Measurements and Experimentation

The Measurements and Experimentation is foundational, establishing the critical role of measurement in science. It begins by defining a physical quantity as the product of a numerical value and a unit. This leads to the distinction between fundamental units (like those for length, mass, and time, which are independent) and derived units (like speed or density, which are combinations of fundamental units). The chapter emphasizes the importance of the internationally accepted S.I. System (Système International d’Unités) as the standard for scientific communication, contrasting it with older systems like CGS and FPS. Furthermore, it details the necessary characteristics of a good unit, ensuring it is invariable, reproducible, and unambiguous.

A major focus of the chapter is the practical aspect of measuring length and achieving high accuracy. It introduces the concept of Least Count (LC)—the smallest value an instrument can measure. While a metre scale is used for basic measurements, the chapter delves into precision instruments: the Vernier Callipers and the Screw Gauge. Students learn the principles of these instruments, how to calculate their respective least counts, and the essential procedure for dealing with zero error (positive and negative). A correct reading must always compensate for this error. Beyond length, the chapter also addresses the measurement of time, introducing the Simple Pendulum as an experimental setup to accurately determine the Time Period of oscillation.

In essence, this chapter equips students with the skills to conduct accurate experiments. It covers the theoretical framework of units and systems, the practical methods for precise measurements using specialised instruments, and the techniques for correcting experimental errors. This understanding of measurement, accuracy, and precision is vital, as it forms the basis for all subsequent experimental work in physics.

Exercise 1 (A)

Question 1.

What is meant by measurement?

Ans: 

Measurement is the fundamental action in science where you determine the size or magnitude of a physical property (like length, mass, or time). This is done by a process of comparison: the quantity being measured (the unknown) is compared against a fixed, universally accepted reference of the same nature, which we call a unit.

The outcome of this comparison is a complete physical quantity, expressed as a product: a numerical value that indicates how many times the standard unit is contained within the quantity, multiplied by the unit itself. For instance, stating an object has a mass of 5 kilograms quantifies that mass as five times the defined standard kilogram unit. This systematic quantification provides the necessary objective data for all scientific analysis and experimentation.

Question 2.

What do you understand about the Term Unit?

Ans: 

A unit is a definite magnitude of a physical quantity, such as length or mass, that is fixed by convention or law and used as a standard of comparison for all other quantities of the same type.

In simple terms, a unit is the reference point that gives meaning to a numerical value in a measurement.

Measurement=Numerical Value×Unit

For example, if a rope is 10 meters long, the unit is the meter (m), which is the universally accepted standard for measuring length. The number 10 indicates that the rope is 10 times the length of the standard unit.

Essential Characteristics of a Standard Unit

For a unit to be useful in science and everyday life, it must possess several key characteristics, which are now largely embodied by the International System of Units (SI):

1. Universally Accepted (Invariable): It must be the same everywhere in the world and accepted by the international community.
2. Well-Defined: It must be defined precisely, ideally in terms of a fundamental physical constant, so that its value does not change.
3. Reproducible: It must be possible to accurately reproduce the standard in any laboratory.
4. Incorruptible: Its value should not change with time or with changes in physical conditions like temperature or pressure.
5. Appropriate Size: It should be a suitable size for measuring the quantity in question (e.g., using a meter for a desk instead of a kilometer).

Importance of Units

Units are absolutely necessary because a numerical value alone is meaningless without a reference scale.

• Clarity and Meaning: They provide context. Saying a trip took “2” is meaningless; saying it took “2 hours” is clear.
• Consistency: They ensure that measurements are the same regardless of who performs the measurement or where it is done. This allows scientists and engineers worldwide to share, compare, and reproduce data reliably (the core of the scientific method).
• Dimensional Analysis: In calculations, units are manipulated algebraically to verify that the final answer has the correct physical meaning (e.g., velocity must have units of length divided by time, like m/s).

Question 3.

What are the three requirements for selecting a unit of a physical quantity?

Ans: 

The three primary requirements for selecting a reliable and universally accepted unit of a physical quantity are Invariance, Unambiguous Definition, and Reproducibility.

These characteristics ensure the unit serves as a consistent standard for scientific and commercial use worldwide.

1. Invariance (Stability and Universality)

The value of the unit must not change with respect to changes in physical conditions, location, or time.

• Stability: The unit should remain constant over a long period. For example, the definition of the second (the SI unit of time) is now based on a specific, unchanging transition of the cesium-133 atom, making it inherently stable.
• Universality (or Invariance with Space): The unit must remain the same regardless of where the measurement is performed (on Earth, in space, or in different labs). This ensures measurements are comparable everywhere.

2. Unambiguous Definition (Clarity)

The unit must be clearly and precisely defined so that everyone can have the exact same understanding of what it represents.

• The definition should be precise enough that it can be measured with extreme accuracy.
• Ideally, the unit should be defined in terms of fundamental physical constants of nature (like the speed of light or the Planck constant), rather than a physical artifact that can be damaged or lost.

3. Reproducibility (Practicality)

It must be possible to reproduce the unit easily and accurately in any standard laboratory, allowing for consistent measurement and calibration.

• This means that the standard quantity can be easily accessed or the procedure for generating the unit can be repeated to verify the measurement.
• A unit should also be of a convenient size for practical measurements (e.g., the kilometer is used for road distances, while the millimeter is used for mechanical drawings).

Question 4.

Name the three fundamental quantities.

