Units and Measurement: A Comprehensive Guide for High School Physics, Summaries of Physics

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APNI KAKSHA

Physical Quantities:

All the quantities in terms of which laws of physics are described, and whose measurement is necessary are called

physical quantities.

Units:

• Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen,

internationally accepted reference standard called Unit.

• The standard unit should be easily reproducible, internationally accepted.

For Example: Unit of length is meter, Centimeter, Milimeter, etc.

Measurement:

The comparison of any physical quantity with its

standard unit is called Measurement.

For Example: 5 kg of oranges means mass of oranges is 5 times of 1 kg.

Fundamental/Base Quantities:

Those physical quantities which are independent to each other and all other quantities may be expressed in terms of

these quantities, are called Fundamental Quantities.

Derived Quantities:

Their are infinite number of physical quantities out of which only seven are fundamental quantities and rest of the

quantities may be derived from these fundamental quantities by multiplication and division these quantities are

called as Derived Quantities.

For Example: To derive speed one may take length and time as a fundamental quantities hence speed is a derived

quantities.

Fundamental & Derived Units:

The unit defined for fundamental quantities are called fundamental units & the units defined for derived quantities

are called derived units.

For Example: Fundamental unit for length is meter and derived unit of speed is m/s.

UNITS AND MEASUREMENT

APNI KAKSHA

Definations of Fundamental Units:

• Meter: It is the unit of length. The distance travelled by light in vacuum is

second

is called 1 𝑚𝑚.

• Kilogram: The mass of a cylinder made of platinum-iridium alloy kept at International Bureau of Weights and

Measures is defined as 1 𝑘𝑘𝑘𝑘.

• Second: Second is the time in which cesium atom vibrates 9192631770 times in an atomic clock.
• Kelvin: Kelvin is the (1/273.16) part of the thermodynamics temperature of the triple point of water.
• Candela: The SI unit of luminous intensity is 1 cd which is the luminous intensity of a

blackbody of surface area

2

placed at the temperature of freezing platinum and

at a pressure of 101,325 N/m2, in the direction perpendicular to its surface.

• Ampere: Ampere is the electric current which it maintained in two straight parallel conductor of infinite length

and of negligible cross-section area placed on metre apart in vacuum will produce between them a force

2 × 10

−

𝑁𝑁 per metre length.

• Mole: Mole is the amount of substance of a system which contains a many elementary entities

(may be atoms, molecules, ions, electrons or group of particles, as this and atoms in 0.012 kg

of carbon isotope 𝐶𝐶

12

6

Some Important Conversion of Plane Angle:

• 1° =

radian

• 1° = 60

′

′

= 1 minute of arc)

• 1′ = 60′′(

′

= 60 second of arc)

SI Prefixes:

Power of 10 Prefix Symbol

18 Exa E

15 peta P

12 tera T

9 giga G

6 mega M

3 kilo k

2 hecto h

1 deka da

• 1 deci d
• 2 centi c
• 3 milli m
• 6 micro 𝜇𝜇
• 9 nano n
• 12 pico p
• 15 femto f
• 18 atto a

APNI KAKSHA

For Example: 1 decimeter = 10-1 meter

1 kilogram = 10

3

gram

1 micrometer = 10

meter

Dimension:

When a quantity is expressed in terms of the base quantities, it is written as a product of different powers of the base

quantities. The exponent of a base quantity that enters into the expression.

For Example: Work = Force × Displacement

Force = Mass × Acceleration

Acceleration =

Velocity

Time

Length/Time

Time

Work = Mass ×

Length/Time

Time

× Length

Hence, the dimension of work are 1 in mass, 2 in length & -2 in time.

• For convenience the base quantities are represented by one symbol

Quantity Symbol

Mass M

Time T

Length L

Electric Current I

Amount of Substance mol

Temperature K

Luminous intensity cd

For Example: Dimensional formula of density is [𝑀𝑀𝐿𝐿

−

], Dimensional formula of force is [𝑀𝑀𝐿𝐿𝑇𝑇

−

]

• Homogeneity Principle: If a equation contains several terms separated by the symbol of equality, plus or

minus then dimension of each term should be same

For Example: 𝑠𝑠 = 𝑢𝑢𝑢𝑢 +

2

each term in the above equation is having same dimension that is equal to L

APNI KAKSHA

Order of Magnitude:

If a number is expressed as 𝑎𝑎 × 10

𝑏𝑏

where 1 ≤ 𝑎𝑎 < 10 and 𝑏𝑏 is a positive or negative integer, then 10

𝑏𝑏

is the order of

magnitude of that number.

