22.4 \times 10^{-3}\ \mathrm{m^3} = 2.24 \times 10^{-2}\ \mathrm{m^3}$$
2. Atomic (molecular) volume. "Size about 1 Å" means a diameter of 1 Å, so the radius is 0.5 Å $= 0.5 \times 10^{-10}$ m — exactly the figure from Example 9. Therefore $$V_{\text{atomic}} = 3.15 \times 10^{-7}\ \mathrm{m^3}$$
3. Take the ratio: $$\frac{V_{\text{molar}}}{V_{\text{atomic}}} = \frac{2.24 \times 10^{-2}}{3.15 \times 10^{-7}}$$
4. Divide the mantissas, subtract the exponents: $\dfrac{2.24}{3.15} = 0.711$ and $10^{-2-(-7)} = 10^{5}$, so the ratio is $0.711 \times 10^{5} = 7.1 \times 10^{4}$.
5. Why so large? In a gas the molecules are far apart compared with their own size. The inter-molecular separation is roughly the cube root of the volume per molecule, which is about $(7.1 \times 10^4)^{1/3} \approx 41$ times the molecular diameter. So a gas is essentially empty space with occasional molecules in it, which is exactly why gases are compressible and liquids are not.

Final Answer: The ratio is about $7.1 \times 10^{4}$, i.e. of order $10^{5}$. It is large because in the gaseous state the inter-molecular separation is roughly 40 times the size of a molecule.

Takeaway: This is a numerical answer that carries real physics. [NEET Important] The examiner's favourite follow-up is "estimate the mean free path" or "why is a liquid nearly incompressible" — both are answered by the same fact: in a liquid the molecules are already touching, in a gas they are $10^4$ to $10^5$ volumes apart.

Example 11: A Fermi estimate — the air in your classroom

Estimate, to the nearest order of magnitude, (a) the mass of air in a classroom measuring roughly 5 m $\times$ 4 m $\times$ 3 m, and (b) the number of gas molecules in it. Take the density of air as $1.2\ \mathrm{kg\ m^{-3}}$ and the molar volume as 22.4 L.

Solution:

1. (a) Volume of the room: $V = 5 \times 4 \times 3 = 60\ \mathrm{m^3}$.
2. Mass: $m = \rho V = 1.2 \times 60 = 72$ kg.
3. Order of magnitude. Write it as $7.2 \times 10^{1}$ kg. Since the mantissa $7.2 > 5$, we round the power up: the order of magnitude is $10^{2}$ kg.
4. (b) Number of moles. Using the molar volume $22.4\ \mathrm{L} = 2.24 \times 10^{-2}\ \mathrm{m^3}$, $$n = \frac{60}{2.24 \times 10^{-2}} = 2.68 \times 10^{3}\ \mathrm{mol}$$
5. Number of molecules: $$N = n N_A = 2.68 \times 10^{3} \times 6.022 \times 10^{23} = 1.61 \times 10^{27}$$
6. Order of magnitude. The mantissa $1.61 \le 5$, so the power stays: the order of magnitude is $10^{27}$.

Final Answer: (a) About 72 kg, order of magnitude $10^{2}$ kg. (b) About $1.6 \times 10^{27}$ molecules, order of magnitude $10^{27}$.

Takeaway: Most students guess "a few kilograms" for the air in a room and are out by a factor of 20 — the air above you weighs about as much as you do. Note also the two different rounding decisions in steps 3 and 6: order of magnitude uses the rule round the power up if the mantissa exceeds 5, and 7.2 does while 1.61 does not.

Example 12: The density of the Sun

The Sun is a hot plasma with an inner core above $10^7$ K and a surface at about 6000 K, so no substance in it can be a solid or a liquid. In what range do you expect the mass density of the Sun to lie — that of solids and liquids, or that of gases? Check your guess: mass of the Sun $= 2.0 \times 10^{30}$ kg, radius $= 7.0 \times 10^8$ m.

