Selina Concise Solutions for ICSE Class 9 Physics Chapter 1 Measurements And Experimentation

Exercise 1(A)

Question 1S. Define the term measurement.
Answer: Measurement is the process of comparing a given physical quantity with a known standard quantity of the same nature.
In simple words: Measurement is like using a ruler to see how long a desk is; you are comparing the unknown length of the desk to the known length of the ruler.

📝 Teacher's Note: Use a simple example like measuring the length of a classroom with a meter scale to explain the concept of comparison.

🎯 Exam Tip: Always mention "same nature" in the definition to ensure you score full marks.

Question 2S. What is a unit?
Answer: Unit is a quantity of constant magnitude which is used to measure the magnitudes of other quantities of the same manner.
In simple words: A unit is a fixed "standard slice" (like one centimeter) that we use to count how big other things are.

📝 Teacher's Note: Explain that a unit must be accepted by everyone so that measurements are consistent everywhere.

🎯 Exam Tip: The keyword here is "constant magnitude."

Question 3S. What are the three requirements for selecting a unit of a physical quantity?
Answer: The three requirements for selecting a unit of a physical quantity are
1. It should be possible to define the unit without ambiguity.
2. The unit should be reproducible.
3. The value of units should not change with space and time.
In simple words: A good unit should be clear to everyone, easy to copy, and should stay the same size whether you are in India or the USA, today or next year.

📝 Teacher's Note: Ask students why using a "hand-span" as a unit is a bad idea (it changes from person to person).

🎯 Exam Tip: Listing all three points is usually required for a 3-mark question.

Question 4S. Define the three fundamental S.I. units: length, mass, and time.
Answer: Definitions of three fundamental quantities:
1. S.I. unit of length (m): A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy made of 90% platinum and 10% iridium) rod kept at 0°C in the International Bureau of Weights and Measures at serves near Paris.
2. S.I. unit of mass (kg): In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at International Bureau of Weights and Measures at serves near Paris.
3. S.I. unit of time (s): A second is defined as 1/86400th part of a mean solar day, i.e.
\[ 1s = \frac{1}{86400} \times \text{one mean solar day} \]
In simple words: These are the standard "master" measurements kept in a safe in France that the whole world uses to make sure our rulers, scales, and clocks are correct.

📝 Teacher's Note: Explain that "Platinum-Iridium" is used because it is very stable and doesn't rust or change much with temperature.

🎯 Exam Tip: Be precise with the year (1889) and the alloy composition (90% platinum, 10% iridium) for high-scoring answers.

Question 5S. Name three systems of unit and their fundamental units.
Answer: Three systems of unit and their fundamental units:
1. C.G.S. system (or French system): In this system, the unit of length is centimeter (cm), unit of mass is gramme (g) and unit of time is second (s).
2. F.P.S. system (or British system): In this system, the unit of length is foot (ft), unit of mass is pound (lb) and unit of time is second (s).
3. M.K.S. system (or metric system): In this system, the unit of length is metre (m), unit of mass is kilogramme (kg) and unit of time is second (s).
In simple words: Different countries used to have different "languages" for measuring. For example, the British used feet and pounds, while the French used centimeters and grams.

📝 Teacher's Note: Help students memorize the names by looking at the initials: C-G-S stands for Centimeter-Gram-Second.

🎯 Exam Tip: Note that "Second" is the fundamental unit of time in all three systems.

Question 6S. What is a fundamental unit?
Answer: A fundamental (or basic) unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit.
In simple words: A fundamental unit is like a "primary color" in measurement; it doesn't need any other units to make it.

📝 Teacher's Note: Contrast fundamental units with derived units to make the concept clearer.

🎯 Exam Tip: The definition must include the word "independent."

Question 7S. List the fundamental quantities, units and symbols in the S.I. system.
Answer: Fundamental quantities, units and symbols in S.I. system are

In simple words: This is the official list of the 7 basic building blocks (and 2 extra ones) for all measurements in science.

📝 Teacher's Note: Remind students that Radian and Steradian are called "Supplementary units."

🎯 Exam Tip: Pay attention to capitalization—Kelvin (K) and Ampere (A) are capitalized because they are named after scientists, while metre (m) is not.