Ans: 

The three most commonly referenced fundamental quantities in the field of physics, particularly in classical mechanics, are:

1. Length (Unit: meter, m)
2. Mass (Unit: kilogram, kg)
3. Time (Unit: second, s)

These three quantities are considered the basic building blocks from which all other mechanical quantities (like velocity, force, and energy) can be mathematically derived.

Question 5.

Name the three systems of units and state the various fundamentals units in them.

Ans: 

System of Units	Fundamental Quantity	Unit Name	Unit Symbol
CGS System	Centimeter	Centimeter	cm
	Gram	Gram	g
	Second	Second	s
\hline
MKS System	Meter	Meter	m
	Kilogram	Kilogram	kg
	Second	Second	s
\hline
FPS System	Foot	Foot	ft
	Pound	Pound (for mass)	lb
	Second	Second	s

Question 6.

Define a fundamental unit.

Ans:

A fundamental unit (or base unit) is a standard unit of measurement for a basic physical quantity that is:

1. Independent: It is defined on its own and does not rely on any other units for its definition.
2. Basic: It cannot be broken down into simpler units.
3. Foundation: All other units of measure (derived units) are created by algebraically combining these fundamental units.

In the globally accepted International System of Units (SI), there are seven fundamental units that form the basis for all scientific measurements.

The Seven SI Fundamental Units

Base Quantity	Fundamental Unit	Symbol
Length	Meter	m
Mass	Kilogram	kg
Time	Second	s
Electric Current	Ampere	A
Thermodynamic Temperature	Kelvin	K
Amount of Substance	Mole	mol
Luminous Intensity	Candela	cd

Key Characteristics

• Non-Derivable: You cannot express a fundamental unit (like the meter) as a combination of other fundamental units (like the second and kilogram).
• Universal Definition: Their definitions are based on invariant physical constants or phenomena (e.g., the speed of light) to ensure they are the same everywhere.
• Building Blocks: Units for complex quantities like velocity (meter/second) or force (Newton, which is kg⋅m/s2) are derived from these fundamental units.

Question 7.

What are the fundamental units in the S.I system? Name them along with their symbols.

Ans: 

The fundamental units of measurement in the International System of Units (SI) are the seven Base Units. These units are independent of each other and form the foundation from which all other SI units (derived units) are created.

7 Fundamental SI Units (Base Units)

Base Quantity	SI Unit (Name)	SI Unit (Symbol)
Length	metre (or meter)	m
Mass	kilogram	kg
Time	second	s
Electric Current	ampere	A
Thermodynamic Temperature	kelvin	K
Amount of Substance	mole	mol
Luminous Intensity	candela	cd

What are Fundamental Units?

Fundamental (or Base) Units are the set of seven basic units that are not defined in terms of any other units. They are the essential building blocks for measuring physical quantities. For example, the unit for force, the Newton (N), is a derived unit because it is defined using three fundamental units: N=kg⋅m/s2.

Question 8.

Explain the meaning of the derived unit with the help of one example?

Ans: 

A derived unit is a unit of measurement created by mathematically combining two or more fundamental (base) units. These units are used to quantify physical quantities whose definitions rely on formulas involving the base quantities. While fundamental units (such as the meter, kilogram, or second) stand on their own, derived units are built upon them using multiplication, division, or exponents.

For instance, consider the unit for Speed. Speed is defined by the formula Distance/Time. By replacing these quantities with their respective fundamental SI units, we derive the unit for speed:

Unit of Speed=Unit of Time /Unit of Distance​=second (s)meter (m)​=m/s

The m/s (meters per second) is therefore a derived unit. Similarly, the unit for Force, the Newton (N), is derived from the units of mass, length, and time: N=kg⋅m/s2. The concept of derived units allows the entire system of physical measurement to be consistently and coherently linked back to a small set of defined base units.

Question 9.

Define Standard metre.

Ans: 

The modern definition of the metre (m), the SI base unit for length, is tied to a fundamental constant of nature to guarantee extreme precision and universality. The Standard Metre is currently defined as:

This definition is unique and highly stable for the following key reasons:

1. Fixed Speed of Light: It effectively fixes the numerical value of the speed of light (c) at exactly 299,792,458 meters per second (m/s). The metre is thus not defined by a physical object (like an old metal bar) but by this universal constant.
2. Universal Reproducibility: Since the speed of light is constant everywhere, the definition ensures that the standard for the metre is universally available and highly reproducible with high accuracy.
3. Dependence on Time: The accuracy of the metre is now inherently linked to the accuracy of the SI base unit for time, the second. The second is itself defined using another unchanging atomic constant (the frequency of the cesium-133 atom).

Question 10.

Name two units of length which are bigger than a metre. How are they related to the metre?

Ans:

Two units of length that are significantly bigger than a metre are the Kilometre and the Astronomical Unit.

1. Kilometre (km):
   • The kilometre is the standard unit used for measuring large terrestrial distances, such as the distance between cities or the length of a road.
   • Relation to the Metre: A kilometre is equal to one thousand metres.
     
     1 km=1000 m
2. Astronomical Unit (AU):
   • The Astronomical Unit is a unit of distance used primarily for measuring distances within our solar system. It is defined as the average distance between the Earth and the Sun.
   • Relation to the Metre: One Astronomical Unit is approximately equal to 149.6 billion metres.
     
     1 AU≈1.496×1011 m

Question 11.

Write the names of two units of length smaller than a metre. Express their relationship with the metre.