Solution:

1. The guess. Since the Sun is entirely gaseous (a plasma), the instinctive guess is a gas-like density, around $1\ \mathrm{kg\ m^{-3}}$.
2. Volume of the Sun, treating it as a sphere: $$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (7.0 \times 10^{8})^{3}$$
3. Cube first: $(7.0)^3 = 343$ and $(10^8)^3 = 10^{24}$, so $R^3 = 343 \times 10^{24} = 3.43 \times 10^{26}\ \mathrm{m^3}$.
4. Multiply by $\frac{4}{3}\pi = 4.19$: $$V = 4.19 \times 3.43 \times 10^{26} = 1.44 \times 10^{27}\ \mathrm{m^3}$$
5. Density: $$\rho = \frac{M}{V} = \frac{2.0 \times 10^{30}}{1.44 \times 10^{27}} = 1.39 \times 10^{3}\ \mathrm{kg\ m^{-3}}$$
6. To 2 significant figures (the data has 2), $\rho = 1.4 \times 10^{3}\ \mathrm{kg\ m^{-3}}$.
7. Compare. Water is $1.0 \times 10^3\ \mathrm{kg\ m^{-3}}$ and air is about $1.2\ \mathrm{kg\ m^{-3}}$. The Sun's mean density is 1.4 times that of water and about a thousand times that of air.

Final Answer: $\rho \approx 1.4 \times 10^{3}\ \mathrm{kg\ m^{-3}}$ — the range of solids and liquids, not gases. The initial guess is wrong.

Takeaway: The resolution of the paradox is gravity. The Sun's enormous mass compresses its interior so hard that the plasma reaches liquid-like densities even at $10^7$ K, and the core is far denser than this average. [Board Important] This is a favourite two-mark reasoning question: the phase (gas/plasma) and the density are set by different things — temperature decides the phase, pressure decides the density.

Example 13: The grocer, the box and the gold

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, and (b) the difference in the masses of the pieces, to correct significant figures?

Solution:

1. (a) Put everything in one unit first. $20.15\ \mathrm{g} = 0.02015$ kg and $20.17\ \mathrm{g} = 0.02017$ kg.
2. Add: $$2.30 + 0.02015 + 0.02017 = 2.34032\ \mathrm{kg}$$
3. Apply the addition rule — the result keeps as many decimal places as the term with the fewest. Counting decimal places: 2.30 has 2, the gold pieces have 5 each. The least is 2.
4. Rounding 2.34032 to 2 decimal places: the first digit dropped is 0, which is less than 5, so the preceding digit is unchanged. Total mass $= 2.34$ kg.
5. (b) Difference: $20.17 - 20.15 = 0.02$ g. Both have 2 decimal places, so the answer keeps 2 decimal places: $0.02$ g.
6. Notice what subtraction did to the precision. Both inputs had 4 significant figures; the difference has just 1.

Final Answer: (a) 2.34 kg (b) 0.02 g (1 significant figure).

Takeaway: [Board Important] Some solution keys quote the answer to (a) as 2.3 kg, reading the grocer's balance as trustworthy only to 0.1 kg. Follow the printed data: 2.30 kg states 2 decimal places, so the decimal-place rule gives 2.34 kg. The deeper lesson is part (b) — adding a tiny precise number to a big rough one gains you nothing, and subtracting two nearly equal numbers destroys significant figures wholesale.

Example 14: Rounding off, including the awkward cases

Round off as instructed: (i) 3.765 to 3 s.f. (ii) 3.755 to 3 s.f. (iii) 0.04653 to 2 s.f. (iv) 87.65 to 3 s.f. (v) 1.2345 to 4 s.f. (vi) 9.996 to 3 s.f.

Solution:

1. Look only at the first digit you are about to drop. More than 5, round up. Less than 5, leave alone. Exactly 5 with nothing after it, use the round-to-even convention.
2. (i) 3.765 $\to$ 3 s.f. The dropped digit is exactly 5, and the preceding digit 6 is even, so the 5 is simply dropped: 3.76.
3. (ii) 3.755 $\to$ 3 s.f. The dropped digit is exactly 5, but the preceding digit 5 is odd, so it is raised by 1: 3.76.
4. (iii) 0.04653 $\to$ 2 s.f. The two significant digits are 4 and 6. The part being dropped is "53", which is more than 5, so the 6 rounds up: 0.047.
5. (iv) 87.65 $\to$ 3 s.f. Dropped digit is exactly 5, preceding digit 6 is even, so drop it: 87.6.
6. (v) 1.2345 $\to$ 4 s.f. Dropped digit is exactly 5, preceding digit 4 is even, so drop it: 1.234.
7. (vi) 9.996 $\to$ 3 s.f. Dropped digit 6 is more than 5, so 9 rounds up, which carries all the way: $9.996 \to 10.0$. Write it as 10.0, keeping the trailing zero so that it still shows 3 significant figures.

