NCERT Solutions Class 11 Physics Chapter 1 — Units and Measurement

NCERT Class 11 Physics — Units and Measurement, Chapter 1 Exercises. All 17 questions solved.

The exercises at the end of Chapter 1 fall into four groups. Questions 1.1 to 1.3 and 1.5 convert quantities from one set of units to another. Questions 1.6 to 1.12 are about measuring — the precision of instruments, and how many significant figures a measured or computed result deserves. Question 1.13 uses dimensions to repair a half-remembered formula. Questions 1.4 and 1.14 to 1.17 are estimates and explanations, where the point is to compare a quantity with a sensible standard. The whole exercise runs on three rules:

• Units are algebra — replace each unit by its value in the new units ($1\ \text{km} = 10^3\ \text{m}$, $1\ \text{h} = 3600\ \text{s}$) and simplify, raising the conversion factor to the same power as the unit: $\text{s}^{-2}$ brings in $3600^2$, not $3600$.
• Significant figures — in multiplication and division, the result keeps as many significant figures as the factor with the fewest; in addition and subtraction, it keeps as many decimal places as the term with the fewest.
• Dimensional consistency — every term in a correct equation has the same dimensions, and a quantity added to a pure number, as in $1 – x$, must itself be dimensionless.

The book’s own instruction heads the exercise: in stating numerical answers, take care of significant figures. Every answer below is rounded accordingly.

Key insight. A unit is a factor you can cancel, and a digit is a claim about precision. Almost every question here is one or the other: the conversions in 1.1 to 1.3 come from substituting units and cancelling them like symbols, and the significant-figure answers in 1.1, 1.9 and 1.10 to 1.12 come from asking how precisely the least precise input is known, and never quoting more than that.

Question 1.1

Fill in the blanks. (a) The volume of a cube of side $1\ \text{cm}$ is equal to $\ldots\ \text{m}^3$. (b) The surface area of a solid cylinder of radius $2.0\ \text{cm}$ and height $10.0\ \text{cm}$ is equal to $\ldots\ (\text{mm})^2$. (c) A vehicle moving with a speed of $18\ \text{km h}^{-1}$ covers $\ldots\ \text{m}$ in $1\ \text{s}$. (d) The relative density of lead is $11.3$. Its density is $\ldots\ \text{g cm}^{-3}$ or $\ldots\ \text{kg m}^{-3}$.

Question 1.1 (a)

Solution. Since $1\ \text{cm} = 10^{-2}\ \text{m}$, cubing both sides gives the volume: $(1\ \text{cm})^3 = (10^{-2}\ \text{m})^3 = 10^{-6}\ \text{m}^3$. The factor is cubed along with the unit.

Question 1.1 (b)

Solution. A solid cylinder has two flat ends as well as its curved side, so its surface area is $2\pi rh + 2\pi r^2 = 2\pi r(r + h)$. Working in millimetres from the start, $r = 20\ \text{mm}$ and $h = 100\ \text{mm}$:

$$A = 2\pi(20)(20 + 100) = 1.508\times10^{4}\ \text{mm}^2$$

The radius, $2.0\ \text{cm}$, is known to only two significant figures, so the answer is quoted to two.

Question 1.1 (c)

Solution. Convert the speed to metres per second by substituting $1\ \text{km} = 1000\ \text{m}$ and $1\ \text{h} = 3600\ \text{s}$:

$$18\ \text{km h}^{-1} = \frac{18\times1000\ \text{m}}{3600\ \text{s}} = 5\ \text{m s}^{-1}$$

so in one second the vehicle covers $5\ \text{m}$.

Question 1.1 (d)

Solution. Relative density is the density of a substance divided by the density of water, which is $1\ \text{g cm}^{-3}$. So the density of lead is $11.3\ \text{g cm}^{-3}$. To convert, $1\ \text{g cm}^{-3} = \dfrac{10^{-3}\ \text{kg}}{10^{-6}\ \text{m}^3} = 10^{3}\ \text{kg m}^{-3}$, which gives $11.3\times10^3 = 1.13\times10^4\ \text{kg m}^{-3}$.

Question 1.2

Fill in the blanks by suitable conversion of units. (a) $1\ \text{kg m}^2\,\text{s}^{-2} = \ldots\ \text{g cm}^2\,\text{s}^{-2}$ (b) $1\ \text{m} = \ldots\ \text{ly}$ (c) $3.0\ \text{m s}^{-2} = \ldots\ \text{km h}^{-2}$ (d) $G = 6.67\times10^{-11}\ \text{N m}^2\,(\text{kg})^{-2} = \ldots\ (\text{cm})^3\,\text{s}^{-2}\,\text{g}^{-1}$.