Ans:

Two units of length smaller than a metre and their relationship with the metre are:

1. Centimetre (cm): One metre contains one hundred centimetres.
   1 cm=10−2 mor1 m=100 cm
2. Millimetre (mm): One metre contains one thousand millimetres.
   1 mm=10−3 mor1 m=1000 mm

Note: Other commonly used units smaller than a metre include the micrometre (μm) and the nanometre (nm).

Question 12.

How is nanometer related to Angstrom?

Ans:

The nanometer (nm) and the Angstrom (A˚) are both metric units of length used to measure extremely small distances, such as atomic dimensions and wavelengths of light. The relationship between them is straightforward: one nanometer is equal to ten Angstroms.

The conversion factors based on the standard unit, the meter (m), are:

• 1 nanometer (nm)=10−9 m
• 1 Angstrom (A˚)=10−10 m

Therefore, mathematically:

1 nm=10−10 m/A˚10−9 m​=10(−9)−(−10) A˚=101 A˚

This means the Angstrom is a smaller unit than the nanometer. While the nanometer is an official SI unit prefix, the Angstrom is a non-SI unit, though it is still commonly used in fields like crystallography and atomic physics because it conveniently expresses bond lengths and atomic radii.

Question 13.

Name the three convenient units used to measure length ranging from very short to very long value. How are they related to the S.I unit?

Ans:

The three convenient units of length used to measure values ranging from very short to very long, along with their relation to the S.I. unit (the meter, m), are:

Range	Unit Name (Symbol)	Used to Measure	Relation to S.I. Unit (Meter, m)
Very Short	Angstrom (A˚)	Atomic sizes, chemical bond lengths, and wavelengths of X-rays.	1 A˚=10−10 m
Intermediate	Centimeter (cm)	Everyday objects like the length of a pencil or a person’s height.	1 cm=10−2 m
Very Long	Light Year (ly)	Interstellar and astronomical distances (e.g., distance to stars).	1 ly≈9.46×1015 m

Relation to the S.I. Unit

The S.I. unit of length is the meter (m). All other units of length are defined in terms of the meter, either as decimal multiples/submultiples (using S.I. prefixes) or as separate non-S.I. units with a defined conversion factor.

1. Angstrom (A˚): This unit is a non-S.I. unit used for extremely small measurements. It is defined as exactly one ten-billionth of a meter.
   1 Angstrom=1×10−10 m
2. Centimeter (cm): This is a decimal submultiple of the S.I. unit, where the prefix centi- means one hundredth (10−2).
   1 Centimeter=1×10−2 m
3. Light Year (ly): This is a non-S.I. unit used for astronomical distances. It is defined as the distance light travels in a vacuum in one Julian year. When converted, its value is an extremely large multiple of the meter.
   
   1 Light Year≈9,460,730,472,580,800 m
   or 1 ly≈9.46×1015 m

Question 14.

Name the S.I unit of mass and define it.

Ans:

The kilogram (kg), the S.I. unit of mass, is now defined by fixing the exact numerical value of the Planck constant (h).

This modern definition sets the value of h as precisely 6.62607015×10−34 when expressed in the unit J⋅s (joule-second), which is equivalent to kg⋅m2⋅s−1. Because the values for the meter (m) and the second (s) are already established using other fundamental constants, the kilogram’s value is derived from this fixed constant. This connection is physically realized using a highly sophisticated weighing apparatus known as the Kibble balance . By basing the kilogram on a universal constant of nature, the standard is now universally stable and reproducible, overcoming the historical issue of relying on a fragile physical object (the International Prototype of the Kilogram) whose mass could change over time.

Question 15.

(a) Complete the following: 1 light year =……….m

(b) Complete the following: 1 m =……….Å

(c) Complete the following: 1 m  =……….. µ

(d) Complete the following: 1 micron =………..Å

(e) Complete the following: 1 fermi = ……….. M

Ans:

(a) 1 light year = 9.46×1015 m 

(b) 1 m = 1010 A˚ (Angstrom) 

(c) 1 m = 106 μ (micron)

(d) 1 micron (μ) = 104 A˚ (Angstrom) 

(e) 1 fermi (f) = 10−15 m

Question 16. 

State two units of mass smaller than a kilogram. How are they related to kilogram?

Ans:

Two units of mass that are smaller than a kilogram (kg) are the gram (g) and the milligram (mg).

Relation to Kilogram

1. Gram (g): The gram is one thousandth of a kilogram. This is based on the metric prefix ‘kilo’ which means 1000. The relationship is:
   1 gram=10001​ kilogramor1 g=10−3 kg
2. Milligram (mg): The milligram is one millionth of a kilogram. It is also one thousandth of a gram (the prefix ‘milli’ means 10−3). The relationship is:
   1 milligram=1,000,0001​ kilogramor1 mg=10−6 kg

Question 17. 

State two units of mass bigger than a kilogram. Give their relationship with the kilogram?

Ans:

Two units of mass larger than a kilogram (kg) are the Quintal and the Metric Tonne.

Here is their relationship with the kilogram:

1. Quintal (q)
   • Relationship: One quintal is equivalent to 100 kilograms.
   • Formula: 1 q=100 kg
2. Metric Tonne (t)
   • Relationship: One metric tonne is equivalent to 1000 kilograms.
   • Formula: 1 t=1000 kg

The Metric Tonne is often used to measure the mass of large, bulk commodities in trade and industry, such as grains, metals, or vehicle cargo.

Question 18. 