Final Answer: (i) 3.76 (ii) 3.76 (iii) 0.047 (iv) 87.6 (v) 1.234 (vi) 10.0.

Takeaway: Parts (i) and (ii) land on the same answer from opposite sides — that is the whole point of the round-to-even convention, and the fastest way to remember it. [JEE Tip] Part (iii) is the trap: the round-to-even rule applies only when the dropped part is exactly 5 with nothing after it. "53" is more than 5, so you round up normally. And never write part (vi) as "10" — that would claim 2 significant figures.

Example 15: The projector

The photograph of a house occupies an area of $1.75\ \mathrm{cm^2}$ on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is $1.55\ \mathrm{m^2}$. What is the linear magnification of the projector-screen arrangement?

Solution:

1. Distinguish the two magnifications. If the linear magnification is $m$, every length is multiplied by $m$, so every area is multiplied by $m^2$. That is the key relation: $$\text{areal magnification} = m^{2} = \frac{A_{\text{screen}}}{A_{\text{slide}}}$$
2. Put both areas in the same unit. Working in $\mathrm{cm^2}$: $1\ \mathrm{m^2} = 10^4\ \mathrm{cm^2}$, so $$A_{\text{screen}} = 1.55\ \mathrm{m^2} = 1.55 \times 10^{4}\ \mathrm{cm^2}$$
3. Form the ratio: $$m^{2} = \frac{1.55 \times 10^{4}}{1.75} = 8857$$
4. Take the square root: $$m = \sqrt{8857} = 94.11$$
5. Significant figures. Both areas are given to 3 significant figures, and the operations are division and a square root, so the answer keeps 3: $m = 94.1$.
6. The "35 mm slide" in the question is a red herring — it names the film format and plays no part in the calculation.

Final Answer: The linear magnification is about 94.1.

Takeaway: Areas scale as the square of the linear magnification. Students who divide $1.55\ \mathrm{m^2}$ by $1.75\ \mathrm{cm^2}$ and stop get 8857 and call it the magnification — off by a factor of 94. [JEE Tip] The same square-and-cube logic runs through the whole chapter: lengths $\times m$, areas $\times m^2$, volumes $\times m^3$.

Example 16: The thickness of a human hair

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 the hair?

Solution:

1. What magnification means. The microscope makes the hair appear 100 times wider than it really is: $$\text{observed width} = 100 \times \text{true thickness}$$
2. Invert it: $$\text{true thickness} = \frac{\text{observed width}}{100} = \frac{3.5\ \mathrm{mm}}{100} = 0.035\ \mathrm{mm}$$
3. In SI: $0.035\ \mathrm{mm} = 0.035 \times 10^{-3}\ \mathrm{m} = 3.5 \times 10^{-5}$ m.
4. Significant figures. The magnification 100 is an exact number (a counted specification, not a measurement), so it has infinite significant figures and does not limit the answer. Only 3.5 mm counts, with 2 significant figures, so the answer keeps 2: $3.5 \times 10^{-5}$ m.
5. Why 20 observations? Averaging many readings reduces the random error in the mean. It does not change the method, only the reliability of the 3.5 mm.

Final Answer: The thickness of the hair is about $0.035\ \mathrm{mm} = 3.5 \times 10^{-5}\ \mathrm{m}$, i.e. roughly 35 $\mu$m.

Takeaway: [Board Important] The magnification 100 is exact, which is why the answer still has 2 significant figures and not 1. This is the same principle as dividing by the number of oscillations when you find a time period — a counted number never degrades your precision.

Example 17: Percentage error in a resistance

In an experiment the potential difference across a resistor is measured as $V = 5.0 \pm 0.2$ V and the current through it as $I = 2.00 \pm 0.05$ A. Find the resistance and the percentage error in it, and quote the result properly.