Question 1.2 (a)

Solution. Substitute $1\ \text{kg} = 10^3\ \text{g}$ and $1\ \text{m} = 10^2\ \text{cm}$; the metre is squared, so its factor is squared too:

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

Question 1.2 (b)

Solution. A light year is the distance light travels in one year: $1\ \text{ly} = (3\times10^8\ \text{m s}^{-1})(3.156\times10^7\ \text{s}) = 9.46\times10^{15}\ \text{m}$. Turning this round,

$$1\ \text{m} = \frac{1}{9.46\times10^{15}}\ \text{ly} = 1.06\times10^{-16}\ \text{ly}$$

The light year is a unit for astronomical distances, so a metre is a vanishingly small fraction of it.

Question 1.2 (c)

Solution. Substitute $1\ \text{m} = 10^{-3}\ \text{km}$ and $1\ \text{s} = \dfrac{1}{3600}\ \text{h}$. Because the second appears as $\text{s}^{-2}$, the factor $3600$ is squared:

$$3.0\ \text{m s}^{-2} = 3.0\times10^{-3}\ \text{km}\times(3600)^2\ \text{h}^{-2} = 3.0\times10^{-3}\times1.296\times10^{7} = 3.9\times10^{4}\ \text{km h}^{-2}$$

The data, $3.0$, has two significant figures, so the answer has two.

Question 1.2 (d)

Solution. First reduce the newton to base units, $1\ \text{N} = 1\ \text{kg m s}^{-2}$, so that $\text{N m}^2\,\text{kg}^{-2} = \text{m}^3\,\text{s}^{-2}\,\text{kg}^{-1}$. Now substitute $1\ \text{m} = 10^2\ \text{cm}$ and $1\ \text{kg} = 10^3\ \text{g}$:

$$1\ \text{m}^3\,\text{s}^{-2}\,\text{kg}^{-1} = \frac{(10^2\ \text{cm})^3}{10^3\ \text{g}}\,\text{s}^{-2} = 10^{3}\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$$

so $G = 6.67\times10^{-11}\times10^{3} = 6.67\times10^{-8}\ \text{cm}^3\,\text{s}^{-2}\,\text{g}^{-1}$.

Question 1.3

A calorie is a unit of heat (energy in transit) and it equals about $4.2\ \text{J}$ where $1\ \text{J} = 1\ \text{kg m}^2\,\text{s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha\ \text{kg}$, the unit of length equals $\beta\ \text{m}$, the unit of time is $\gamma\ \text{s}$. Show that a calorie has a magnitude $4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}$ in terms of the new units.

Solution. The new unit of mass is $\alpha$ kg, so one kilogram is $1/\alpha$ of it: $1\ \text{kg} = \alpha^{-1}$ new mass units. In the same way $1\ \text{m} = \beta^{-1}$ new length units and $1\ \text{s} = \gamma^{-1}$ new time units. Substituting these into the calorie, and raising each to the power its unit carries,

$$1\ \text{cal} = 4.2\ \text{kg m}^2\,\text{s}^{-2} = 4.2\,(\alpha^{-1})(\beta^{-1})^{2}(\gamma^{-1})^{-2} = 4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}$$

in the new units. The direction of each power is the thing to check: a bigger unit of mass ($\alpha > 1$) should make the number smaller, and $\alpha^{-1}$ does exactly that.

Question 1.4

Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary: (a) atoms are very small objects (b) a jet plane moves with great speed (c) the mass of Jupiter is very large (d) the air inside this room contains a large number of molecules (e) a proton is much more massive than an electron (f) the speed of sound is much smaller than the speed of light.

Solution. A quantity with dimensions has a size only relative to some unit or some other quantity of the same kind. A metre is large beside an atom and tiny beside the Earth, so the word “large” says nothing until we say large compared with what. A statement becomes meaningful once it names the standard. Parts (e) and (f) already compare two quantities of the same kind, so they need no change; the others do.

Question 1.4 (a)

Solution. An atom is about $10^{-10}\ \text{m}$ across; the comparison has to be with something whose size we know.

Question 1.4 (b)

Solution. A jet at about $250\ \text{m s}^{-1}$ is fast beside road traffic but very slow beside light, so the standard has to be stated.

Question 1.4 (c)

Question 1.4 (d)

Question 1.4 (e)

Solution. This already compares two masses: a proton is about $1836$ times as massive as an electron.

Question 1.4 (f)

Solution. This too compares two speeds: sound travels at about $340\ \text{m s}^{-1}$ in air and light at $3\times10^8\ \text{m s}^{-1}$, nearly a million times faster.

