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PhysicsCh 1 Units and Measurement

Units and Measurement: Class 11 Physics · Chapter 1

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 80 min
  • 1 simulation
  • 25 practice questions
  • 25 MCQs

What you will learn in Units and Measurement

  • Explain what a unit is, and tell base quantities, derived quantities and systems of units apart
  • Name the seven SI base quantities and units, and measure plane and solid angles in radians and steradians
  • Count significant figures, write values in scientific notation and estimate orders of magnitude
  • Apply the significant-figure rules for products, quotients, sums and differences, and round off correctly
  • Estimate the uncertainty of a result calculated from measured values
  • Write dimensional formulae and dimensional equations for physical quantities
  • Use dimensional analysis to check equations, convert units and deduce relations, and state its limits

Original companion text aligned to NCERT Physics Class 11, Chapter 1, pp. 1–12. Not affiliated with or endorsed by NCERT.

1.1 Introduction

A tailor in Kolkata notes that a sleeve should be 58 cm long. A pharmacist in Pune weighs out 250 mg of a powder. Both are doing the same thing: holding a quantity up against an agreed standard and counting how many times the standard fits into it. That standard is a unit (मात्रक). All of physics rests on such comparisons, so we begin with units, and with how far a measured number can be trusted.

A measured result therefore has two parts: a pure number, its numerical measure, and a unit. “The corridor is 40” tells you nothing; “the corridor is 40 m” is a measurement. If you switch to a smaller unit, the number grows in proportion, so that the product of the two stays the same (Fig. 1.1):

Changing the unit
n1u1=n2u2
(1.1)
where
n1,n2
numerical measures
u1,u2
the two units

For example, 2.5 m is also 250 cm: the unit became 100 times smaller and the number 100 times larger.

3/3Unit B is half as long and fits 6 times: 3uA=6uB.

Fig. 1.1The same length measured in two units. Halving the unit doubles the number.

A plank lies across the top of the figure. In the first step it stands alone with the question how many units long it is. In the second step three equal blocks marked A lie end to end under it and exactly match its length, so n equals 3. In the third step six shorter blocks marked B, each half as long as A, also match the plank, so n equals 6, and the label 3 u A equals 6 u B shows that the number times the unit is unchanged.

Speed is distance divided by time. Density is mass divided by volume. Force is mass times acceleration. Because physical quantities are linked like this, a handful of independent standards is enough for all of them. We pick a few base quantities (also called fundamental quantities), such as length, mass and time, and agree on a base unit for each. Every other unit is then built by multiplying and dividing base units; these are derived units (व्युत्पन्न मात्रक), such as the metre per second. A chosen set of base units, with all the derived units that follow from them, is called a system of units (Fig. 1.2).

4/4Mass × acceleration gives force: 1 kg m s−2 is called a newton (N).

Fig. 1.2Derived units are built from base units by multiplying and dividing.

Three boxes on the left hold the base quantities length in metres, time in seconds and mass in kilograms. In the second step an arrow from length leads to a box for area, in square metres. In the third step arrows from length and time lead to speed, in metres per second, and a further arrow marked divide by time leads to acceleration, in metres per second squared. In the fourth step arrows from mass and acceleration lead to force, in kilogram metre per second squared, which is named the newton.

Definition
Unit

A fixed amount of a physical quantity, agreed worldwide, used as the yardstick for that quantity. To measure is to find how many such units the quantity contains.

ExampleIn 1.5 kg of rice, 1.5 is the numerical measure and the kilogram is the unit.

Quick check
1 of 3
A rod is 3.6 m long. What is its numerical measure in a unit that is 4 times longer than a metre?
Show answer

Answer (B) 0.9

A bigger unit fits fewer times: n2=n1u1/u2=3.6×1/4=0.9. Multiplying by 4 (giving 14.4) treats the new unit as if it were smaller.

1.2 The international system of units

Until the middle of the twentieth century, different countries, and even different branches of science, measured in different systems. Three were especially common. They differed in the base units chosen for length, mass and time (Table 1.1).

Table 1.1Base units of length, mass and time in four systems. The SI keeps the MKS three and adds four more base units.
SystemLengthMassTime
CGScentimetregramsecond
FPS (British)footpoundsecond
MKSmetrekilogramsecond
SImetrekilogramsecond

Today almost the whole world uses one system, the Système International d'Unités (French for International System of Units), or SI. It was adopted in 1960 at a meeting of the General Conference on Weights and Measures (CGPM), an intergovernmental body served by the BIPM, the International Bureau of Weights and Measures, near Paris. The SI reached its present seven base units in 1971, when the mole was added. In November 2018 the CGPM redefined all seven in terms of fixed values of constants of nature, and the new definitions came into force on 20 May 2019. SI multiples go in powers of ten, so converting between millimetres, metres and kilometres only moves the decimal point. From here on, every quantity in this book is in SI units unless we say otherwise.

Table 1.2The seven SI base quantities and units. Since 2019 each unit is fixed by declaring an exact value for a constant of nature. You need not memorise these values.
Base quantityUnitSymbolDefining constant (exact value)
Lengthmetremspeed of light in vacuum, c=299 792 458 m s−1
MasskilogramkgPlanck constant, h=6.626 070 15×10−34 J s
Timesecondscaesium-133 hyperfine frequency, ΔνCs=9 192 631 770 Hz
Electric currentampereAelementary charge, e=1.602 176 634×10−19 C
Thermodynamic temperaturekelvinKBoltzmann constant, k=1.380 649×10−23 J K−1
Amount of substancemolemolAvogadro constant, NA=6.022 140 76×1023 mol−1
Luminous intensitycandelacdluminous efficacy of 540×1012 Hz light, Kcd=683 lm W−1

How can a constant define a unit? Take the metre. Light in vacuum always travels at the same speed, and the SI simply declares that speed to be exactly 299 792 458 metres per second. A metre is then the distance light covers in 1/299 792 458 of a second. Any well-equipped laboratory can realise it, with no master rod to consult. The other units are fixed in the same spirit: the second by a frequency of the caesium atom, the kilogram through the Planck constant, and so on.

One caution about the mole: always name the entities you are counting. A mole of hydrogen atoms and a mole of hydrogen molecules both contain about 6.022×1023 particles, but the second has twice the mass of the first.

Every other unit is derived from the seven. Several derived units have special names, such as the newton (force), joule (energy), watt (power), pascal (pressure) and hertz (frequency). A few units outside the SI are so convenient that they stay in use beside it (Table 1.3): farmland is sold by the hectare, grain prices are quoted per quintal, and diamonds are weighed in carats.

Table 1.3Some units used alongside the SI.
UnitSymbolValue in SI units
minutemin60 s
hourh3600 s
dayd86 400 s
degree (angle)°(π/180) rad
litreL10−3 m3
tonnet103 kg
quintalq100 kg
hectareha104 m2
carat (gems)ct2×10−4 kg
barbar105 Pa
standard atmosphereatm101 325 Pa
astronomical unitau1.496×1011 m
electronvolteV1.602×10−19 J

Prefixes scale a unit by powers of ten (Table 1.4). Symbols follow a few fixed rules, which keep a written value unambiguous everywhere in the world:

Table 1.4The SI prefixes you will use most often.
PrefixSymbolFactor
teraT1012
gigaG109
megaM106
kilok103
centic10−2
millim10−3
microµ10−6
nanon10−9
picop10−12
femtof10−15
  • Unit symbols are printed upright and never take a plural s or a full stop: 5 kg, not 5 kgs.
  • Leave a space between the number and the symbol, as in 3 m or 25 °C; the angle signs are the only exception, as in 30°.
  • Symbols of units named after people are capitals (N, Pa, W), but the names themselves are written in lower case (newton, pascal, watt).
  • Use one prefix at a time: write 1 ng, not 1 mµg; a prefix and its unit form a single symbol, so km² means (km)².