For Example: Diameter of sun is 13.9 × 10

8

𝑚𝑚, then order of magnitude is 10

9

∵ Diameter of sun is 1.39 × 10

9

Q. Write down the dimensional formulas of the following?

a) Force b) Work c) kinetic energy

d) momentum e) Angular momentum f) Velocity

g) Acceleration h) Torque i) Angular frequency

Sol. a) Force = ma ⇒

[

][

][

−

] = [

−

]

b) Work = Force × displacement ⇒ [𝑀𝑀𝐿𝐿𝑇𝑇

−

][𝐿𝐿] = [𝑀𝑀𝐿𝐿

2

−

]

c) Unit of energy and work is same so their dimensions must be same ⇒

[

2

−

]

d) Momentum = 𝑚𝑚 × 𝑣𝑣 ⇒ [𝑀𝑀][𝐿𝐿𝑇𝑇

−

] = [𝑀𝑀𝐿𝐿𝑇𝑇

−

]

e) Angular momentum = mvr ⇒

[

][

−

][

] = [

2

−

]

f) Unit of velocity = m/s ⇒ [velocity] =

[𝐿𝐿]

[𝑇𝑇]

= [𝐿𝐿𝑇𝑇

−

]

g) unit of acceleration = 𝑚𝑚/𝑠𝑠

2

[Acceleration] =

[ 𝐿𝐿

]

[ 𝑇𝑇

2

]

= [𝐿𝐿𝑇𝑇

−

]

h) Torque = 𝑟𝑟 × 𝑓𝑓 ⇒ [𝐿𝐿][𝑀𝑀𝐿𝐿𝑇𝑇

−

] = 𝑀𝑀𝐿𝐿

2

−

i) Angular frequency

2𝜋𝜋

𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝𝑇𝑇𝑝𝑝𝑇𝑇𝑝𝑝𝑝𝑝

[

] =

1

[ 𝑇𝑇

]

= [𝑇𝑇

−

]

Q. Find out the dimensional formula of universal gravitational constant 𝐶𝐶.

Sol. We know that

1

2

2

[

] =

[𝐶𝐶][𝑀𝑀

1

][𝑀𝑀

2

]

[

2

]

[

−

][

2

]

[𝑀𝑀

2

]

= [𝐶𝐶] ⇒ [𝐶𝐶] = [𝑀𝑀

−

3

−

]

Q. The distance covered by a particle in time (𝑢𝑢) is given by 𝑥𝑥 = 𝑎𝑎 + 𝑏𝑏𝑢𝑢 + 𝑐𝑐𝑢𝑢

2

3

find the dimensions of

𝑎𝑎, 𝑏𝑏, 𝑐𝑐 and 𝑑𝑑

Sol. Since all the terms in equations have same dimensions

[𝑥𝑥} = [𝑎𝑎] = [𝑏𝑏𝑢𝑢] = [𝑐𝑐𝑢𝑢

2

] = [𝑑𝑑𝑢𝑢

3

]

[

] = [

] = [

][

] = [

][

2

] = [

][

3

]

[𝑎𝑎] = [𝐿𝐿]

[𝑏𝑏] = [𝐿𝐿𝑇𝑇

−

]

[𝑐𝑐] = [𝐿𝐿𝑇𝑇

−

]

[

] = [

−

]

APNI KAKSHA

Significant Figure:

In the measured value of physical quantity, the number of digits about the correctness of which we are sure plus the

next doubtful digit, are called the significant figures.

Rules for Finding Significant Figure:

• All non-zeros digits are significant figures, e.g., 4362 𝑚𝑚 has 4 significant figures.
• All zeros occuring between two significant digit are significant figures, e.g., 1005 has 4 significant figures.
• All zeros to the right of the last non-zero digit are not significant, e.g., 6250 has only 3 significant figures.
• In a digit less than one, all zeros to the right of the decimal point and to the left of a non-zero digit are not

significant, e.g., 0.00325 has only 3 significant figures.