Solution:

1. Central value. By Ohm's law, $$R = \frac{V}{I} = \frac{5.0}{2.00} = 2.5\ \Omega$$
2. Percentage error in each measurement: $$\frac{\Delta V}{V} \times 100 = \frac{0.2}{5.0} \times 100 = 4.0\%, \qquad \frac{\Delta I}{I} \times 100 = \frac{0.05}{2.00} \times 100 = 2.5\%$$
3. Combine. For a quotient the percentage errors add (they never subtract, no matter that the quantity is being divided): $$\frac{\Delta R}{R} \times 100 = 4.0 + 2.5 = 6.5\%$$
4. Absolute error: $$\Delta R = \frac{6.5}{100} \times 2.5 = 0.1625 \approx 0.2\ \Omega$$
5. Quote it. The absolute error is rounded to one significant figure, and the central value is then rounded to the same decimal place: $$R = 2.5 \pm 0.2\ \Omega$$

Final Answer: $R = 2.5\ \Omega$ with a percentage error of $6.5\%$, quoted as $R = 2.5 \pm 0.2\ \Omega$.

Takeaway: [JEE/NEET] For $Z = A/B$ or $Z = AB$, the relative errors add. The instinct to subtract for a division is wrong — errors do not cancel, they accumulate, because you must plan for the worst case. Also note step 5: an error is quoted to one significant figure, and the answer is then matched to it.

Example 18: Error in a kinetic energy

A body of mass $m = 2.0 \pm 0.1$ kg is moving with speed $v = 10.0 \pm 0.5\ \mathrm{m\ s^{-1}}$. Find its kinetic energy and the percentage error in it.

Solution:

1. Central value. $$K = \frac{1}{2}mv^{2} = \frac{1}{2}(2.0)(10.0)^{2} = 100\ \mathrm{J}$$
2. Percentage errors in the inputs: $$\frac{\Delta m}{m} \times 100 = \frac{0.1}{2.0} \times 100 = 5\%, \qquad \frac{\Delta v}{v} \times 100 = \frac{0.5}{10.0} \times 100 = 5\%$$
3. Apply the power rule. For $Z = A^{p}B^{q}$, the relative errors add with their exponents as weights: $$\frac{\Delta K}{K} = \frac{\Delta m}{m} + 2\,\frac{\Delta v}{v}$$ The factor $\frac{1}{2}$ is a pure number and contributes nothing.
4. Substitute: $$\frac{\Delta K}{K} \times 100 = 5 + 2(5) = 15\%$$
5. Absolute error: $\Delta K = 0.15 \times 100 = 15$ J, so $K = 100 \pm 15$ J.

Final Answer: $K = 100$ J with a percentage error of $15\%$, i.e. $K = 100 \pm 15$ J.

Takeaway: The speed and the mass were each measured to the same $5\%$, yet the speed contributes twice as much error, because it enters squared. [JEE Tip] When a question asks "which quantity should be measured most carefully?", the answer is always the one with the largest exponent, not the one with the largest value.

Example 19: Dimensional formulae of thermal quantities

Write the dimensional formulae of (i) heat $Q$, (ii) specific heat capacity $c$, (iii) latent heat $L$, (iv) coefficient of thermal conductivity $K$, (v) the universal gas constant $R$, and (vi) Boltzmann's constant $k$.

Solution:

1. (i) Heat. Heat is energy in transit, so it has exactly the dimensions of energy: $$[Q] = [\mathrm{M\,L^{2}\,T^{-2}}]$$
2. (ii) Specific heat capacity. From $Q = mc\,\Delta T$, we get $c = \dfrac{Q}{m\,\Delta T}$: $$[c] = \frac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{M}][\mathrm{K}]} = [\mathrm{M^{0}\,L^{2}\,T^{-2}\,K^{-1}}]$$
3. (iii) Latent heat. From $Q = mL$, so $L = Q/m$ — no temperature appears, because a phase change happens at constant temperature: $$[L] = \frac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{M}]} = [\mathrm{M^{0}\,L^{2}\,T^{-2}}]$$
4. (iv) Thermal conductivity. From the conduction equation $Q = \dfrac{K A (\Delta T) t}{x}$, rearranged as $K = \dfrac{Q x}{A (\Delta T) t}$: $$[K] = \frac{[\mathrm{M\,L^{2}\,T^{-2}}][\mathrm{L}]}{[\mathrm{L^{2}}][\mathrm{K}][\mathrm{T}]} = \frac{[\mathrm{M\,L^{3}\,T^{-2}}]}{[\mathrm{L^{2}\,K\,T}]} = [\mathrm{M\,L\,T^{-3}\,K^{-1}}]$$
5. (v) Gas constant. From $PV = nRT$, so $R = \dfrac{PV}{nT}$. Note that $[PV] = [\mathrm{M\,L^{-1}\,T^{-2}}][\mathrm{L^{3}}] = [\mathrm{M\,L^{2}\,T^{-2}}]$, which is an energy — as it must be: $$[R] = \frac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{mol}][\mathrm{K}]} = [\mathrm{M\,L^{2}\,T^{-2}\,K^{-1}\,mol^{-1}}]$$
6. (vi) Boltzmann's constant. $k = R/N_A$, and $N_A$ carries the dimension $[\mathrm{mol^{-1}}]$, so dividing removes the mole: $$[k] = [\mathrm{M\,L^{2}\,T^{-2}\,K^{-1}}]$$