Question 8S. What are derived units? Give an example.
Answer: The units of quantities other than those measured in fundamental units can be obtained in terms of the fundamental units, and thus the units so obtained are called derived units.
Example:
Speed = Distance/time
Hence, the unit of speed = fundamental unit of distance/fundamental unit of time
Or, the unit of speed = metre/second or \( ms^{-1} \).
As the unit of speed is derived from the fundamental units of distance and time, it is a derived unit.
In simple words: Derived units are "combo units." For example, speed is made by mixing distance and time together.

📝 Teacher's Note: Use the area of a square (\( m \times m = m^2 \)) as another simple example of a derived unit.

🎯 Exam Tip: When giving an example, always show the formula to demonstrate how it is derived from base units.

Question 9S. How was the metre originally defined in 1889?
Answer: A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy with 90% platinum and 10% iridium) rod kept at 0° C in the International Bureau of Weights and Measures at serves near Paris.
In simple words: It was the distance between two scratches on a very special metal stick kept in a cold room in France.

📝 Teacher's Note: Explain that 0° C is used to prevent the metal from expanding or contracting, which would change the distance.

🎯 Exam Tip: "Platinum-iridium" and "0° C" are the two key technical details to include.

Question 10S. Name two units of length bigger than a metre and state their values.
Answer: Astronomical unit (A.U.) and kilometer (km) are units of length which are bigger than a metre.
1 km = 1000 m
1 A.U. = \( 1.496 \times 10^{11} \text{ m} \)
In simple words: Kilometers are used for trips between cities, and Astronomical Units are used for the massive distances between the Earth and the Sun.

📝 Teacher's Note: One A.U. is roughly the average distance between the Earth and the Sun.

🎯 Exam Tip: Remember the power of 11 for the Astronomical Unit value.

Question 11S. Name two units of length smaller than a metre and state their values.
Answer: Centimeter (cm) and millimeter (mm) are units of length smaller than a metre.
\( 1 \text{ cm} = 10^{-2} \text{ m} \)
\( 1 \text{ mm} = 10^{-3} \text{ m} \)
In simple words: These are the tiny lines you see on your school ruler.

📝 Teacher's Note: Use a standard ruler to show the difference between 1 cm and 1 mm.

🎯 Exam Tip: Use scientific notation (\( 10^{-2} \)) to show you are comfortable with standard physics formatting.

Question 12S. Convert 1 nm into Angstrom (\( \text{\AA} \)).
Answer: 1 nm = 10 \( \text{\AA} \)
In simple words: One nanometer is made up of 10 even smaller units called Angstroms.

📝 Teacher's Note: Explain that these units are used for measuring things like the size of atoms or the wavelength of light.

🎯 Exam Tip: This is a very common 1-mark fill-in-the-blank question.

Question 13S. Name three convenient units of length and their relation with S.I. units.
Answer: Three convenient units of length and their relation with the S.I. unit of length:
1. 1 Angstrom (\( \text{\AA} \)) = \( 10^{-10} \text{ m} \)
2. 1 kilometre (km) = \( 10^3 \text{ m} \)
3. 1 light year (ly) = \( 9.46 \times 10^{15} \text{ m} \)
In simple words: We use Angstroms for atoms, kilometers for roads, and light years for the stars.

📝 Teacher's Note: Point out that despite having "year" in the name, a "light year" is a unit of distance, not time.

🎯 Exam Tip: Light year is a very popular exam topic. It is the distance light travels in one year.

Question 14S. What is the S.I. unit of mass? How was it defined in 1889?
Answer: S.I. unit of mass is ‘kilogramme’.
In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at the International Bureau of Weights and Measures at serves near Paris.
In simple words: A kilogram is based on the weight of one specific metal cylinder kept safely in France.

📝 Teacher's Note: Mention that mass is a measure of the "amount of matter" in an object.

🎯 Exam Tip: Spell it as "kilogramme" in formal exams if that is what your textbook uses.