• All zeros to the right of a non-zero digit in the decimal part are significant, e.g., 1.4750 has 5 significant figures.
• Order of magnitude is never significant, e.g. 1.63 × 10

9

has 3 significant figure.

• While changing units number of significant figure remains same, e.g. 2.0 𝑘𝑘𝑘𝑘 can’t be written as 2000 g because

2.0 has 2 significant figure but 2000 has 1 significant figure. 2.0 kg can be written as 2.0 × 10

3

𝑘𝑘 because both

2.0 and 2.0 × 10

3

has 2 significant figure.

Significant Figures in Algebric Operations:

1. In Addition or Subtraction in addition or subtraction of the numerical values the final result should retain the

least decimal place as in the various numerical values e.g.,

If 𝑙𝑙

1

= 4.326 𝑚𝑚 and 𝑙𝑙

2

Then, 𝑙𝑙

1

2

As 𝑙𝑙

2

has measured upto two decimal places, therefore

1

2

2. In Multiplication or Division of the numerical values, the final result should retain the least significant figures

as the various numerical values e.g., If length 𝑙𝑙 = 12.5 𝑚𝑚 and breadth 𝑏𝑏 = 4.125 𝑚𝑚.

Then, area 𝑑𝑑 = 𝑙𝑙 × 𝑏𝑏 = 12.5 × 4.125 = 51.5625 𝑚𝑚

2

As 1 has only 3 significant figures, therefore

2

Rules of Rounding Off Significant Figures:

• If the digit to be dropped is less than 5, then the preceding digit is left unchanged. e.g., 1.54 is rounded off to 1.5.
• If the digit to be dropped is greater than 5, then the preceding digit is raised by one. e.g., 2.49 is rounded off to

• If the digit to be dropped is 5 followed by digit other than zero, then the preceding digit is raised by one. e.g.,

3.55 is rounded off to 3.6.

• If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd and left

unchanged if it is even. e.g., 3.750 is rounded off to 3.8 and 4.650 is rounded off to 4.6.

Error:

The lack in accuracy in the measurement due to the limit of accuracy of the instrument or due to any other cause is

called an error.

(Practice Question in the End, Q.3)

(Practice Question in the End, Q.5)

(Practice Question in the End, Q.9)

APNI KAKSHA

1. Error of a sum or a difference:

When two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute

errors in the individual quantities.

2. Error of a product or a quotient:

When two quantities are multiplied or divided, the relative error in the result is the sum of the relative error in

the multipliers.

3. Error in case of a measured quantity raised to a power:

The relative error in a physical quantity raised to the power k is the k times the relative error in the individual

quantity.

If general (if 𝑍𝑍 = 𝑑𝑑

𝑝𝑝

𝑞𝑞

𝑝𝑝

Then, ∆𝑍𝑍/𝑍𝑍 = 𝑝𝑝

Q. Two resistors of resistances 𝑅𝑅

1

= 100 ± 3 𝑜𝑜ℎ𝑚𝑚 and 𝑅𝑅

2

= 200 ± 4 𝑜𝑜ℎ𝑚𝑚 are connected in parallel. Find the

equivalent resistance of the parallel combination. Use the relation

′

1

2

and

′

′

1

1

2

2

2

2

[NCERT]

Sol. Percentage

′

1

2

′

′

Partial Differentiation both side

′

′

1

1

2

2

2

2

′

′

1

1

2

2

2

2

2

× �

2

2

′

Q. The resistance 𝑅𝑅 = 𝑉𝑉/𝐼𝐼 where 𝑉𝑉 = (100 ± 5)𝑉𝑉 and 𝐼𝐼 = (10 ± 0.2)𝑑𝑑. Find the percentage error in R.

[NCERT]