Plane angle and solid angle

Two more SI units measure angles. A plane angle is the opening between two lines that meet at a point. Draw a circle of radius r about that point; the lines cut off an arc of length s. The angle is defined as the ratio

Plane angle
θ=sr
(1.2)
where
s
length of the arc
r
radius of the circle

Its unit is the radian (rad). One radian is the angle whose arc equals the radius. A full turn has an arc of 2πr, so a full turn is 2π rad, and 1 rad=180∘/π≈57.3∘. Because θ is a ratio, it does not matter how big a circle you draw (Fig. 1.3).

3/3When the arc equals the radius, the angle is 1 rad≈57.3∘.

Fig. 1.3An angle is the ratio of arc to radius, so it does not depend on the size of the circle.NCERT Fig. 1.1 · p. 2

On the left two rays leave a point at an angle theta. In the first step a blue arc of radius r is marked, with arc length s. In the second step an orange arc of radius 2r appears between the same rays; its length is 2s, so the ratio of arc to radius is unchanged. In the third step a separate drawing on the right shows an angle whose arc length equals its radius r; the label says this angle is 1 radian, about 57.3 degrees.

SimulationArc, radius and angle

A circle with two points A and B on it and radii drawn from the centre O. You can drag B around the circle in steps of one degree and drag a handle to change the radius. Readouts show the angle AOB in degrees, the arc length AB and the chord AB in centimetres, and the radius r, together with the formula for the arc length.

Fig. 1.4Drag point B to change the angle, and drag the radius handle to resize the circle. Compare arc AB with the radius r.
Try this
  • Drag B until arc AB is equal to the radius. What does the angle read? Compare it with 180° divided by π.
  • Set the angle to 90°, then drag the radius handle. Divide arc AB by r each time: does the ratio change?
  • Make AB a diameter. How many radii long is the arc, and how many radians is the angle?

A solid angle is the three-dimensional version: the opening of a cone, such as the beam of a torch. Put the tip of the cone at the centre of a sphere of radius r; the cone cuts a small patch of area dA out of the sphere's surface. The solid angle is

Solid angle
dΩ=dAr2
(1.3)
where
dA
area of the patch on the sphere
r
radius of the sphere

Its unit is the steradian (sr). A sphere twice as large has a patch four times larger in the same cone, so the ratio stays fixed (Fig. 1.5). The whole sphere has area 4πr2, so all directions around a point make up 4π sr≈12.6 sr. The radian and the steradian are ratios of two lengths and of two areas, so both are dimensionless: they carry a name but no dimension.

2/3At radius 2r the patch has area 4 dA, and (2r)2=4r2: the same solid angle.

Fig. 1.5Solid angle is area divided by radius squared, so it does not depend on the size of the sphere.NCERT Fig. 1.1 · p. 2

A sphere is drawn with its centre O. In the first step two lines from O mark a narrow cone that cuts a small blue patch of area dA on the sphere of radius r. In the second step a sphere of radius 2r surrounds it, and the same cone cuts an orange patch with four times the area, so the solid angle is unchanged. In the third step the whole small sphere is shaded and the label shows that the full sphere subtends 4 pi steradians.

Quick check
1 of 4
Which of these is not an SI base unit?
Show answer

Answer (A) newton

The newton is a derived unit, 1 N=1 kg m s−2. The candela and the mole may look unfamiliar, but both are base units.

1.3 Significant figures

Measure the same pencil against two rulers and you will write down different numbers of digits (Fig. 1.6). On a scale marked only in whole centimetres, the end of the pencil lies between 7 and 8, a little before halfway, so you record 7.4 cm: the 7 is read off a mark and the 4 is judged by eye. On a scale with millimetre marks, the end lies between 7.4 and 7.5 cm, and you can judge one more digit: 7.46 cm. In each reading every digit except the last comes straight from the scale, and the last is a careful guess.

That is the rule for writing any measured value: all the digits you are sure of, then one estimated digit, and nothing after it. Together these are the significant figures (सार्थक अंक) of the measurement. How many you can honestly claim depends on the smallest division of the instrument, its least count (अल्पतमांक). Writing 7.4613 cm for the pencil would claim a precision that neither ruler can give.

3/3Significant figures = all the sure digits + the first estimated digit.

Fig. 1.6The finer the scale, the more digits you can read. The last digit you write is always an estimate.

A yellow pencil lies along the top, starting at zero, with a dashed line dropping from its tip. In the first step a ruler marked only in whole centimetres shows the tip between 7 and 8, and the reading 7.4 centimetres appears with the 4 coloured as the estimated digit. In the second step a ruler with millimetre marks appears below; the tip lies between 7.4 and 7.5, and the reading becomes 7.46 centimetres, with the 6 now the estimated digit. In the third step a bracket under 7.4 is labelled sure and a box round the 6 is labelled estimated.

Because the count is fixed by the measurement, it cannot depend on the unit you happen to write. A coin's mass of 3.79 g is also 0.00379 kg: the extra zeros only shift the decimal point, and both forms have three significant figures. With that in mind, here are the rules for counting significant figures in a written number:

  1. Every non-zero digit is significant. 3.47 has three significant figures.
  2. Zeros between non-zero digits are significant, wherever the decimal point is. 40.08 and 4008 both have four.
  3. In a number less than 1, the zeros ahead of the first non-zero digit do not count, and neither does the zero usually written before the decimal point. They only locate the decimal point: 0.0036 has two significant figures.
  4. Trailing zeros in a whole number with no decimal point are not significant. By this convention 5600 m has two significant figures.
  5. Trailing zeros after a decimal point are significant. 2.50 has three significant figures and 0.07100 has four.

4/5A zero at the end, after the decimal point, was measured: it counts. Total: 4.

Fig. 1.7Leading zeros never count. Sandwiched zeros always count. Trailing zeros count only after a decimal point.

The number 0.005070 is written in large digits, one per box. In the second step the first three zeros fade and are crossed out with the label place-holders, because they only locate the decimal point. In the third step the boxes holding 5, 0 and 7 turn green with the label that they count, since the zero sits between non-zero digits. In the fourth step the final zero turns green too, because it comes after the decimal point, and the total reads 4 significant figures. In the fifth step the whole number 5070 metres is shown with only 3 figures, beside 5.070 times 10 cubed metres with 4 figures.

Rule 4 can hide real information. A groundsman measures a boundary rope with a tape as 6500 cm, correct to the nearest centimetre, so all four digits were measured. Yet written as 6500 cm the value seems to have only two significant figures, because its zeros sit at the end of a whole number. Change the unit to metres and the same length reads 65.00 m, where rule 5 makes all four digits count. A whole number ending in zeros simply cannot tell you whether those zeros were measured.

Scientific notation removes the doubt. Write the value as a×10b, with 1≤a<10 and b a positive or negative whole number. Every digit of a is significant, and the power of ten only sets the scale. The rope is 6.500×103 cm, or 6.500×10−2 km: four significant figures in any unit. Had the tape been read only to the nearest metre, you would write 6.5×103 cm. Without scientific notation, fall back on rules 4 and 5.