Final Answer: $[Q] = [\mathrm{M\,L^{2}\,T^{-2}}]$, $[c] = [\mathrm{L^{2}\,T^{-2}\,K^{-1}}]$, $[L] = [\mathrm{L^{2}\,T^{-2}}]$, $[K] = [\mathrm{M\,L\,T^{-3}\,K^{-1}}]$, $[R] = [\mathrm{M\,L^{2}\,T^{-2}\,K^{-1}\,mol^{-1}}]$, $[k] = [\mathrm{M\,L^{2}\,T^{-2}\,K^{-1}}]$.

Takeaway: Every one of these is "energy divided by something". Once you see that, the whole family is one step of division away from $[\mathrm{M\,L^2\,T^{-2}}]$. [JEE Tip] Two traps here: latent heat has no $[\mathrm{K}]$ in it (constant temperature), and $[c] = [L]/[\mathrm{K}]$ exactly — specific heat and latent heat differ by one power of temperature and nothing else.

Example 20: Dimensional formulae of modern-physics constants

Find the dimensional formulae of (i) Planck's constant $h$, (ii) Stefan's constant $\sigma$, (iii) Wien's constant $b$, (iv) the Rydberg constant $R_{\infty}$. Then (v) use your answer to (i) to check the de Broglie relation $\lambda = h/p$.

Solution:

1. (i) Planck's constant. The defining relation is $E = h\nu$, so $h = E/\nu$. Frequency has $[\nu] = [\mathrm{T^{-1}}]$: $$[h] = \frac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{T^{-1}}]} = [\mathrm{M\,L^{2}\,T^{-1}}]$$
2. (ii) Stefan's constant. Stefan's law says the energy radiated per unit area per unit time is $\dfrac{E}{At} = \sigma T^{4}$, so $\sigma = \dfrac{E}{A t T^{4}}$: $$[\sigma] = \frac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{L^{2}}][\mathrm{T}][\mathrm{K^{4}}]} = [\mathrm{M\,L^{0}\,T^{-3}\,K^{-4}}]$$
3. (iii) Wien's constant. Wien's displacement law is $\lambda_m T = b$, a wavelength times a temperature: $$[b] = [\mathrm{L}][\mathrm{K}] = [\mathrm{M^{0}\,L\,T^{0}\,K}]$$
4. (iv) Rydberg constant. It appears as $\dfrac{1}{\lambda} = R_{\infty}\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$. The bracket is a pure number, so $R_{\infty}$ has the dimensions of a reciprocal wavelength: $$[R_{\infty}] = [\mathrm{M^{0}\,L^{-1}\,T^{0}}]$$
5. (v) Check de Broglie. Momentum has $[p] = [\mathrm{M\,L\,T^{-1}}]$, so $$\left[\frac{h}{p}\right] = \frac{[\mathrm{M\,L^{2}\,T^{-1}}]}{[\mathrm{M\,L\,T^{-1}}]} = [\mathrm{L}]$$ which is indeed a length. The relation is dimensionally consistent.

Final Answer: $[h] = [\mathrm{M\,L^{2}\,T^{-1}}]$, $[\sigma] = [\mathrm{M\,T^{-3}\,K^{-4}}]$, $[b] = [\mathrm{L\,K}]$, $[R_{\infty}] = [\mathrm{L^{-1}}]$, and $\lambda = h/p$ checks out.