Sol. Percentage error in 𝑉𝑉 =

× 100 = 5%

Percentage error in 𝐼𝐼 =

× 100 = 2%

Total Percentage error in 𝑅𝑅 = 5% + 2% = 7%

Q. The period of oscillation of a simple pendulum is 𝑇𝑇 = 2𝜋𝜋�𝐿𝐿/𝑘𝑘. Measured value of 𝐿𝐿 is 20.0 cm known to 1

mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s

resolution. What is the accuracy in the determination of 𝑘𝑘? [NCERT]

Sol. 𝑇𝑇

2

2

2

2

+ 2 ×

× 100 =

+ 2 ×

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Practice Questions

Q1. Fill in the blanks by suitable conversion of units [NCERT Exercise]

a) 1 𝑘𝑘𝑘𝑘 𝑚𝑚

2

−

2

−

b) 1 𝑚𝑚 =.... 𝑙𝑙𝑦𝑦

c) 3.0 𝑚𝑚 𝑠𝑠

−

−

d) 𝐶𝐶 = 6.67 × 10

−

2

−

3

−

−

Sol. a) = 1 𝑘𝑘𝑘𝑘 = 10

3

2

2

2

4

2

2

−

= 1 𝑘𝑘𝑘𝑘 × 1 𝑚𝑚

2

× 1 𝑠𝑠

−

3

𝑘𝑘 × 10

4

2

× 1 𝑠𝑠

−

7

2

−

b) Light year is the total distance travelled by light in one year.

1 ly = Speed of light × One year

= (3 × 10

8

) × (365 × 24 × 60 × 60

= 9.46 × 10

15

9.46×

15

= 1.057 × 10

−

ly

c) 1 𝑚𝑚 = 10

−

1 ℎ = 60 × 60 𝑠𝑠 = 3600 𝑠𝑠 ⇒ 1 𝑠𝑠 =

−

−

−

2

−

−

= (3 × 10

−

𝑘𝑘𝑚𝑚) × ((3600)

2

−

) = 3.88 × 10

−

−

d) 1 𝑁𝑁 = 1 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠

−

3

3

6

3

∴ 6.67 × 10

−

2

−

= 6.67 × 10

−

× (1 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠

−

2

−

= 6.67 × 10

−

−

3

−

) = 6.67 × 10

−

× (

−

−

6

3

−

= 6.67 × 10

−

3

−

−

Q2. A calorie is a unit of heat or energy and it equals about 4.2 J where 1 𝐽𝐽 = 1 𝑘𝑘𝑘𝑘 𝑚𝑚

2

−

. Suppose we employ a

system of units in which the unit of mass equals 𝛼𝛼 𝑘𝑘𝑘𝑘, the unit of length equals 𝛽𝛽 𝑚𝑚, the unit of time is 𝛾𝛾 𝑠𝑠.

Show that a calorie has a magnitude 4.2 𝛼𝛼

−

−

2

in terms of the new units. [NCERT Exercise]

Sol. Given that,

1 calorie = 4.2(1 𝑘𝑘𝑘𝑘

2

−

New unit of mass = 𝛼𝛼 𝑘𝑘𝑘𝑘

Hence, in terms of the new unit, 1 𝑘𝑘𝑘𝑘 =

1

𝛼𝛼

−

In terms of the new unit of length,

1

𝛽𝛽

−

or 1 𝑚𝑚

2

−

And, in terms of the new unit of time,

1

𝛾𝛾

−

2

−

−

2

∴ 1 calorie = 4.2(1𝛼𝛼

−

−

2

−

−

2

APNI KAKSHA

Q5. The mass and volume of a body are 4.237 𝑘𝑘 and 2.5 𝑐𝑐𝑚𝑚

3

, respectively. The density of the material of the

body in correct significant figures is [NCERT Exemplar]

a) 1.6948 𝑘𝑘 𝑐𝑐𝑚𝑚

−

b) 1.69 𝑘𝑘 𝑐𝑐𝑚𝑚

−

c) 1.7 𝑘𝑘 𝑐𝑐𝑚𝑚

−

d) 1.695 𝑘𝑘 𝑐𝑐𝑚𝑚

−

Sol. The correct option is C, 1.7 𝑘𝑘/𝑐𝑐𝑚𝑚

3

We know that, the density of a material is given by, 𝜌𝜌 =

, where m is mass and V is volume.