(a)Complete the following: 1 g =………kg 

(b) Complete the following : 1 mg =………kg 

(c) Complete the following: 1 quintal =………kg 

(d)Complete the following : 1 a.m.u (or u) =……..kg

Ans:

Unit	Conversion to Kilogram (kg)
(a) 1 g	10−3 kg
(b) 1 mg	10−6 kg
(c) 1 quintal	100 kg
(d) 1 a.m.u (or u)	1.6605×10−27 kg

Question 19. 

Name the S.I unit of time and define it.

Ans:

The modern definition of the second (s), the S.I. unit of time, is based on a fundamental and unchanging atomic property, which ensures its exceptional accuracy and universal consistency.

One second is formally defined as the duration of exactly 9,192,631,770 periods of the radiation produced by the transition between the two specific hyperfine energy levels of an unperturbed Caesium-133 atom in its ground state. This definition effectively fixes the numerical value of the Caesium-133 transition frequency (ΔνCs​) at 9,192,631,770 Hertz (Hz), where Hz is equivalent to s−1. By referencing a constant of nature—the resonant frequency of the Caesium-133 atom—the second is made the most stable and accurately measurable of all the fundamental S.I. units.

Question 20.

Name two units of time bigger than a second. How are they related to the Second?

Ans:

Two units of time that are bigger than a second (s) are the minute (min) and the hour (h).

Their relationship to the second, which is the SI base unit of time, is as follows:

1. Minute (min):
   • One minute is defined as a duration of 60 seconds.
   • Relation to Second: 1 min=60 s
2. Hour (h):
   • One hour is defined as a duration of 60 minutes.
   • Since 1 minute is 60 seconds, one hour is equal to 60×60=3600 seconds.
   • Relation to Second: 1 h=3600 s

Question 21. 

What is a leap year?

Ans:

A leap year is a calendar year that contains an extra day, making it 366 days long instead of the usual 365. This additional day, known as a leap day, is added to the month of February, extending it to February 29th. The purpose of a leap year is to keep our civil calendar synchronized with the solar year (the actual time it takes the Earth to complete one full orbit around the Sun).

The Earth’s orbit takes approximately 365.2422 days (or about 365 days and 6 hours). If we only used a 365-day year, we’d lose roughly six hours every year. Over a century, the calendar would drift by about 24 days, causing seasons to gradually fall out of alignment with the calendar dates. By adding a leap day almost every four years, we largely correct this accumulated discrepancy, ensuring that the seasons begin consistently around the same time each year.

The rules for determining a leap year under the Gregorian calendar (the one used by most of the world) are:

1. A year is a leap year if it is evenly divisible by 4 (e.g., 2024, 2028).
2. Exception 1: If the year is evenly divisible by 100, it is not a leap year (e.g., 1700, 1800, 1900).
3. Exception 2: If the year is evenly divisible by 400, it is a leap year (e.g., 1600, 2000, 2400).

This set of rules results in an average year length of 365.2425 days, which is extremely close to the true solar year, keeping the calendar accurate for centuries.

Question 22.

The year 2020 will have February of 29 days’. Is this statement true?

Ans:

The statement, “The year 2020 will have February of 29 days,” is True. 

Explanation

The year 2020 was a Leap Year, which means it had 366 days instead of the usual 365. This extra day is added to the calendar by extending the month of February from 28 days to 29 days.

The rules of the Gregorian calendar determine a leap year:

1. A year is typically a leap year if it is evenly divisible by 4 (which 2020÷4=505, a whole number).
2. Exception 1: Years divisible by 100 (like 1900 or 2100) are not leap years…
3. Exception 2: …unless they are also divisible by 400 (like 2000 or 2400), in which case they are leap years.

Since 2020 is divisible by 4 but not by 100, it satisfied the primary rule for a leap year, confirming that its February contained the extra day, February 29th.

Question 23.

What is a lunar month?

Ans:

A lunar month is a unit of time that is based on the Moon’s motion and its cycle of phases.

The most common and practical definition of a lunar month is the synodic month (or lunation), which is the average time it takes for the Moon to complete one cycle of its visible phases—for example, from one New Moon to the next.

Key Facts about the Lunar Month (Synodic Month)

• Duration: The average length of a synodic month is approximately 29.53 days (specifically, 29 days, 12 hours, 44 minutes, and 3 seconds).
• Basis: This duration is determined by the alignment of the Moon, Earth, and Sun. Because the Earth is constantly orbiting the Sun, the Moon has to travel an extra distance in its orbit after completing a full revolution around the Earth to “catch up” and return to the same alignment (i.e., the same phase, like New Moon).
• Calendars: Lunar calendars (like the Islamic Hijri calendar) and lunisolar calendars (like the Chinese and Jewish calendars) base their months on this cycle, typically alternating between 29 and 30 days to closely match the average 29.5 days.

Other Astronomical Lunar Months

In astronomy, there are several other precise definitions of a “month” based on different reference points:

• Sidereal Month: The time it takes for the Moon to complete one full orbit around the Earth when measured against the background of fixed stars. This is shorter than the synodic month, lasting about 27.32 days.
• Anomalistic Month: The time it takes for the Moon to return to its closest point to Earth (perigee). This cycle is important for determining the size and intensity of tides and phenomena like a Supermoon, lasting about 27.55 days.
• Draconic Month: The time it takes for the Moon to pass through the same orbital node (the points where the Moon’s orbit crosses the Earth’s orbital plane). This period is crucial for predicting eclipses and lasts about 27.21 days.