Question 1.5

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes $8\ \text{min}$ and $20\ \text{s}$ to cover this distance?

Solution. If the speed of light is to be $1$ new unit of length per second, the new unit must be the distance light covers in one second, $3\times10^8\ \text{m}$. Light takes $8\ \text{min}\ 20\ \text{s} = 500\ \text{s}$ to reach us from the Sun, so the distance is

$$d = ct = (1\ \text{new unit s}^{-1})(500\ \text{s}) = 500\ \text{new units}$$

Question 1.6

Which of the following is the most precise device for measuring length:

Solution. The most precise device is the one with the smallest least count. For the vernier callipers, with the usual main scale in millimetres, $20$ vernier divisions give a least count of $\dfrac{1\ \text{mm}}{20} = 0.05\ \text{mm}$. For the screw gauge, least count $=$ pitch $\div$ number of circular divisions $= \dfrac{1\ \text{mm}}{100} = 0.01\ \text{mm}$. A wavelength of visible light is about $6\times10^{-7}\ \text{m} = 0.0006\ \text{mm}$, far smaller than either.

Question 1.7

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 $3.5\ \text{mm}$. What is the estimate on the thickness of hair?

Solution. A magnification of $100$ means the image is $100$ times as wide as the object. Dividing the average observed width by the magnification recovers the real thickness:

$$\text{thickness} = \frac{3.5\ \text{mm}}{100} = 0.035\ \text{mm}$$

Question 1.8

Answer the following: (a) You are given a thread and a metre scale. How will you estimate the diameter of the thread? (b) A screw gauge has a pitch of $1.0\ \text{mm}$ and $200$ divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of $100$ measurements of the diameter expected to yield a more reliable estimate than a set of $5$ measurements only?

Question 1.8 (a)

Solution. The thread is far thinner than the smallest division on a metre scale ($1\ \text{mm}$), so one thickness cannot be read directly. Instead, wind the thread round a pencil or rod in closely packed turns, with no gaps and no overlaps, until the coil is a few centimetres long. Measure the length $L$ of the coil with the scale and count the number of turns $n$. Each turn contributes one diameter, so $d = L/n$. The error of about $1\ \text{mm}$ in reading $L$ is shared among all $n$ turns, which is what makes the method work.

Question 1.8 (b)

Solution. On paper, yes: the least count is $\dfrac{\text{pitch}}{\text{divisions}} = \dfrac{1.0\ \text{mm}}{200} = 0.005\ \text{mm}$, and more divisions would make it smaller still. In practice, no. Beyond a point the divisions become too close for the eye to tell apart, and the screw itself limits the reading: its thread is never perfectly uniform, and backlash and wear in the screw introduce errors larger than the least count. So the accuracy cannot be raised without limit this way.

Question 1.8 (c)

Solution. Every reading carries a random error, sometimes too high and sometimes too low. When many readings are averaged these errors largely cancel, so the mean of $100$ readings lies much closer to the true value than the mean of $5$ — the uncertainty of the mean falls roughly as $1/\sqrt{n}$. A large set also samples the rod at many places and orientations, so any variation in its diameter is averaged in rather than missed.

Question 1.9

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

Solution. Linear magnification is a ratio of lengths. Areas scale as the square of lengths, so the area magnification is the square of the linear magnification. With both areas in square metres ($1.75\ \text{cm}^2 = 1.75\times10^{-4}\ \text{m}^2$):

$$\text{area magnification} = \frac{1.55}{1.75\times10^{-4}} = 8857, \qquad m = \sqrt{8857} = 94.1$$

Both data have three significant figures, so the answer has three.

Question 1.10

State the number of significant figures in the following: (a) $0.007\ \text{m}^2$ (b) $2.64\times10^{24}\ \text{kg}$ (c) $0.2370\ \text{g cm}^{-3}$ (d) $6.320\ \text{J}$ (e) $6.032\ \text{N m}^{-2}$ (f) $0.0006032\ \text{m}^2$.

Solution. Three rules decide every case. Zeros before the first non-zero digit only locate the decimal point and are not significant. Zeros between non-zero digits are significant. Trailing zeros after a decimal point are significant, because someone chose to write them. And the power of ten in scientific notation carries no significant figures.

(a) In $0.007$ only the $7$ counts: $1$. (b) In $2.64\times10^{24}$ the digits $2$, $6$, $4$ count: $3$. (c) In $0.2370$ the leading zero does not count, but the final zero does: $4$. (d) In $6.320$ the trailing zero counts: $4$. (e) In $6.032$ the zero lies between non-zero digits: $4$. (f) In $0.0006032$ the four leading zeros do not count, and the zero inside $6032$ does: $4$.