Some numbers in formulas are not measurements at all. The 4 in the surface area of a sphere, 4πr2, the 12 in 12mv2, and the 60 in 1 min=60 s are exact. Treat such numbers as having unlimited significant figures (4 = 4.000…), so they never limit the precision of a result. Constants such as π=3.14159… can be taken to as many figures as the data need; 3.14 or 3.142 is usually enough.

Scientific notation also gives a quick sense of size. Round a to 1 if a≤5, or to 10 if a>5. The value is then roughly a single power of ten, 10n, and we say it is of the order of 10n; the exponent n is its order of magnitude (परिमाण की कोटि).Some books round at 10≈3.16 instead of 5. We follow the NCERT rule; the two agree except when a lies between 3.16 and 5. Mount Everest rises 8.849×103 m; since 8.849>5, it is of the order of 104 m. The Moon is 3.84×108 m away, of the order of 108 m. So roughly 108/104=104 Everests, stacked end to end, would reach the Moon. Fig. 1.8 places some other lengths on the same ladder of powers of ten.

3/3The Earth (107 m) and the Earth–Sun distance (1011 m): 15 orders of magnitude beyond a hair.

Fig. 1.8Orders of magnitude of some lengths. Each tick is ten times the one before.

A horizontal axis of sizes in metres runs from ten to the minus ten on the left to ten to the eleven on the right, with one tick per power of ten. In the first step an atom, a virus and a human hair are marked at ten to the minus ten, minus seven and minus four. In the second step a person and a cricket pitch appear at ten to the zero and ten to the one. In the third step the Earth's diameter and the Earth–Sun distance appear at ten to the seven and ten to the eleven, and a bracket from the hair to the Earth–Sun distance is labelled ten to the fifteen times.

Quick check
1 of 4
How many significant figures does 0.03060 kg have?
Show answer

Answer (B) 4

The leading zeros (0.0) only place the decimal point. The 3, the sandwiched 0, the 6 and the final 0 after the decimal point all count: four. Counting every zero (6) is the usual slip.

1.3.1 Rules for arithmetic operations with significant figures

You time a toy car over 1.25 m and get 3.8 s. A calculator reports the speed as 0.328947368 m/s. Only the first two digits mean anything: the time was known to two figures, so the speed is 0.33 m/s. A calculated result can never be more precise than the measurements that went into it. Two rules make sure it is reported that way.

PropertyMultiplying and dividing

The result keeps as many significant figures as the measured value with the fewest significant figures.

PropertyAdding and subtracting

The result keeps as many decimal places as the measured value with the fewest decimal places.

A shopkeeper puts a 12.6 kg sack of rice, weighed on a platform scale that reads to 0.1 kg, into a carton with a 0.845 kg packet of dal, weighed on a kitchen scale that reads to 1 g. The calculator gives 13.445 kg. But the sack's mass says nothing about hundredths or thousandths of a kilogram, so those columns of the sum are unknown too (Fig. 1.9). The total is 13.4 kg.

3/3Keep one decimal place: 13.4 kg.

Fig. 1.9In a sum, any column that contains an unknown digit makes the same column of the answer unknown.

Two masses, 12.6 and 0.845 kilograms, are written one below the other with the decimal points lined up, and their sum 13.445 is written under a line. In the second step question marks fill the empty hundredths and thousandths places of 12.6, and those two columns are shaded and labelled unreliable. In the third step the last two digits of the sum fade and the answer 13.4 kilograms is shown.

The rule really is about decimal places, not significant figures. On one balance a bottle weighs 152.37 g and a tablet 0.42 g. Both are known to a hundredth of a gram, so the total 152.79 g is reliable to its last digit, even though 0.42 g has only two significant figures; rounding the sum to two figures would throw away three good digits. Subtraction can go the other way. Two clinical thermometers read 36.8∘C and 36.6∘C, three significant figures each, yet their difference, 0.2∘C, has only one. The small difference of two nearly equal measurements is always far less certain, relatively, than either measurement.

1.3.2 Rounding off the uncertain digits

A calculation usually produces more digits than you can keep, so the result must be rounded. When the part you drop is clearly more or less than half a unit in the last kept place, the rule is the familiar one: to three significant figures, 4.837 becomes 4.84 and 4.832 becomes 4.83.

The interesting case is a dropped digit of exactly 5, which lies precisely halfway. Always rounding such values up would push results upwards on average. The convention is: if the digit before the 5 is even, leave it; if it is odd, raise it by one. So 6.125 rounds to 6.12 (2 is even) and 6.135 rounds to 6.14 (3 is odd). Over many roundings, half the ties go down and half go up. The tie rule applies only when the dropped part is exactly 5; 6.1251 is more than halfway and rounds up to 6.13 (Fig. 1.10).

3/36.125→6.12 (2 is even) but 6.135→6.14 (3 is odd).

Fig. 1.10Rounding to three significant figures. Ties go to the even neighbour.

A number line runs from 6.12 to 6.14 with small ticks every thousandth and dashed halfway marks at 6.125 and 6.135. In the first step the value 6.123 hops left to 6.12, because less than half is dropped. In the second step 6.128 hops right to 6.13, because more than half is dropped. In the third step the two halfway values are shown: 6.125 hops down to 6.12 because 2 is even, and 6.135 hops up to 6.14 because 3 is odd.

Two habits complete the picture. In a calculation with several steps, carry one extra digit in the intermediate results and round only the final answer; Section 1.3.3 shows what goes wrong otherwise. And use constants to a sensible number of figures: the Avogadro constant 6.02214076×1023 mol−1 is usually taken as 6.02×1023 mol−1, and π as 3.14 or 3.142, to match the precision of the data. The inspector below counts significant figures and rounds any value you type, using exactly these rules.

InteractiveSignificant-figure inspector

A text box for a measured value, with example buttons such as 0.005070, 3200 and 6.125. Each digit appears in its own box: digits that count have a solid outline and the word counts, leading zeros are dashed and labelled place, and doubtful trailing zeros are dotted with a question mark. Cards show the number of significant figures, the value in scientific notation and its order of magnitude, with a note explaining the rules used. A menu rounds the value to fewer significant figures and explains the rounding, including the even-digit rule for an exact 5.

Fig. 1.11Type a measured value or pick an example. The inspector marks the digits that count, writes the value in scientific notation, finds its order of magnitude and rounds it.
Example 1.1NCERT Example 1.1 · p. 6

Surface area and volume of a ball bearing

A steel ball bearing has a radius of 0.6150 cm, measured with a screw gauge. Find its surface area and volume to the correct number of significant figures.

Try it yourself first. Then reveal the solution one step at a time.

Example 1.2NCERT Example 1.2 · p. 6

Average speed of a marathon runner

A runner completes a marathon, 42.195 km, in 2.1 h (timed only to the nearest tenth of an hour). Express her average speed with the correct significant figures.

Try it yourself first. Then reveal the solution one step at a time.

1.3.3 Rules for determining the uncertainty in the results of arithmetic calculations

Significant figures are a rough guide to precision. When we know the uncertainty of each measurement, we can do better and estimate the uncertainty of a calculated result. Three observations cover what you need.

(1) In a product, relative errors add. A trolley of mass m=2.50±0.01 kg rolls at v=1.20±0.02 m/s. The relative error (आपेक्षिक त्रुटि) of each value is its uncertainty divided by the value: 0.01/2.50=0.4% for the mass and 0.02/1.20=1.7% for the speed. The momentum p=mv=3.00 kg m/s then carries a relative error of about 0.4%+1.7%=2.1%, which is 0.06 kg m/s. So we report p=3.00±0.06 kg m/s. Fig. 1.12 shows where the rule comes from, using the simplest product to picture, the area of a rectangle.