Question 15S. State the values for the following:
(a) 1 light year
(b) 1 m in \(\text{\AA}\)
(c) 1 m in micron
(d) 1 micron in \(\text{\AA}\)
(e) 1 fermi in m
Answer:
(a) 1 light year = \( 9.46 \times 10^{15} \text{ m} \)
(b) 1 m = \( 10^{10} \text{ \AA} \)
(c) 1 m = \( 10^6 \text{ \mu} \) (micron)
(d) 1 micron = \( 10^4 \text{ \AA} \)
(e) 1 fermi = \( 10^{-15} \text{ m} \)
In simple words: These are the "conversion factors" used to switch between very big and very tiny measurements.

📝 Teacher's Note: Explain that "fermi" is used in nuclear physics to measure the size of a nucleus.

🎯 Exam Tip: Sub-part (b) is a common trap—remember it's \( 10^{10} \), not \( 10^{-10} \), because a metre is much larger than an Angstrom.

Question 16S. Name two units of mass smaller than a kilogramme.
Answer: The units ‘gramme’ (g) and ‘milligramme’ (mg) are two units of mass smaller than ‘kilogramme’.
1 g = \( 10^{-3} \text{ kg} \)
1 mg = \( 10^{-6} \text{ kg} \)
In simple words: These are the units used for kitchen ingredients (grams) or tiny pills (milligrams).

📝 Teacher's Note: Remind students that "milli" always means 1/1000th.

🎯 Exam Tip: Always show the conversion to the base SI unit (kg) in your answer.

Question 17S. Name two units of mass bigger than a kilogramme.
Answer: The units ‘quintal’ and ‘metric tonne’ are two units of mass bigger than ‘kilogramme’.
1 quintal = 100 kg
1 metric tonne = 1000 kg
In simple words: These are used for very heavy things, like bags of rice (quintals) or whole cars (tonnes).

📝 Teacher's Note: Use the example of truck loads to help students grasp the scale of a metric tonne.

🎯 Exam Tip: Remember: 10 Quintals = 1 Metric Tonne.

Question 18S. State the S.I. values for:
(a) 1 g
(b) 1 mg
(c) 1 quintal
(d) 1 a.m.u. (or u)
Answer:
(a) 1 g = \( 10^{-3} \text{ kg} \)
(b) 1 mg = \( 10^{-6} \text{ kg} \)
(c) 1 quintal = 100 kg
(d) 1 a.m.u (or u) = \( 1.66 \times 10^{-27} \text{ kg} \)
In simple words: A.M.U. is an incredibly small unit used only for the weight of single atoms.

📝 Teacher's Note: a.m.u. stands for Atomic Mass Unit. It's the standard for chemistry and atomic physics.

🎯 Exam Tip: The a.m.u. conversion is frequently used in modern physics numericals.

Question 19S. What is the S.I. unit of time? Define a second.
Answer: The S.I. unit of time is second (s).
A second is defined as 1/86400th part of a mean solar day, i.e.
\[ 1s = \frac{1}{86400} \times \text{one mean solar day} \]
In simple words: One second is what you get if you take a full day and divide it into 86,400 equal slices.

📝 Teacher's Note: Show the math: \( 24 \text{ hours} \times 60 \text{ minutes} \times 60 \text{ seconds} = 86,400 \text{ seconds} \) in a day.

🎯 Exam Tip: Always include the fraction "1/86400" in the definition.

Question 20S. Name two units of time bigger than a second.
Answer: The units ‘minute’ (min) and ‘year’ (yr) are two units of time bigger than second(s).
1 min = 60 s
1 yr = \( 3.1536 \times 10^7 \text{ s} \)
In simple words: Everyone knows minutes and years, but in science, we often need to turn years into a giant number of seconds.

📝 Teacher's Note: Explain that scientific notation \( 10^7 \) makes it easier to write 31 million seconds.

🎯 Exam Tip: If you forget the year conversion, calculate it: \( 365 \times 24 \times 3600 \).

Question 21S. What is a leap year?
Answer: A leap year is the year in which the month of February has 29 days.
In simple words: Every four years, we add one extra day to February to keep our calendar aligned with Earth's orbit around the sun.

📝 Teacher's Note: Mention that a leap year has 366 days instead of 365.

🎯 Exam Tip: This is a simple logic question often used as a warm-up.