Question 24. 

(a) Complete the following: 1 nano second =……….s.

(b) Complete the following: 1 µs =……….s 

(c) Complete the following: 1 mean solar day = _______ s.

(d) Complete the following: 1 year =……….s.

Ans:

Conversion	Value in Seconds (s)
(a) 1 nano second (ns)	10−9 s
(b) 1 microsecond (μs)	10−6 s
(c) 1 mean solar day	86,400 s
(d) 1 year (≈365 days)	31,536,000 s (≈3.15×107 s)

Question 25. 

(a) Name the physical quantities which are measured in the given units:- u 

(b) Name the physical quantities which are measured in the given units:- ly 

(c) the physical quantities which are measured in the given units:- ns

(d) Name the physical quantities which are measured in the given units:- nm

Ans: 

Unit	Physical Quantity Measured
(a) u (or amu)	Mass (specifically, atomic/molecular mass)
(b) ly (Light Year)	Length or Distance (specifically, astronomical distances)
(c) ns (nanosecond)	Time
(d) nm (nanometer)	Length

Question 26. 

(a) Write the derived unit of speed

(b) Write the derived unit of force

(c) Write the derived unit of work

(d) Write the derived unit of the pressure

Ans:

The derived units in the International System of Units (SI) for the given quantities are:

Quantity	Derived Unit (Special Name)	Derived Unit (in Base Units)
(a) Speed	None (N/A)	metre per second (m/s or m⋅s−1)
(b) Force	Newton (N)	kilogram metre per second squared (kg⋅m/s2 or kg⋅m⋅s−2)
(c) Work	Joule (J)	kilogram metre squared per second squared (kg⋅m2/s2 or N⋅m)
(d) Pressure	Pascal (Pa)	kilogram per metre per second squared (kg/(m⋅s2) or N/m2)

Explanation

Derived units are defined by multiplying or dividing SI base units (like metre, kilogram, second) according to the physical formula:

• Speed: Speed=Distance/Time. The unit is metre/second.
• Force: Force=Mass×Acceleration. Since Acceleration is m/s2, the unit is kg⋅m/s2, which is named the Newton (N).
• Work: Work=Force×Distance. The unit is Newton×metre (N⋅m), which is named the Joule (J).
• Pressure: Pressure=Force/Area. The unit is Newton/metre2 (N/m2), which is named the Pascal (Pa).

Question  27.

(a) How are the following derived units related to the fundamental unit?

Newton

(b) How are the following derived units related to the fundamental unit?

watt

(c) How are the following derived units related to the fundamental unit?

joule

(d) How are the following derived units related to the fundamental unit?

Pascal

Ans: 

Derived Unit	Quantity	Relationship to Fundamental Units	Formula Basis
(a) Newton (N)	Force	kg⋅m⋅s−2	Force=mass×acceleration
(b) Watt (W)	Power	kg⋅m2⋅s−3	Power=timework​
(c) Joule (J)	Energy or Work	kg⋅m2⋅s−2	Work=force×distance
(d) Pascal (Pa)	Pressure	kg⋅m−1⋅s−2	Pressure=areaforce​

Question  28. 

(a) Name the physical quantity related to the following unit :- km2

(b) Name the physical quantity related to the following unit:-Newton

(c) Name the physical quantity related to the following unit :- Joule

(d) Name the physical quantity related to the following unit: Pascal

(e) Name the physical quantity related to the following unit:- watt

Ans:

(a) km2 is the unit for Area (km2 stands for square kilometre). 

(b) Newton (N) is the unit for Force.

(c) Joule (J) is the unit for Energy or Work. 

(d) Pascal (Pa) is the unit for Pressure. 

(e) Watt (W) is the unit for Power.

Exercise 1 (A)

Question 1. 

The fundamental unit is:-

1. Newton
2. Pascal
3. hertz
4. second

Question 2.

Which of the following unit is not a fundamental unit:

1. metre
2. Litre
3. second
4. kilogram

Question 3.

The unit of time is :

1. light year
2. parsec
3. leap year
4. angstrom

Question 4.

1 Å is equal to :

1. 0.1 nm
2. 10-10 cm
3. 10-8 m
4. 104 µ

Question 5.

ly is the unit of:-

1. time
2. length
3. mass
4. none of this

Exercise 1 (A)

Question 1.

(a) The wavelength of light of a particular colour is 5800 Å. Express it in nanometre.

(b)The wavelength of light of a particular colour is 5800 Å. Express it in metre.

Ans:

The wavelength of light is 5800 A˚. Here are the conversions:

(a) Expressed in Nanometres (nm)

The relationship is 1 A˚=0.1 nm.

Wavelength=5800 A˚=580 nm

(b) Expressed in Metres (m)

The relationship is 1 A˚=10−10 m.

Wavelength=5800 A˚=5.8×10−7 m

(Calculation Note: 5800×10−10 m=5.8×103×10−10 m=5.8×10(3−10) m=5.8×10−7 m.)

Question 2.

The size of bacteria is 1 µ. Find the number of bacteria present in 1 m length.

Ans:

The total number of bacteria that fit along the 1 m length is 106, or one million.

Question 3.

The distance of a galaxy is 5·6 × 1025 m . Assuming the speed of light to be 3 × 108 ms−1. Find the time taken by light to travel this distance . 