Question 1.11

The length, breadth and thickness of a rectangular sheet of metal are $4.234\ \text{m}$, $1.005\ \text{m}$, and $2.01\ \text{cm}$ respectively. Give the area and volume of the sheet to correct significant figures.

Solution. Convert the thickness first: $t = 2.01\ \text{cm} = 0.0201\ \text{m}$. The sheet is a thin cuboid, and its area is the whole surface — two large faces plus the four edges:

$$A = 2(lb + bt + tl) = 2(4.25517 + 0.0202 + 0.0851) = 8.7209\ \text{m}^2$$

$$V = lbt = 4.234\times1.005\times0.0201 = 0.085529\ \text{m}^3$$

The length and breadth have four significant figures, but the thickness has only three, so both results are quoted to three: $A = 8.72\ \text{m}^2$ and $V = 0.0855\ \text{m}^3$.

If only the area of one face is wanted, it is $lb = 4.234\times1.005 = 4.255\ \text{m}^2$, to four significant figures, since both factors have four. The book’s answer, $8.72\ \text{m}^2$, is the whole surface.

Question 1.12

The mass of a box measured by a grocer’s balance is $2.30\ \text{kg}$. Two gold pieces of masses $20.15\ \text{g}$ and $20.17\ \text{g}$ are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Question 1.12 (a)

Solution. Put everything in kilograms: $20.15\ \text{g} = 0.02015\ \text{kg}$ and $20.17\ \text{g} = 0.02017\ \text{kg}$. The arithmetic sum is

$$2.30 + 0.02015 + 0.02017 = 2.34032\ \text{kg}$$

In addition, the result can be no more precise than the least precise term, so it keeps as many decimal places as that term has. The box’s mass, $2.30\ \text{kg}$, is known to two decimal places of a kilogram — to the nearest $10\ \text{g}$ — so the total is rounded to two decimal places: $2.34\ \text{kg}$. The precise weighings of the gold do not help, because the box’s mass is uncertain by far more than the gold’s.

Note on the book’s answer. The answer key prints $2.3\ \text{kg}$. That would be right if the box’s mass were given as $2.3\ \text{kg}$, known to one decimal place. The question gives $2.30\ \text{kg}$, and the trailing zero is significant, so by the book’s own rule for addition the total is $2.34\ \text{kg}$.

Question 1.12 (b)

Solution. Both masses are known to two decimal places of a gram, so their difference is too:

$$20.17 – 20.15 = 0.02\ \text{g}$$

Subtraction has cost almost all the precision: the difference has only one significant figure, even though each mass had four.

Question 1.13

A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_0$ of a particle in terms of its speed $v$ and the speed of light, $c$. (This relation first arose as a consequence of special relativity due to Albert Einstein.) A boy recalls the relation almost correctly but forgets where to put the constant $c$. He writes:

$$m = \frac{m_0}{(1 – v^2)^{1/2}}$$

Guess where to put the missing $c$.

Solution. The left side, $m$, has the dimensions of mass, and so does $m_0$; the bracket must therefore be dimensionless. But $1$ is a pure number and $v^2$ has dimensions $[\text{L}^2\text{T}^{-2}]$, and a pure number cannot be added to a squared speed. The only way to make $v^2$ dimensionless using $c$ is to divide it by $c^2$, which has the same dimensions:

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

This also behaves sensibly: when $v \ll c$ the bracket is nearly $1$ and $m \approx m_0$.

Question 1.14

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

Solution. Treat each atom as a sphere of radius $r = 0.5\ \text{Å} = 0.5\times10^{-10}\ \text{m}$. One atom occupies

$$V_1 = \frac43\pi r^3 = \frac43\pi(0.5\times10^{-10})^3 = 5.24\times10^{-31}\ \text{m}^3$$

and a mole contains Avogadro’s number of atoms, $N_A = 6.023\times10^{23}$:

$$V = N_A V_1 = (6.023\times10^{23})(5.24\times10^{-31}) = 3.15\times10^{-7}\ \text{m}^3$$

Since the atomic size is given only roughly, the answer is an estimate: about $3\times10^{-7}\ \text{m}^3$, a third of a cubic centimetre.

Question 1.15

One mole of an ideal gas at standard temperature and pressure occupies $22.4\ \text{L}$ (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about $1\ \text{Å}$). Why is this ratio so large?