3/3The corner Δl Δb is negligible, so ΔA/A=Δl/l+Δb/b.

Fig. 1.12The extra area comes almost entirely from the two strips; the corner is a product of two small quantities.

A green rectangle with length l along the bottom and breadth b up the side has area A equal to l b. In the second step a thin strip of width delta l is added along the right side, with area b delta l, and a thin strip of height delta b along the top, with area l delta b. In the third step the tiny corner square of area delta l times delta b is highlighted as negligible, and the result delta A over A equals delta l over l plus delta b over b is shown.

DerivationRelative error of a product

Let the true length and breadth lie within l±Δl and b±Δb. The largest the area can be is

(l+Δl)(b+Δb)=lb+b Δl+l Δb+Δl Δb

The last term is a product of two small quantities. For errors of 0.1 cm it is 0.01 cm2, negligible beside the others. So the greatest error in the area is

ΔA=b Δl+l Δb
(1.4)

Dividing by A=lb gives the rule in its most useful form:

Relative errors add
ΔAA=Δll+Δbb
(1.5)

The same result holds for a quotient l/b, because the worst case is one value too large and the other too small. For a square, l2=l×l, so the relative error doubles; for a cube it triples.

(2) In sums and differences, absolute errors add. A rod of length 25.4±0.1 cm has 3.86±0.01 cm cut off. The remaining piece could be as long as 25.5−3.85=21.65 cm or as short as 25.3−3.87=21.43 cm, so it is 21.54±0.11 cm: only 21.5 cm is worth writing. The larger uncertainty dominates, which is why sums and differences are rounded to the fewest decimal places, and why a difference can have fewer significant figures than either value.

(3) Equal significant figures do not mean equal precision. A stopwatch reads 2.05 s for one sprint start and 8.75 s for a short race, both to ±0.01 s and both with three significant figures. The first has a relative error of 0.01/2.05≈0.5%, the second only 0.01/8.75≈0.1%. Counting significant figures is a rough guide; the relative error is the exact measure of precision.

Finally, here is why you carry an extra digit through intermediate steps. Find the volume of a cube of edge 1.27 m in two steps. The area of a face is 1.6129 m2; round it to 1.61 m2, multiply by the edge, and you get 2.0447≈2.04 m3. But the true volume, 1.273=2.048 m3, rounds to 2.05 m3. Keep four figures in the middle step, 1.613 m2, and the product 2.0485 rounds correctly to 2.05 m3.

Quick check
1 of 4
Give 3.6 cm×1.452 cm to the correct number of significant figures.
Show answer

Answer (A) 5.2 cm2

3.6 has two significant figures, the fewest, so the product 5.2272 is reported as 5.2 cm2. Keeping three figures (5.23) follows the more precise factor instead of the less precise one.

1.4 Dimensions of physical quantities

Two quantities can differ enormously in size and units and still be the same kind of thing. The wavelength of green light and the height of Mount Everest are both lengths. The speed of a snail and the speed of light are both a length divided by a time. The idea of dimensions captures this kind, and ignores the size.

Every physical quantity can be built from the seven SI base quantities, so we treat these seven as the basic dimensions and write each in square brackets: mass [M], length [L], time [T], amount of substance [mol], electric current [A], temperature [K] and luminous intensity [cd]. Write a quantity as a product of base quantities; the exponent each one carries is its dimension (विमा) in that base quantity, and together these exponents are the dimensions of the quantity. Square brackets around a symbol, as in [v], are read as “the dimensions of v”.

In mechanics three of them are enough: [M], [L] and [T]. To find the dimensions of a mechanical quantity, trace its definition back to mass, length and time. Momentum is mass times velocity, and velocity is a length per unit time, so

Dimensions of momentum
[p]=[M] [L][T]=[MLT−1]

Read the exponents straight off the brackets: momentum is of the first power in mass and in length, and of power −1 in time. Force, mass times acceleration, works out to [MLT−2]. Area, a length times a length, is [L2]; to stress that mass and time play no part, we may write it as [M0L2T0]. Two things are deliberately left out. Size does not appear: a glacier creeping a metre a day and a rifle bullet both move with dimensions [LT−1]. Nor does the particular meaning of the quantity, as you will see in Section 1.6.2. Fig. 1.13 pictures the exponents for four common quantities.

3/4Multiply by mass: [MLT−2].

Fig. 1.13Dimensions are powers. Bars above the line are positive powers, bars below are negative.

Three bars, labelled M, L and T, stand on a zero line with grid lines at plus and minus one and two. In the first step, for speed, M is 0, L is plus 1 and T is minus 1. In the second step, acceleration, the T bar grows down to minus 2. In the third step, force, the M bar rises to plus 1 while L stays plus 1 and T minus 2. In the fourth step, pressure, the L bar flips down to minus 1, giving M plus 1, L minus 1 and T minus 2. The quantity's name and dimensional formula are shown at the top in each step.

Quick check
1 of 3
What are the dimensions of the work done by a force, W=Fd?
Show answer

Answer (B) [ML2T−2]

[F][d]=[MLT−2][L]=[ML2T−2]. The first option is force itself; the third is power, work per unit time.

1.5 Dimensional formulae and dimensional equations

Every derived quantity is defined by a relation, and the relation fixes its dimensions. Take power, the rate of doing work: P=W/t, where the work W is force times distance. Replace each symbol by its dimensions and add up the exponents of each base quantity (Fig. 1.14). The result, [ML2T−3], is the dimensional formula (विमीय सूत्र) of power: a compact list of the base quantities involved, each with its exponent. Written as an equation, [P]=[ML2T−3], it becomes a dimensional equation (विमीय समीकरण). Here are three more:

Dimensional equations
[a]=[M0LT−2][p]=[MLT−1][ρ]=[ML−3T0]
(1.6)

Table 1.5 collects the dimensional formulae of the quantities you will use most often in this book. You need not memorise it: each row follows from its defining relation in a line or two.

3/3Add the powers of M, L and T: [P]=[ML2T−3].

Fig. 1.14Writing a dimensional equation: relation, substitute, collect powers.

Three boxes are stacked with arrows between them. In the first step the top box shows the relation P equals W over t equals F d over t. In the second step the middle box replaces each symbol by its dimensions: M L T to the minus 2 times L, divided by T. In the third step the bottom box gives the result M L squared T to the minus 3, and a side panel adds the powers: M is 1, L is 1 plus 1 equals 2, and T is minus 2 minus 1 equals minus 3.

Table 1.5Dimensional formulae of some common quantities. Ratios of like quantities are dimensionless.
QuantityDefining relationSI unitDimensional formula
Arealength × breadthm2[L2]
Volumelength × breadth × heightm3[L3]
Densitymass ÷ volumekg m−3[ML−3]
Velocity, speeddisplacement ÷ timem s−1[LT−1]
Accelerationvelocity ÷ timem s−2[LT−2]
Momentum, impulsemass × velocitykg m s−1[MLT−1]
Forcemass × accelerationN[MLT−2]
Work, energy, torqueforce × distanceJ (N m for torque)[ML2T−2]
Powerwork ÷ timeW[ML2T−3]
Pressure, stressforce ÷ areaPa[ML−1T−2]
Frequency1 ÷ periodHz[T−1]
Angle, strain, refractive indexratio of like quantitiesrad, or none[M0L0T0]
Spring constantforce ÷ extensionN m−1[MT−2]
Gravitational constantFr2/(m1m2)N m2 kg−2[M−1L3T−2]
Planck constant, angular momentumenergy ÷ frequencyJ s[ML2T−1]
Electric chargecurrent × timeC[AT]
Example

A dimensional formula from a defining relation

Honey is hard to stir because its layers resist sliding past each other. The force needed is F=ηA ΔvΔx, where A is the area of a layer, Δv the difference in speed between two layers a distance Δx apart, and η a property of the liquid called its viscosity. Find the dimensional formula and the SI unit of η.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Match each quantity with its dimensional formula.