[Hint : Time taken = Distance travelled speed

Ans:

The time (t) light takes to cover a specific distance (D) is fundamentally calculated by dividing that distance by the speed of light (v), following the basic formula:

t=vD​

Given Parameters

• Distance to the Galaxy (D): 5.6×1025 m
• Speed of Light (v): 3×108 m/s

Step-by-Step Calculation

The values are substituted into the formula for time:

t=3×108 m/s5.6×1025 m​

To simplify, the numerical coefficients are separated from the powers of ten:

t=(35.6​)×10(25−8) s

Performing the division and simplifying the exponent yields:

t≈1.866…×1017 s

Final Conclusion

The calculated time required for light to travel the distance of 5.6×1025 m is approximately 1.87×1017 seconds.

This incredibly long duration is immense, translating to nearly 5.9 billion years, highlighting the vast scale of astronomical distances.

Question 4.

The wavelength of light is 589 nm. What is its wavelength in Å?

Ans:

The wavelength of the light, λ, is 589 nm. To find its value in A˚ (Angstrom), we use the established conversion factor between nanometers (nm) and Angstroms (A˚).

Conversion Factor

Both the nanometer and the Angstrom are units used to measure extremely small lengths, such as the wavelength of light. Their relationship is:

1 nm=10 A˚

Calculation

To convert the wavelength from nm to A˚, we multiply the value by 10:

λ=589 nmλ=589×10 A˚λ=5890 A˚

Result

The wavelength of light is 5890 A˚.

Question 5.

The mass of an oxygen atom is 16.0 u. Express it in kg.

Ans:

The mass of the oxygen atom, 16.0 u, can be expressed in kilograms (kg) using the conversion factor for the unified atomic mass unit (u).

Conversion

The relationship between the atomic mass unit (u) and the kilogram (kg) is:

1 u≈1.661×10−27 kg

Calculation

To find the mass in kilograms, multiply the given mass in u by the conversion factor:

Mass in kg=16.0 u×(1.661×10−27 ukg​)

Mass in kg≈2.658×10−26 kg

Rounding to three significant figures, the same as the input mass:

Mass in kg≈2.66×10−26 kg

The mass of an oxygen atom is approximately 2.66×10−26 kilograms.

Question 6. 

It takes 8 min for light to reach from the sun to the earth surface . If the speed of light is taken to be 3 ×108  m/s−1 , find the distance from the sun to the earth in km .

Ans:

Here is the calculation to determine the distance from the Sun to the Earth, based on the travel time of light.

The fundamental relationship governing this calculation is: Distance=Speed×Time.

1. Standardize the Time Unit (Minutes to Seconds) 

To ensure dimensional consistency with the speed of light (which is given in meters per second), the travel time must be converted from minutes to seconds:

Time (t)=8 minutes×60 minute seconds​=480 seconds

2. Compute the Distance in Meters (m)

Using the established values: the time (t=480 s) and the speed of light (v=3×108 m/s):

D=v×tD=(3×108 m/s)×480 sD=1440×108 m

D=1.44×1011 m

3. Convert Distance to Kilometers (km)

The final distance is required in kilometers, so the value in meters is divided by 1000 (since 1 km=1000 m):

Dkm​=103 m/km1.44×1011 m​Dkm​=1.44×10(11−3) kmDkm​=1.44×108 km

The distance between the Sun and the Earth is 1.44×108 kilometers (or 144 million kilometers).

Question 7. 

‘The distance of a star from the earth is 8.33 light minutes.’ What do you mean by this statement? Express the distance in meters.

Ans:

This statement, ‘The distance of a star from the earth is 8.33 light minutes,’ means that light, traveling at its tremendous speed, takes 8.33 minutes to cover the distance between that star and the Earth.

It is defined as the distance light travels in a vacuum over the course of one minute. This unit is used to express vast astronomical distances in a way that is easier to grasp than using excessively large numbers of kilometers or miles.

Distance in Meters

To express this distance in meters, we use the formula:

Distance(D)=Speed of Light(v)×Time(t)

We need to use the standard SI units for this calculation, so we’ll use:

• Speed of light (v): 3.00×108 m/s
• Time (t): 8.33 minutes

1. Convert Time to Seconds

First, convert the time from minutes to seconds:

t=8.33 min×60 mins​=499.8 s

2. Calculate Distance

Now, substitute the values into the distance formula:

D=(3.00×108 m/s)×499.8 s

D=1499.4×108 m

3. Express in Scientific Notation

Finally, convert the result to standard scientific notation:

D=1.50×1011 m

The distance of the star from the Earth is approximately 1.50×1011 meters.

Exercise 1 (B)

Question 1. 

Explain the meaning of the term ‘least count of an instrument’ by taking a suitable example.

Ans:

The Least Count (LC) of an instrument is the smallest measurement that can be accurately measured or resolved by that instrument. It represents the precision of the measuring device.

In simpler terms, it is the value of the smallest division marked on the instrument’s scale. Any measurement taken using that instrument can only be considered reliable up to that minimum value.

Example: A Standard Meter Ruler

A common plastic or wooden ruler is a good example of how least count is determined.

Instrument Details

• Main Scale Divisions: A standard meter ruler is typically marked with centimeters (cm) and millimeters (mm).
• The space between two consecutive centimeter markings is 1 cm.
• Each centimeter is further divided into smaller markings, which are millimeters. The smallest interval between two adjacent markings on the ruler is 1 mm.