Solution. Take the molecule as a sphere whose size, $1\ \text{Å}$, is its radius, in the same sense as the $0.5\ \text{Å}$ of Question 1.14. The volume of a mole of molecules is then

$$V_{\text{atomic}} = N_A\cdot\frac43\pi r^3 = (6.023\times10^{23})\cdot\frac43\pi(10^{-10})^3 = 2.52\times10^{-6}\ \text{m}^3$$

The molar volume is $22.4\ \text{L} = 22.4\times10^{-3}\ \text{m}^3$, so

$$\frac{V_{\text{molar}}}{V_{\text{atomic}}} = \frac{22.4\times10^{-3}}{2.52\times10^{-6}} = 8.9\times10^{3} \approx 10^{4}$$

If the $1\ \text{Å}$ is read as the diameter instead, the ratio comes out as $7.1\times10^4$. Either way it is of the order of $10^4$, which is all the rough data can support.

The ratio is large because a gas is mostly empty space. The molecules themselves fill only about one part in $10^4$ of the volume the gas occupies; the rest is the space between them. The separation between neighbouring molecules is correspondingly larger than their size, by a factor of about $(10^4)^{1/3} \approx 20$.

Question 1.16

Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Solution. What the eye judges is not an object’s distance but the direction in which it is seen. As the train moves a distance $s$, the line of sight to an object at distance $d$ turns through an angle of about $s/d$. For a tree $20\ \text{m}$ away, one second at $30\ \text{m s}^{-1}$ turns the line of sight through more than a radian — the tree sweeps backwards across the view. For a hill $5\ \text{km}$ away the same second turns it by less than $0.01\ \text{rad}$, and for the Moon or a star the change is effectively zero. So the near objects appear to rush backwards while the distant ones hold their direction. Since we know we are moving, an object whose direction does not change seems to be keeping pace with us.

Question 1.17

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7\ \text{K}$, and its outer surface at a temperature of about $6000\ \text{K}$. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun $= 2.0\times10^{30}\ \text{kg}$, radius of the Sun $= 7.0\times10^{8}\ \text{m}$.

Solution. The temperature suggests a gas, so the natural guess is a gas-like density. The data say otherwise. The Sun’s volume is

$$V = \frac43\pi R^3 = \frac43\pi(7.0\times10^8)^3 = 1.44\times10^{27}\ \text{m}^3$$

so its mean density is

$$\rho = \frac{M}{V} = \frac{2.0\times10^{30}}{1.44\times10^{27}} = 1.4\times10^{3}\ \text{kg m}^{-3}$$

That is about one and a half times the density of water — in the range of liquids and solids, most of which lie between about $7\times10^2$ and $2\times10^4\ \text{kg m}^{-3}$ — and about a thousand times the density of air. The plasma is squeezed to this density by gravity: the enormous weight of the outer layers presses down on the inner ones.

Common mistakes

• Question 1.1(b): using only the curved surface. A solid cylinder has two flat ends, so the area is $2\pi r(r + h)$, not $2\pi rh$. The other slip is converting $\text{cm}^2$ to $\text{mm}^2$ with a factor of $10$ instead of $100$.
• Question 1.2(c): multiplying by $3600$ once. The unit is $\text{s}^{-2}$, so the conversion factor for time is squared: $3600^2 = 1.296\times10^7$. Using $3600$ alone gives an answer smaller by that factor.
• Question 1.3: getting the powers upside down. A larger new unit makes the number smaller, so each quantity enters as the reciprocal of its new unit: $\alpha^{-1}$ for mass, $\beta^{-2}$ for length squared, and $\gamma^{+2}$ for time, because time appears as $\text{s}^{-2}$.
• Question 1.9: reporting the area ratio. $8857$ is how much larger the area is. Magnification is a ratio of lengths, so the square root is needed: $94.1$.
• Question 1.10: counting leading zeros, or dropping trailing ones. The zeros in $0.007$ and $0.0006032$ only place the decimal point, while the final zeros in $0.2370$ and $6.320$ are significant because they were deliberately written.
• Question 1.12(a): applying the multiplication rule to a sum. In addition it is decimal places, not significant figures, that decide the result — and $2.30\ \text{kg}$ has two of them, so the total is $2.34\ \text{kg}$, not $2.3\ \text{kg}$.
• Question 1.14: forgetting Avogadro’s number, or using $0.5\ \text{Å}$ as a diameter. The question asks for a mole of atoms, so one atom’s volume must be multiplied by $6.02\times10^{23}$. Halving the radius would shrink the volume eightfold.
• Question 1.17: trusting the temperature over the data. “No substance remains solid or liquid” tempts students to answer “gas density”, but the calculation gives $1.4\times10^3\ \text{kg m}^{-3}$, the density of a liquid.

Practise next

• Chapter 2, Motion in a Straight Line — the unit conversions and significant figures of this exercise, put to work on displacement, velocity and acceleration.