Tap an item on the left, then its match on the right.

Show answer
Answer
  • power : [ML2T−3]
  • pressure : [ML−1T−2]
  • momentum : [MLT−1]
  • frequency : [T−1]
Power is energy per time, pressure is force per area, momentum is mass × velocity and frequency is the reciprocal of a time.

1.6 Dimensional analysis and its applications

Dimensions are more than labels. Only quantities of the same kind can be added, subtracted or compared, so dimensions put strict limits on what a correct equation can look like. Using these limits is called dimensional analysis (विमीय विश्लेषण). It lets us check equations, convert units, and sometimes predict the form of a relation before we have a full theory.

In any calculation, carry the units along with the numbers and let them cancel as symbols would: hours in a numerator cancel hours in a denominator, so (km/h)×h=km. Dimensions behave in the same way, [MLT−2]/[M]=[LT−2]. Handling units like this is also the safest way to convert them: replace each unit by its value in the new units, then simplify.

Example

Converting a pressure to CGS units

Atmospheric pressure is about 1.0×105 Pa. Express it in the CGS unit of pressure, the dyne per square centimetre, where 1 dyn=1 g cm s−2.

Try it yourself first. Then reveal the solution one step at a time.

The same idea gives a general recipe. Suppose a quantity has dimensions [MaLbTc] and numerical measure n1 in a system whose units of mass, length and time are M1, L1 and T1. Applying n1u1=n2u2 ((1.1)) to each base unit, its measure in a second system is

Converting between systems
n2=n1(M1M2)a×(L1L2)b(T1T2)c
(1.7)

1.6.1 Checking the dimensional consistency of equations

Five kilograms plus three seconds is not eight of anything. A sum, a difference or an equality makes sense only between quantities of the same kind, so every term in a correct physical equation must have the same dimensions.

PrinciplePrinciple of homogeneity of dimensions

In a physically meaningful equation, every term on both sides has the same dimensions.

Applying the principle of homogeneity is quick. For a body moving with constant acceleration a, a student remembers v2=u2+2as. Every term should be a squared speed, [L2T−2]. The terms v2 and u2 clearly are, and [as]=[LT−2][L]=[L2T−2], so the equation passes. Had she written v=u+at2, the last term would have dimensions [L] while the others have [LT−1], and the equation fails (Fig. 1.15).

3/3In v=u+at2 the last term is [L], not [LT−1]: inconsistent, so it must be wrong.

Fig. 1.15Every term in an equation must have the same dimensions.

The equation v squared equals u squared plus 2 a s is written with each term in a box. In the second step the dimensions L squared T to the minus 2 appear under all three terms, the boxes turn green and the label reads consistent. In the third step a second equation, v equals u plus a t squared, appears below; its first two terms have dimensions L T to the minus 1 but the last has dimensions L, so its box is dashed and orange and the label reads inconsistent.

Checking dimensions is really checking units, but in a form that works in every system at once: whether you think in metres or feet, grams or kilograms, a length is [L]. No conversion factors are needed, which makes it the quickest first test of any new result.

Some quantities have no dimensions at all. A ratio of two like quantities is a pure number: an angle (arc ÷ radius), a strain (extension ÷ original length), a relative density (one density ÷ another). In the same way, the argument of a function such as sin⁡x, log⁡x or ex must be dimensionless. Such functions are built from sums of powers of their argument, for example ex=1+x+x2/2+…, and x, x2 and x3 can be added only if x is a pure number. In y=Asin⁡ωt, therefore, ωt has no dimensions, so [ω]=[T−1].

The test has a blind spot: pure-number factors are invisible to it. The equations v2=u2+2as, v2=u2+as and v2=u2+4as all pass, though only the first is right. Failing the test proves an equation wrong; passing it shows only that the equation might be right.

Example 1.3NCERT Example 1.3 · p. 8

Checking a relation for flowing liquids

For a liquid flowing smoothly along a pipe, a book gives P+12ρv2+ρgh=constant, where P is the pressure, ρ the density, v the speed, g the acceleration due to gravity and h the height. Is the relation dimensionally consistent?

Try it yourself first. Then reveal the solution one step at a time.

Example 1.4NCERT Example 1.4 · p. 9

Ruling out formulas for the speed of sound

Sound travels through a gas at a speed v that depends on the gas pressure P and its density ρ. Five formulas are proposed, where γ is a pure number:

  • (a) v=P/ρ
  • (b) v=ρ/P
  • (c) v=P/ρ
  • (d) v=γP/ρ
  • (e) v=P/ρ+P/ρ

Which can you rule out using dimensions alone?

Try it yourself first. Then reveal the solution one step at a time.

1.6.2 Deducing relation among the physical quantities

Dimensions can do more than check. Suppose we know, from experiment or a good physical guess, which quantities a result depends on. Assume the result is a product of powers of those quantities, times a dimensionless constant, and demand that both sides have the same dimensions. In mechanics this gives three equations, one each for M, L and T, so the method can fix at most three unknown powers.

Example 1.5NCERT Example 1.5 · p. 9

Speed of a wave on a stretched string

Pluck a sitar string and a wave runs along it. Suppose the wave's speed v depends on the tension F in the string, its mass per unit length μ, and its length l. Use dimensions to find the form of the relation.

Try it yourself first. Then reveal the solution one step at a time.

The exponent finder below lets you try the method on five more situations. Change the powers until the dimensions match, and notice which quantities turn out not to matter.

InteractiveExponent finder

A menu offers five situations: a wave on a stretched string, a stone dropped from a height, a mass bouncing on a spring, ripples on deep water and air drag on a fast ball. For each, three quantities are listed with their dimensions, and buttons raise or lower each power in steps of one half. A table compares the powers of M, L and T wanted with those of your product, the current formula is shown, and a message says what still differs or confirms a match and names any quantity that drops out. A button reveals the three equations for the powers.

Fig. 1.16Choose a situation, then raise or lower the power of each quantity until both sides have the same dimensions.