Least Count

The least count of the ruler is the value of its smallest division:

Least Count (LC)=1 mm or 0.1 cm

Implication

This means that when using a standard ruler:

1. You can confidently measure a length, say, to 15.3 cm.
2. You cannot accurately measure a length like 15.35 cm because the instrument lacks the resolution (the smallest division) to resolve the second decimal place (the 0.05 cm or 0.5 mm value).
3. Any recorded measurement will have an uncertainty equal to the least count (e.g., 15.3 cm±0.1 cm).

A smaller least count indicates a more precise instrument (e.g., a Vernier caliper has a LC of 0.1 mm, making it ten times more precise than a standard ruler).

Question 2. 

A boy makes a ruler with graduations in cm on it, (i.e. 100 divisions in 1 m). To what accuracy  can this ruler measure? How can this accuracy be increased?

Ans:

The accuracy (specifically, the precision) to which a ruler can measure is determined by its Least Count (LC), which is the value of its smallest graduation.

1. Measurement Accuracy

The ruler has 100 divisions in 1 m (or 100 cm).

• Length of the smallest division (Least Count):
  LC=Number of DivisionsTotal Length​=100100 cm​=1 cm
• Accuracy: The ruler can measure with an accuracy of ±1 cm. This means any measurement taken with it can only be reliably estimated to the nearest centimeter.

2. Increasing Accuracy

The accuracy of this ruler can be increased by reducing its least count.

• Method: To reduce the least count, the boy must increase the number of graduations (divisions) within the 1 m length.
• Example: If the boy marks 1000 divisions in 1 m (i.e., marking every millimeter), the new least count would be:
  New LC=1000100 cm​=0.1 cm (or 1 mm)

This new ruler would measure to an accuracy of ±0.1 cm, which is ten times greater than the original ruler.

Question 3.

A boy measures the length of a pencil and expresses it to be 2.6 cm. What is the accuracy of his measurement? Can he write it as 2.60 cm?

Ans:

The accuracy of the boy’s measurement of 2.6 cm is determined by the precision of the instrument he used, which is indicated by the number of decimal places recorded.

Accuracy of Measurement

The measurement 2.6 cm has two significant figures, with the last digit (the 6) being the estimated one. Assuming he used a standard centimetre ruler or scale (which typically has divisions up to the 1 mm mark, or 0.1 cm), the least count of the instrument would be 0.1 cm.

The accuracy (or more precisely, the absolute error or uncertainty) is usually taken as half the least count for readings taken between the marks, or ± the full least count for readings taken directly on a mark. In this context, the reading 2.6 cm implies that the measurement is accurate to the first decimal place:

Accuracy=±0.1 cm

This means the actual length of the pencil lies between 2.5 cm and 2.7 cm.

Can he write it as 2.60 cm?

No, he cannot write the measurement as 2.60 cm.

• Writing 2.60 cm implies an accuracy of ±0.01 cm (one-hundredth of a centimetre), meaning the measuring instrument has a least count of 0.01 cm (or 0.1 mm), such as a Vernier Callipers.
• Since his initial reading of 2.6 cm only expresses accuracy to the 0.1 cm place, adding the final zero (2.60) would incorrectly suggest that he used a more precise instrument, like a Vernier Callipers, and that he is certain the length is exactly 2.60 cm and not, for example, 2.59 cm or 2.61 cm.

Question 4. 

Define the least count of vernier callipers. How do you determine it?

Ans:

The Least Count (LC) of a Vernier Callipers is the smallest distance or length that can be accurately measured using the instrument. It represents the limit of the instrument’s precision. 

Determining the Least Count

The Least Count is calculated using the difference between the smallest division on the main scale and the smallest division on the Vernier scale.

The formula to determine the Least Count is:

LC=1 MSD−1 VSD

Where:

• MSD is the value of one Main Scale Division (the smallest marking on the main scale).
• VSD is the value of one Vernier Scale Division (the value corresponding to the smallest marking on the Vernier scale).

In practice, VSD is often not directly known. The LC is more practically calculated using the following derived formula:

LC=Total number of divisions on the Vernier scaleValue of the smallest Main Scale Division​

If, for example, 10 Vernier Scale Divisions (VSD) are equal in length to 9 Main Scale Divisions (MSD), and the MSD value is 1 mm, the calculation is:

LC=101 mm​=0.1 mm or 0.01 cm

Question 5.

Define the term ‘Vernier constant’.

Ans:

The term ‘Vernier constant’ is synonymous with the Least Count (LC) of a Vernier Callipers.

The Vernier Constant is defined as the difference between the value of one Main Scale Division (1 MSD) and the value of one Vernier Scale Division (1 VSD).

It represents the smallest distance or length that can be accurately measured by the Vernier Callipers. 

Formula

The Vernier Constant (VC) is mathematically expressed as:

VC=1 MSD−1 VSD

Alternative Definition

In practical terms, it is the ratio of the smallest division on the main scale to the total number of divisions on the Vernier scale:

VC=Total number of divisions on Vernier ScaleSmallest value on Main Scale​

Question 6.

When is a vernier Callipers said to be free from zero error?

Ans:

A Vernier Callipers is said to be free from zero error when the zero mark of the Vernier scale perfectly coincides with the zero mark of the Main scale when the jaws of the callipers are brought into gentle contact.

In this ideal condition, the instrument registers a reading of exactly zero when the measured object’s size is zero. If any other Vernier scale mark coincides, or if the zeros do not align, a zero error is present.

Question 7.

What is meant by zero error of vernier callipers ? How is it determined? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement?