Takeaway: You do not need to understand Stefan's law or the Rydberg formula to get their dimensions — you only need the defining equation and the courage to divide. [JEE Tip] $[h] = [\mathrm{M\,L^2\,T^{-1}}]$ is the same as angular momentum, which is not a coincidence: $h$ is an angular momentum, and that fact is worth two marks in any "identical dimensions" question.

Example 21: Which of these share dimensions?

From the list below, group together every quantity that has the same dimensional formula: work, torque, impulse, linear momentum, pressure, energy density, angular momentum, Planck's constant, frequency, velocity gradient, gravitational potential, latent heat, surface tension, spring constant.

Solution:

1. Work down the list writing each formula. Do not try to spot pairs by intuition; compute, then sort.
2. Work $= Fs \to [\mathrm{M\,L\,T^{-2}}][\mathrm{L}] = [\mathrm{M\,L^{2}\,T^{-2}}]$. Torque $= Fr \to$ the same, $[\mathrm{M\,L^{2}\,T^{-2}}]$.
3. Impulse $= Ft \to [\mathrm{M\,L\,T^{-2}}][\mathrm{T}] = [\mathrm{M\,L\,T^{-1}}]$. Momentum $= mv \to [\mathrm{M}][\mathrm{L\,T^{-1}}] = [\mathrm{M\,L\,T^{-1}}]$.
4. Pressure $= F/A \to [\mathrm{M\,L^{-1}\,T^{-2}}]$. Energy density $= E/V \to \dfrac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{L^{3}}]} = [\mathrm{M\,L^{-1}\,T^{-2}}]$.
5. Angular momentum $= mvr \to [\mathrm{M}][\mathrm{L\,T^{-1}}][\mathrm{L}] = [\mathrm{M\,L^{2}\,T^{-1}}]$, and $[h] = [\mathrm{M\,L^{2}\,T^{-1}}]$ from Example 20.
6. Frequency $= 1/T \to [\mathrm{T^{-1}}]$. Velocity gradient $= \dfrac{dv}{dx} \to \dfrac{[\mathrm{L\,T^{-1}}]}{[\mathrm{L}]} = [\mathrm{T^{-1}}]$.
7. Gravitational potential $= \dfrac{W}{m} \to \dfrac{[\mathrm{M\,L^{2}\,T^{-2}}]}{[\mathrm{M}]} = [\mathrm{L^{2}\,T^{-2}}]$, and latent heat $= Q/m \to [\mathrm{L^{2}\,T^{-2}}]$.
8. Surface tension $= F/l \to \dfrac{[\mathrm{M\,L\,T^{-2}}]}{[\mathrm{L}]} = [\mathrm{M\,T^{-2}}]$, and spring constant $= F/x \to [\mathrm{M\,T^{-2}}]$.

Final Answer:

Group	Members	Dimensional formula
1	work, torque	$[\mathrm{M\,L^{2}\,T^{-2}}]$
2	impulse, linear momentum	$[\mathrm{M\,L\,T^{-1}}]$
3	pressure, energy density	$[\mathrm{M\,L^{-1}\,T^{-2}}]$
4	angular momentum, Planck's constant	$[\mathrm{M\,L^{2}\,T^{-1}}]$
5	frequency, velocity gradient	$[\mathrm{M^{0}\,L^{0}\,T^{-1}}]$
6	gravitational potential, latent heat	$[\mathrm{M^{0}\,L^{2}\,T^{-2}}]$
7	surface tension, spring constant	$[\mathrm{M\,T^{-2}}]$

Takeaway: [NEET Important] These seven pairs cover almost every "which pair has the same dimensions" question ever set. Learn them as pairs, not as fourteen separate formulae. And remember what identical dimensions do not mean: work and torque have the same dimensions but are completely different physical quantities — one is a scalar, one is a vector, and you cannot add them.

Example 22: Dimensions from the van der Waals equation

The van der Waals equation for one mole of a real gas is $$\left(P + \frac{a}{V^{2}}\right)(V - b) = RT$$ where $P$ is pressure and $V$ is volume. Find the dimensional formulae of the constants $a$ and $b$.