Dimensional analysis is powerful, but it has clear limits:

  • It cannot find dimensionless constants, such as the k above, the 12 in 12mv2 or the 2π in a period.
  • It cannot tell apart quantities with the same dimensions, such as work and torque, both [ML2T−2].
  • It finds only product-of-powers forms. It fails for a sum of terms such as s=ut+12at2, and for relations involving sin⁡, log⁡ or ex.
  • In mechanics it can fix at most three unknown powers, because there are only three base dimensions to compare.
  • It cannot tell you which quantities matter in the first place; that needs physical insight or experiment.
Quick check
1 of 4
Which equation is dimensionally incorrect? Here s is a distance, u and v are speeds, a is an acceleration and t a time.
Show answer

Answer (C) v=u+12at2

In v=u+12at2 the last term has dimensions [LT−2][T2]=[L], a length, while v and u are speeds. Each of the other three is consistent.
Recap

Units and Measurement: key points and summary

  1. A measurement compares a quantity with a chosen standard, its unit; the result is a numerical measure times the unit, and n1u1=n2u2 (Section 1.1).
  2. Base quantities have base units; all other units are derived from them, and together they form a system of units (Section 1.1).
  3. The SI has seven base units (m, kg, s, A, K, mol, cd), defined since 2019 by exact values of constants of nature; CGS, FPS and MKS were earlier systems (Section 1.2).
  4. Plane angle θ=s/r is measured in radians and solid angle dΩ=dA/r2 in steradians; both are dimensionless (Plane angle and solid angle).
  5. Some non-SI units, such as the litre, hectare and electronvolt, remain in use; prefixes and unit symbols follow fixed rules (Section 1.2).
  6. Significant figures are the reliable digits plus the first uncertain one; their number depends on the instrument, not on the unit (Section 1.3).
  7. Scientific notation, a×10b, removes the ambiguity of trailing zeros; the order of magnitude is the exponent of the nearest power of ten (Section 1.3).
  8. Products and quotients keep the fewest significant figures; sums and differences keep the fewest decimal places (Section 1.3.1).
  9. A dropped part above half raises the last digit and below half leaves it; an exact 5 raises only an odd digit. Carry one extra digit through intermediate steps (Section 1.3.2).
  10. In a product or a quotient the relative errors add; subtraction can sharply reduce the number of significant figures (Section 1.3.3).
  11. Dimensions are the powers of the base quantities that make up a quantity; force has the dimensional formula [MLT−2] (Section 1.4, Section 1.5).
  12. Units and dimensions obey ordinary algebra, which makes unit conversion routine (Section 1.6).
  13. By the principle of homogeneity every term of a valid equation has the same dimensions; a dimensionally wrong equation is wrong, but a consistent one may still be wrong (Section 1.6.1).
  14. Dimensional analysis can deduce product-of-powers relations, but not dimensionless constants, sums of terms or the difference between quantities of the same dimensions (Section 1.6.2).
Flash cards · 10

Dimensional formulae to remember

1 / 10
Practice · 25 questions

Practice questions with answers (NCERT “Exercises”)

Give numerical answers to the correct number of significant figures. Questions 1–17 practise the same skills as NCERT Exercises 1.1–1.17, with new situations and numbers; questions 18–25 are extra board and entrance-style practice.

  1. NumericalEasyNCERT Exercise 1.1

    Fill in the values.

    1. A matchbox measures 5.0 cm×3.5 cm×1.5 cm. What is its volume in cubic metres?

      m³
      Show answer

      Answer 0.000026 m³

      V=5.0×3.5×1.5=26.25 cm3, and 1 cm3=10−6 m3, so V=2.6×10−5 m3 to two significant figures.
    2. A cubical die has edges of 1.6 cm. What is its total surface area in square millimetres?

      mm²
      Show answer

      Answer 1500 mm²

      In millimetres the edge is 16, so A=6×162=1536 mm2. The edge has two significant figures, so A=1.5×103 mm2.
    3. A train runs steadily at 90 km/h. How many metres does it cover in 10 seconds?

      m
      Show answer

      Answer 250 m

      90 km/h=90×1000 m/3600 s=25 m/s, so in 10 s it covers 2.5×102 m.
    4. Aluminium is 2.70 times as dense as water. What is its density in kilograms per cubic metre?

      kg/m³
      Show answer

      Answer 2700 kg/m³

      Water has density 1.00 g cm−3=1000 kg m−3, so aluminium has 2.70 g cm−3=2.70×103 kg m−3.
  2. NumericalMediumNCERT Exercise 1.2

    Convert each quantity.

    1. 1 kg m−1 s−2 in g cm−1 s−2

      g cm⁻¹ s⁻²
      Show answer

      Answer 10 g cm⁻¹ s⁻²

      1 kg m−1 s−2=103 g×(102 cm)−1×1 s−2=10 g cm−1 s−2.
    2. one astronomical unit, 1.496×1011 m, in light-years (1 ly=9.46×1015 m)

      ly
      Show answer

      Answer 0.0000158 ly

      1.496×1011/9.46×1015=1.58×10−5 ly.
    3. an acceleration of 1.5 m s−2 in centimetres per minute squared

      cm min⁻²
      Show answer

      Answer 540000 cm min⁻²

      1 m=100 cm and 1 s=(1/60) min, so 1.5 m s−2=1.5×100×602=540 000 cm min−2=5.4×105 cm min−2 (two significant figures).
    4. the Planck constant, 6.63×10−34 J s, in erg seconds (1 erg=1 g cm2 s−2)

      erg s
      Show answer

      Answer 6.63e-27 erg s

      1 J=103 g×(102 cm)2×1 s−2=107 erg, so h=6.63×10−27 erg s.
  3. NumericalMediumNCERT Exercise 1.3

    Standard atmospheric pressure is about 1.01×105 Pa, where 1 Pa=1 kg m−1 s−2. In a new system of units, the unit of mass is α kg, the unit of length β m and the unit of time γ s.

    1. Show that the pressure has the numerical value 1.01×105 α−1βγ2 in the new units.

      Show solution

      Answer1.01×105 α−1βγ2

      One kilogram is α−1 new mass units, one metre is β−1 new length units and one second is γ−1 new time units. Pressure has dimensions [ML−1T−2], so

      1 Pa=α−1 (β−1)−1 (γ−1)−2=α−1βγ2

      new units. Multiplying by 1.01×105 gives the result. This is equation (1.7) with a=1, b=−1, c=−2.

    2. What is the numerical value if α=2, β=0.5 and γ=20?

      new units
      Show answer

      Answer 10100000 new units

      1.01×105×12×0.5×202=1.01×105×100=1.01×107.
  4. Short answerEasyNCERT Exercise 1.4

    Is a 1 km walk long? For a toddler, yes; for a long-distance runner, hardly. Explain why calling a dimensional quantity large or small means nothing without a standard of comparison, and rewrite each statement so that it makes a definite comparison.

    1. A day is a long time.

      Show solution

      A day, 8.64×104 s, is about 105 times as long as a heartbeat (∼1 s), but only about 1/27 000 of a 75-year life (2.4×109 s), and a vanishing fraction of the age of the Earth (∼1.4×1017 s).

    2. Mount Everest is very high.

      Show solution

      At 8.85 km, Everest is about 120 times as tall as the Qutub Minar (73 m), yet it is only 0.14% of the Earth's radius (6371 km): a small wrinkle on the planet.

    3. A mosquito is very light.

      Show solution

      A mosquito, about 2.5 mg, is roughly 40 times lighter than a honeybee (∼0.1 g), but nearly 1020 times heavier than a water molecule (3.0×10−26 kg).

    4. Light travels incredibly fast.

      Show solution

      Light, 3.0×108 m/s, is nearly a million times faster than sound in air (343 m/s), and nothing travels faster; yet it still needs about 8 minutes to reach us from the Sun.

  5. NumericalEasyNCERT Exercise 1.5

    Imagine a new unit of length, picked to make light travel exactly 1 unit per second in vacuum; time is still measured in seconds. At its closest approach, light from Mars takes about 3.0 minutes to reach the Earth.