3

We know that is multiplication or division, the final answer should have as many significant figures as in

given data with minimum number of significant figures.

Here, 2.5 𝑐𝑐𝑚𝑚

3

have the minimum number or significant figures equal to two. Therefore, the final answer

should have two significant figures.

On rounding off 1.6948 𝑘𝑘/𝑐𝑐𝑚𝑚

3

to two significant figures, we get, 1.7 𝑘𝑘/𝑐𝑐𝑚𝑚

3

Q6. Yor measure two quantities as 𝑑𝑑 = 1.0 𝑚𝑚 ± 0.2 𝑚𝑚, 𝐵𝐵 = 2.0 𝑚𝑚 ± 0.2 𝑚𝑚. We should report correct value for

√𝑑𝑑𝐵𝐵 as: [NCERT Exemplar]

a) 1.4 ± 0.4 𝑚𝑚 b) 1.41 𝑚𝑚 ± 0.15 𝑚𝑚

c) 1.4 𝑚𝑚 ± 0.3 𝑚𝑚 d) 1.4 𝑚𝑚 ± 0.2 𝑚𝑚

Sol. Here, 𝑑𝑑 = 1.0 ± 0.2 𝑚𝑚, 𝐵𝐵 = 2.0 𝑚𝑚 ± 0.2 𝑚𝑚

2

Rounding off to two significant figures, we get

× √𝑑𝑑𝐵𝐵 =

× 1.414 = 0.212 𝑚𝑚

Rounding off to one significant figures, we get

The correct value for √𝑑𝑑𝐵𝐵 is 1.4 𝑚𝑚 ± 0.2 𝑚𝑚

Q7. Young’s modulus of steel is 1.9 × 10

11

2

. When expressed in CGS units of dynes/cm

2

, it will be equal to

(1N = 10

5

dyne, 1 m

2

4

cm

2

) [NCERT Exemplar]

a) 1.9 × 10

10

b) 1.9 × 10

11

c) 1.9 × 10

12

d) 1.9 × 10

13

Sol. The correct option is D 1.9 × 10

12

Given, Young’s modulus 𝑌𝑌 = 1.9 × 10

11

2

As we know that 1 𝑁𝑁 = 10

5

dyne and 1 𝑚𝑚 = 10

2

So, converting the value to CGS we get

1.9×

11

×

5

2

�

2

2

⇒ 𝑌𝑌 = 1.9 × 10

12

2

APNI KAKSHA

Q8. If momentum

, area

and time

are taken to be fundamental quantities, then energy has the

dimensional formula [NCERT Exemplar]

a) (𝑃𝑃

1

−

1

) b) (𝑃𝑃

2

1

1

c) �𝑃𝑃

1

−1/ 2

1

� d) �𝑃𝑃

1

1 / 2

−

Sol. Let

[

] = [

]

𝑥𝑥

[

]

𝑦𝑦

[

]

𝑧𝑧

Dimensional Formula of energy is [𝑀𝑀𝐿𝐿

2

−

]

Dimensional Formula of momentum is

[

−

]

Dimensional Formula of Area is [𝐿𝐿

2

]

Dimensional Formula of Time is

[

]

2

−

= [𝑀𝑀𝐿𝐿𝑇𝑇

−

]

𝑥𝑥

[𝐿𝐿

2

]

𝑦𝑦

[𝑇𝑇]

𝑍𝑍

[𝐸𝐸] = �𝑃𝑃𝑑𝑑

1 / 2

−

Q9. We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn

out to be 2.63 𝑠𝑠, 2.56 𝑠𝑠, 2.42 𝑠𝑠, 2.71 𝑠𝑠 and 2.80 𝑠𝑠. Calculate the absolute errors, relative error or percentage

error. [NCERT Solved Example]

Sol. The mean period of oscillation of the pendulum

The absolute errors in the measurement are

1

2

3

4

5

Mean absolute error is

𝑇𝑇𝑇𝑇𝑚𝑚𝑛𝑛

Percentage error is

𝑇𝑇𝑇𝑇𝑚𝑚𝑛𝑛

𝑇𝑇𝑇𝑇𝑚𝑚𝑛𝑛

× 100 =

× 100 ≈ 4%