Solution:

1. Use the principle of homogeneity on the brackets. Only quantities with the same dimensions can be added or subtracted, so each bracket must be dimensionally uniform.
2. The second bracket first, because it is easier. $V - b$ requires $$[b] = [V] = [\mathrm{M^{0}\,L^{3}\,T^{0}}]$$ So $b$ is a volume — physically, the volume actually occupied by the molecules themselves.
3. The first bracket. $P + \dfrac{a}{V^{2}}$ requires $$\left[\frac{a}{V^{2}}\right] = [P] = [\mathrm{M\,L^{-1}\,T^{-2}}]$$
4. Solve for $a$: $$[a] = [P][V^{2}] = [\mathrm{M\,L^{-1}\,T^{-2}}][\mathrm{L^{3}}]^{2} = [\mathrm{M\,L^{-1}\,T^{-2}}][\mathrm{L^{6}}]$$
5. Collect the powers of length: $-1 + 6 = 5$. Therefore $$[a] = [\mathrm{M\,L^{5}\,T^{-2}}]$$

Final Answer: $[a] = [\mathrm{M\,L^{5}\,T^{-2}}]$ and $[b] = [\mathrm{L^{3}}]$.

Takeaway: You never needed to know any thermodynamics. The homogeneity principle alone tells you that anything added to a pressure is a pressure and anything subtracted from a volume is a volume. [JEE Tip] This is the single most reliable technique in the chapter: find the term you already know, and force everything added to it to match.

Example 23: Finding the dimensions of unknown constants

(a) A force varies with time as $F = a\sin(ct) + b\cos(dt)$. Find the dimensional formulae of $a$, $b$, $c$ and $d$. (b) The pressure of a gas is given by $P = \dfrac{a - t^{2}}{b x}$, where $t$ is time and $x$ is distance. Find $[a]$, $[b]$ and $[a/b]$.

Solution:

1. (a) The arguments first. The argument of any trigonometric function must be a pure number — you cannot take the sine of 3 seconds. So $[ct] = [\mathrm{M^{0}L^{0}T^{0}}]$, giving $$[c] = \frac{1}{[\mathrm{T}]} = [\mathrm{T^{-1}}]$$ and by the same argument $[d] = [\mathrm{T^{-1}}]$.
2. The coefficients. Since $\sin(ct)$ and $\cos(dt)$ are pure numbers, $a\sin(ct)$ has the dimensions of $a$ alone. And that sum must equal a force: $$[a] = [b] = [F] = [\mathrm{M\,L\,T^{-2}}]$$
3. (b) The numerator. $a - t^{2}$ is a subtraction, so $a$ must have the dimensions of $t^{2}$: $$[a] = [\mathrm{M^{0}\,L^{0}\,T^{2}}]$$
4. Solve for $b$. Rearranging, $b = \dfrac{a - t^{2}}{P x}$, so $$[b] = \frac{[\mathrm{T^{2}}]}{[\mathrm{M\,L^{-1}\,T^{-2}}][\mathrm{L}]} = \frac{[\mathrm{T^{2}}]}{[\mathrm{M\,T^{-2}}]}$$
5. Dividing: the mass power is $0 - 1 = -1$, the length power is $0 - 0 = 0$, the time power is $2 - (-2) = 4$. Therefore $$[b] = [\mathrm{M^{-1}\,L^{0}\,T^{4}}]$$
6. The ratio: $$\left[\frac{a}{b}\right] = \frac{[\mathrm{T^{2}}]}{[\mathrm{M^{-1}\,T^{4}}]} = [\mathrm{M\,L^{0}\,T^{-2}}]$$ which, incidentally, is the dimension of surface tension.

Final Answer: (a) $[a] = [b] = [\mathrm{M\,L\,T^{-2}}]$, $[c] = [d] = [\mathrm{T^{-1}}]$. (b) $[a] = [\mathrm{T^{2}}]$, $[b] = [\mathrm{M^{-1}\,T^{4}}]$, $[a/b] = [\mathrm{M\,T^{-2}}]$.

Takeaway: Two rules solve every problem of this type, and you apply them in this order: (1) the argument of $\sin$, $\cos$, $\tan$, $\log$ or $e^{x}$ is always dimensionless; (2) anything added to or subtracted from a quantity shares its dimensions. Only after those do you go looking for the unknown by rearranging. [JEE Tip] Watch the sign in step 5 — dividing by $[\mathrm{T^{-2}}]$ adds 2 to the time exponent.