    1. What is the Earth–Mars distance in the new unit of length?

      new units
      Show answer

      Answer 180 new units

      3.0 min=180 s; with light speed 1 new unit per second, the distance is 1.8×102 new units (light-seconds).
    2. What is this distance in metres?

      m
      Show answer

      Answer 54000000000 m

      180 s×3.0×108 m/s=5.4×1010 m, about 54 million kilometres.
  6. Short answerMediumNCERT Exercise 1.6

    Which of these instruments measures length most precisely? (i) a vernier callipers whose 10 vernier divisions equal 9 main-scale millimetres; (ii) a screw gauge of pitch 1.0 mm with 200 divisions on its circular scale; (iii) a travelling microscope that reads to 0.001 cm.

    Show solution

    Precision is set by the least count. One vernier division is 910 mm, so the vernier's least count is 1−0.9=0.1 mm. The screw gauge's is pitch ÷ divisions =1.0/200=0.005 mm. The microscope reads to 0.01 mm. So (ii) is the most precise, twice as fine as the microscope and twenty times finer than the vernier.

  7. NumericalEasyNCERT Exercise 1.7

    Through a microscope that magnifies 400 times, a pollen grain appears 12.0 mm across (the average of 15 readings). Estimate the real diameter of the grain.

    mm
    Show solution

    Real size = apparent size ÷ magnification =12.0 mm/400=0.0300 mm, that is 3.00×10−5 m or 30.0 μm, a typical size for a pollen grain.

  8. Short answerMediumNCERT Exercise 1.8

    Answer the following.

    1. How can you find the thickness of one sheet of your notebook using only a 15 cm ruler marked in millimetres?

      Show solution

      Press together 200 sheets (count sheets, not page numbers: each sheet carries two pages) and measure the thickness of the stack, for example 9 mm. Then one sheet is 9/200=0.045 mm. The ruler's uncertainty of about 1 mm is shared among 200 sheets, so each sheet is known to about 0.005 mm.

    2. A student claims she can make a screw gauge read to 0.0001 mm simply by engraving 5000 divisions on the circular scale of a 0.5 mm-pitch screw. Is she right?

      Show solution

      No. More divisions lower the least count only on paper. Such fine divisions could not be read by eye, and the real limit is set by how uniform the screw thread is, by backlash (slack when the screw reverses direction) and by how firmly the object is gripped.

    3. Why are you asked to repeat a measurement several times and take the mean? Would 50 repetitions be better than 5?

      Show solution

      Each reading has small random errors, sometimes too large and sometimes too small. In the mean these errors partly cancel, so the mean lies closer to the true value than a single reading. The more readings, the better the cancellation, so 50 readings give a more reliable mean than 5 (as long as the error is random, not a fixed error of the instrument).

  9. NumericalEasyNCERT Exercise 1.9

    A photograph on a phone screen covers 2.4 cm2. Projected on a wall, the same picture covers 1.35 m2. What is the linear magnification of the projector?

    Show solution

    1.35 m2=13 500 cm2. Area magnification =13 500/2.4=5625. Lengths are magnified by the square root: m=5625=75. (To two significant figures, set by 2.4 cm2, this is 75.)

  10. Quick answerEasyNCERT Exercise 1.10

    State the number of significant figures in each value.

    1. 0.00450 kg has significant figures.

      Show answer

      Answer 3

      4, 5 and the trailing 0 after the decimal point; the leading zeros do not count.
    2. 3.080×105 m has significant figures.

      Show answer

      Answer 4

      All the digits of 3.080 count, including the sandwiched and the trailing zero.
    3. 0.10070 s has significant figures.

      Show answer

      Answer 5

      1, 0, 0, 7 and the final 0 count; only the zero before the decimal point does not.
    4. 4500 m, written like this, has significant figures.

      Show answer

      Answer 2

      Trailing zeros in a whole number without a decimal point are not counted. Write 4.500×103 m if they were measured.
  11. NumericalMediumNCERT Exercise 1.11

    A swimming-pool lane is 50.00 m long and 2.50 m wide, and is filled with water to a depth of 1.8 m.

    1. Find the area of the water surface in the lane, to the correct significant figures.

      m²
      Show answer

      Answer 125 m²

      50.00×2.50=125.0 m2; the width has three significant figures, so the area is 125 m2.
    2. Find the volume of water in the lane, to the correct significant figures.

      m³
      Show answer

      Answer 220 m³

      125.0×1.8=225.0 m3. The depth has two significant figures, so keep two: the dropped part is exactly 5 and the digit before it, 2, is even, so the volume is 2.2×102 m3.
  12. NumericalMediumNCERT Exercise 1.12

    A carpenter measures a plank roughly as 3.2 m long. With a fine steel tape he marks and cuts off two pieces, 0.875 m and 0.862 m long.

    1. How much of the plank is left, to the correct precision?

      m
      Show answer

      Answer 1.5 m

      3.2−0.875−0.862=1.463 m. The plank is known only to 0.1 m, so the piece left is 1.5 m.
    2. By how much do the two cut pieces differ, to the correct precision?

      m
      Show answer

      Answer 0.013 m

      0.875−0.862=0.013 m, to three decimal places (only two significant figures are left).
  13. Short answerMediumNCERT Exercise 1.13

    A fast-moving rod looks shorter to an observer it passes. A student remembers the formula as L=L01−v2, where L0 is the rod's length at rest and v its speed, but has forgotten where the speed of light c goes. Use dimensions to put c in the right place.

    Show solution

    AnswerL=L01−v2/c2

    L and L0 are both lengths, so the square root must be dimensionless. Inside it, 1 is a pure number, so v2 must be divided by something with dimensions [L2T−2]: that is c2. Hence L=L01−v2/c2, which also makes sense physically, because L→L0 when v≪c.

  14. NumericalMediumNCERT Exercise 1.14

    Treat a carbon atom as a sphere of radius 0.70 Å, where 1 A˚=10−10 m. What is the total volume, in cubic metres, of the atoms in one mole of carbon?

    m³
    Show solution

    Volume of one atom: 43πr3=43π(7.0×10−11 m)3=1.44×10−30 m3. For a mole: 1.44×10−30×6.02×1023=8.7×10−7 m3, under one cubic centimetre (two significant figures, set by the radius).

  15. NumericalHardNCERT Exercise 1.15

    At STP (0 °C and 1 atm), a mole of any ideal gas fills 22.4 L. Take a molecule of air to be a sphere of diameter about 3.0 Å.

    1. Estimate the total volume of the molecules in one mole, in cubic metres.

      m³
      Show answer

      Answer 0.0000085 m³

      r=1.5×10−10 m; 43πr3×NA=1.41×10−29×6.02×1023=8.5×10−6 m3 (about 8.5 cm3).
    2. Find the ratio of the molar volume to this volume.

      Show answer

      Answer 2600

      22.4×10−3 m3/8.5×10−6 m3≈2.6×103.
    3. Why is the ratio so large?

      Show solution

      In a gas the molecules are far apart compared with their own size. The molecules themselves fill only about one part in a few thousand of the volume; the rest is empty space through which they move freely. In a liquid or solid, by contrast, the molecules are packed almost touching.

  16. Short answerMediumNCERT Exercise 1.16

    On a night drive along a highway, the streetlights beside the road sweep past quickly, the lights of a village on a distant hillside slide backwards slowly, and the Moon seems to travel along with the car. Explain these observations.

    Show solution

    Our eyes judge motion by how fast the line of sight to an object turns. For an object at distance d abeam of a car moving at speed v, the line of sight turns at about v/d radians per second. At v=20 m/s: a streetlight 10 m away gives 2 rad/s, a fast sweep; a village 4 km away gives 0.005 rad/s, a slow drift; the Moon, 3.84×108 m away, gives about 5×10−8 rad/s, far too small to notice.