Ans:

The Zero Error (ZE) of a Vernier Callipers is a systematic error that occurs when the jaws are fully closed, and the zero mark of the Vernier scale does not precisely align with the zero mark of the Main scale. This indicates the instrument has a built-in offset.

Types and Determination

Type of Zero Error	Alignment Condition	Determination of n	Zero Error (ZE) Formula
Positive (+ZE)	Vernier zero is to the right of the Main scale zero.	n = The coinciding Vernier Scale Division (VSD)	+ZE=+n×LC
Negative (−ZE)	Vernier zero is to the left of the Main scale zero.	n = The coinciding Vernier Scale Division (VSD)	−ZE=−(N−n)×LC (Where N is total $\text{VSD}$s)

Correction

To obtain the true measurement, a Zero Correction (ZC) is applied. The ZC is always the negative of the ZE:

Zero Correction (ZC)=−(Zero Error)

The Correct Reading is then found by:

Correct Reading=Observed Reading+Zero Correction

• If ZE is Positive (+n×LC), the ZC is Negative, so you subtract the error.
• If ZE is Negative (−(N−n)×LC), the ZC is Positive, so you add the correction.

Question 8.

A vernier caliper has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.

Ans:

A vernier caliper has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.

Question 9.

Draw a neat and labelled diagram of vernier callipers. Name its main parts and state their functions.

Ans:

Here are the main parts of a Vernier Callipers and their functions:

Main Parts and Functions:

1. Main Scale (Fixed):
   • Function: This is a fixed scale, typically graduated in centimeters and millimeters. It provides the primary reading of the measurement, similar to a ruler.
2. Vernier Scale (Movable):
   • Function: This scale slides along the main scale. It has divisions that are slightly smaller than the main scale divisions, allowing for more precise readings than the main scale alone. It helps determine the fractional part of the smallest main scale division.
3. External Measuring Jaws (Lower Jaws):
   • Function: These are the larger, flat jaws used to measure the external diameter or width of an object. The object is placed between these jaws.
4. Internal Measuring Jaws (Upper Jaws):
   • Function: These are the smaller, pointed jaws used to measure the internal diameter of hollow objects (like the inside of a pipe or a beaker).
5. Depth Bar/Rod (or Depth Probe):
   • Function: This thin rod extends from the end of the movable jaw as it slides. It is used to measure the depth of holes or steps in an object.
6. Locking Screw:
   • Function: Once an object is correctly positioned and the jaws are set, this screw is tightened to fix the movable jaw in place, preventing it from slipping and ensuring an accurate reading can be taken.
7. Fine Adjustment Screw:
   • Function: On some Vernier Callipers, this screw is used for making very small, precise movements of the movable jaw to ensure accurate contact with the object being measured.

Question 10.

State three uses of the vernier callipers.

Ans:

Vernier callipers are essential tools for precise dimensional measurement in science, engineering, and workshops. They are highly valued for their versatility, as they can perform three distinct types of measurements on an object: 

1. Outer Dimensions: They measure the external length, width, or diameter of a solid object (like a rod or block) by clamping it between the two large, main jaws.
2. Inner Dimensions: They accurately determine the internal diameter (or bore) of a hollow object (such as a tube or container) using the smaller, upper jaws.
3. Depth: They measure the depth of a recess or a blind hole using the thin measuring strip or tail that extends from the back end of the instrument’s main body.

Question 11.

Name the two scales of vernier callipers and explain how it is used to measure length correctly up to 0.01 cm.

Ans:

The two scales of a Vernier Callipers are the Main Scale and the Vernier Scale.

I. The Two Scales

1. Main Scale (MS): This is the fixed scale on the main body of the instrument. It is marked like a standard ruler, typically graduated in centimeters (cm) with the smallest division being 1 mm (or 0.1 cm). This scale provides the initial reading.
2. Vernier Scale (VS): This is the sliding scale attached to the movable jaw. It is a shorter scale with divisions that are slightly smaller than the divisions on the main scale. Its purpose is to provide an accurate fraction of the smallest Main Scale Division, increasing the instrument’s precision.

II. Measuring Length up to 0.01 cm

The ability to measure accurately up to 0.01 cm is due to the instrument’s Least Count (LC).

Least Count (LC)=0.01 cm

This precision is achieved because the Vernier Scale is constructed so that a specific number of its divisions (N) match (N−1) divisions on the Main Scale. For a standard metric Vernier Callipers, 10 VSD often equal 9 MSD, and since 1 MSD=0.1 cm:

LC=1 MSD−1 VSD=Number of divisions on VSValue of 1 MSD​=100.1 cm​=0.01 cm

The Reading Procedure

The final length measurement (TR) is determined by combining the readings from both scales:

Total Reading (TR)=MSR+(VSC×LC)

1. Main Scale Reading (MSR): This is the reading on the Main Scale immediately before the zero mark of the Vernier Scale. This gives the measurement up to the first decimal place (e.g., 2.3 cm).
2. Vernier Scale Coincidence (VSC): This is the number of the division on the Vernier Scale that exactly aligns with any division on the Main Scale.
3. Fractional Part: The VSC is multiplied by the Least Count (0.01 cm) to get the fractional part of the measurement (e.g., if VSC=4, the fractional part is 4×0.01=0.04 cm).
4. Final Result: The MSR and the fractional part are added to give the final reading with two decimal places of precision (e.g., 2.3 cm+0.04 cm=2.34 cm).

This method allows the user to accurately resolve a distance that is smaller than the smallest division on the main scale.