    So the Moon stays in the same direction as you drive. Since everything near the road slides backwards while the Moon keeps its place in the window, it seems to move along with you.

  17. NumericalMediumNCERT Exercise 1.17

    Jupiter is made mostly of hydrogen and helium, the two lightest elements, and has no solid surface.

    1. Would you guess its mean density to be closer to that of a gas (about 1 kg/m³), or to the range for liquids and solids (10³ to 10⁴ kg/m³)?

      Show solution

      Light gases suggest a low density, but the planet's huge mass squeezes the gas enormously, so it is reasonable to expect a density far above that of ordinary gases. Part (b) checks the guess.

    2. Check your guess using: mass of Jupiter =1.90×1027 kg, mean radius =6.99×107 m.

      kg/m³
      Show answer

      Answer 1330 kg/m³

      V=43π(6.99×107)3=1.43×1024 m3, so ρ=1.90×1027/1.43×1024=1.33×103 kg/m3: in the range of liquids, a little denser than water. The enormous pressure inside compresses the hydrogen to liquid-like densities.
  18. Quick answerEasy

    The torque of a force, 𝐫×𝐅, has the same dimensional formula as

    Show answer

    Answer (A) work

    Torque is force × distance, [ML2T−2], the same as work, though the two are different quantities.
  19. Quick answerMedium

    Assertion (A): The equation v=u+12at is dimensionally consistent. Reason (R): Every dimensionally consistent equation is physically correct.

    Show answer

    Answer (C) A is true, but R is false.

    Every term of v=u+12at is [LT−1], so A is true. But the correct relation is v=u+at: the wrong factor 12 passes the test, so R is false.
  20. Quick answerMedium

    The radius of a ball is measured with a relative error of 0.4%. What is the percentage error in its calculated volume?

    %
    Show answer

    Answer 1.2 %

    V∝r3, and r3=r×r×r, so the relative errors add three times: 3×0.4%=1.2%.
  21. Quick answerEasy

    How many dynes make one newton? (1 dyn=1 g cm s−2)

    dyn
    Show answer

    Answer 100000 dyn

    1 N=1 kg m s−2=103 g×102 cm×1 s−2=105 dyn.
  22. Quick answerHard

    If the speed of light c, the Planck constant h and the gravitational constant G are taken as base quantities, which combination has the dimensions of length?

    Show answer

    Answer (A) hG/c3

    [hG]=[ML2T−1][M−1L3T−2]=[L5T−3]. Dividing by [c3]=[L3T−3] leaves [L2], so hG/c3 is a length (the Planck length).
  23. Quick answerEasy

    Arrange these lengths in increasing order of magnitude.

    1. diameter of the Earth
    2. thickness of a human hair
    3. height of a person
    4. diameter of an atom
    5. distance from the Earth to the Sun
    Show answer
    Answer
    1. diameter of an atom
    2. thickness of a human hair
    3. height of a person
    4. diameter of the Earth
    5. distance from the Earth to the Sun
    Roughly 10−10, 10−4, 100, 107 and 1011 m.
  24. Quick answerMedium

    Match each pair of quantities with the dimensional formula they share.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • work and torque : [ML2T−2]
    • angular momentum and Planck constant : [ML2T−1]
    • pressure and stress : [ML−1T−2]
    • frequency and angular speed : [T−1]
    Each pair shares dimensions though the quantities differ; this is exactly why dimensional analysis cannot tell them apart.
  25. Quick answerMedium

    Evaluate 4.52×0.01302.6 to the correct number of significant figures.

    Show answer

    Answer (D) 0.023

    The calculator gives 0.0226. The value 2.6 has only two significant figures, so the answer is 0.023. Note that 0.0130 has three, because its trailing zero follows the decimal point.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercise 1.1Practice Q1
NCERT Exercise 1.2Practice Q2
NCERT Exercise 1.3Practice Q3
NCERT Exercise 1.4Practice Q4
NCERT Exercise 1.5Practice Q5
NCERT Exercise 1.6Practice Q6
NCERT Exercise 1.7Practice Q7
NCERT Exercise 1.8Practice Q8
NCERT Exercise 1.9Practice Q9
NCERT Exercise 1.10Practice Q10
NCERT Exercise 1.11Practice Q11
NCERT Exercise 1.12Practice Q12
NCERT Exercise 1.13Practice Q13
NCERT Exercise 1.14Practice Q14
NCERT Exercise 1.15Practice Q15
NCERT Exercise 1.16Practice Q16
NCERT Exercise 1.17Practice Q17

Quick check: MCQs on Units and Measurement

  1. 1.1 Introduction3 questions
  2. Plane angle and solid angle4 questions
  3. 1.3 Significant figures4 questions
  4. 1.3.3 Rules for determining the uncertainty in the results of arithmetic calculations4 questions
  5. 1.4 Dimensions of physical quantities3 questions
  6. 1.5 Dimensional formulae and dimensional equations3 questions
  7. 1.6.2 Deducing relation among the physical quantities4 questions
Glossary · 15

Key terms (English–Hindi)

base quantityमूल राशिmool raashi
One of the seven quantities chosen as independent in the SI, such as length, mass and time.First seen in the chapter →
derived unitव्युत्पन्न मात्रकvyutpann maatrak
A unit made by multiplying and dividing base units, such as m s−1 or the newton.First seen in the chapter →
dimensional equationविमीय समीकरणvimiya sameekaran
An equation that sets a quantity equal to its dimensional formula, such as [F]=[MLT−2].First seen in the chapter →
dimensional formulaविमीय सूत्रvimiya sootra
An expression such as [MLT−2] showing which base quantities, with which powers, make up a quantity.First seen in the chapter →
dimensionsविमाएँvimaayen
The exponents that each base quantity carries when a physical quantity is written as a product of base quantities.First seen in the chapter →
least countअल्पतमांकalpatamaank
The smallest value an instrument can read directly, such as 1 mm on a school ruler.First seen in the chapter →
order of magnitudeपरिमाण की कोटिparimaan ki koti
The exponent of the power of ten nearest a value: a×10b has order of magnitude b if a≤5 and b+1 if a>5.First seen in the chapter →
principle of homogeneityविमीय समांगता का सिद्धांतvimiya samaangata ka siddhaant
Every term that is added, subtracted or equated in a valid physical equation has the same dimensions.First seen in the chapter →
radianरेडियनradian
SI unit of plane angle: the angle at the centre of a circle whose arc equals the radius; θ=s/r.First seen in the chapter →
relative errorआपेक्षिक त्रुटिaapekshik truti
The uncertainty of a value divided by the value; multiplied by 100 it is the percentage error.First seen in the chapter →
scientific notationवैज्ञानिक संकेतनvaigyanik sanketan
Writing a number as a×10b with 1≤a<10; every digit of a is significant.First seen in the chapter →
significant figuresसार्थक अंकsaarthak ank
The digits of a measured value that are known reliably, plus the first uncertain digit.First seen in the chapter →
steradianस्टेरेडियनsteradian
SI unit of solid angle: a cone from the centre of a sphere that cuts off an area r2 of its surface; Ω=A/r2.First seen in the chapter →
system of unitsमात्रकों की पद्धतिmaatrakon ki paddhati
A complete set of base units together with the derived units built from them, such as CGS, FPS, MKS or SI.First seen in the chapter →
unitमात्रकmaatrak
A fixed amount of a quantity, agreed worldwide, that other amounts of the same quantity are compared with.First seen in the chapter →

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