Question 22S. Is the value of units affected by changing place?
Answer: No, the value of units should not change with space and time. (The statement "The value of units should not change with space and time" is a requirement for a good unit).
In simple words: A kilogram has to be exactly the same weight on top of Mt. Everest as it is at the bottom of the sea.

📝 Teacher's Note: This is a fundamental property of standard units. If they changed, international trade and science would be impossible.

🎯 Exam Tip: Simply answering "Yes" or "No" might not be enough; explain that consistency is a "requirement" for units.

Question 23S. What is a lunar month?
Answer: One lunar month is the time in which the moon completes one revolution around the earth. A lunar month is made of nearly 4 weeks.
In simple words: It is the time from one full moon to the next full moon.

📝 Teacher's Note: A lunar month is approximately 27.3 days.

🎯 Exam Tip: Define it specifically as the time for "one revolution of the moon around the earth."

Question 24S. Express the following in seconds:
(a) 1 nanosecond
(b) 1 \(\mu\)s
(c) 1 mean solar day
(d) 1 year
Answer:
(a) 1 nanosecond = \( 10^{-9} \text{ s} \)
(b) 1 \( \mu\text{s} \) = \( 10^{-6} \text{ s} \)
(c) 1 mean solar day = 86400 s
(d) 1 year = \( 3.15 \times 10^7 \text{ s} \)
In simple words: These are standard conversions to the base unit of time.

📝 Teacher's Note: Help students with the prefixes: nano = billionth, micro = millionth.

🎯 Exam Tip: Mean solar day (86400) is a high-frequency value in numerical problems.

Question 25S. Which physical quantities do these units measure?
(a) kg
(b) light year
(c) min
(d) \(\text{\AA}\)
Answer: (a) Mass (b) Distance (or length) (c) Time (d) Length
In simple words: Units are just labels that tell us what kind of measurement we are looking at.

📝 Teacher's Note: Remind students again that light year is distance, not time.

🎯 Exam Tip: This is a standard identification task for lower secondary levels.

Question 26S. Give the S.I. symbols for the units of:
(a) Speed
(b) Force
(c) Energy
(d) Pressure
Answer: (a) \( ms^{-1} \) (b) \( kg \cdot ms^{-2} \) (c) \( kg \cdot m^2s^{-2} \) (d) \( kg \cdot m^{-1}s^{-2} \)
In simple words: These look complicated, but they are just fundamental units like mass, length, and time mixed together.

📝 Teacher's Note: These are the units in terms of "Base S.I. units." For example, 1 Newton (Force) is actually \( 1 \text{ kg} \cdot \text{m/s}^2 \).

🎯 Exam Tip: Using negative powers (like \( s^{-1} \)) is the correct way to write "per second" in science papers.

Question 27S. Express the following in terms of base S.I. units:
(a) Newton (Force)
(b) Watt (Power)
(c) Joule (Energy)
(d) Pascal (Pressure)
Answer:
(a) \( kg \cdot ms^{-2} \)
(b) \( kg \cdot m^2s^{-3} \)
(c) \( kg \cdot m^2s^{-2} \)
(d) \( kg \cdot m^{-1}s^{-2} \)
In simple words: Every complex unit in science can be broken down into kilograms, metres, and seconds.

📝 Teacher's Note: Use the formulas \( F = ma \), \( W = Fd \), and \( P = W/t \) to show students where these units come from.

🎯 Exam Tip: This is a high-level question often found in competitive exams.

Question 28S. Identify the derived quantities from the list:
(a) Area (b) Force (c) Energy (d) Pressure (f) Power
Answer: All of the options provided are derived quantities.
(a) Area (b) Force (c) Energy (d) Pressure (f) Power
In simple words: None of these are "base" units; they are all made by combining the 7 fundamental building blocks.

📝 Teacher's Note: Ask students to name the 7 fundamental quantities again to prove these five are not on that list.

🎯 Exam Tip: This list identifies most of the derived quantities covered in the Class 9 curriculum.

Exercise 1(A) Multiple Choice Questions

Question 1M. Which of the following is an S.I. unit of time?
(a) hour
(b) minute
(c) second
(d) day
Answer: (c) second
In simple words: While we use hours and minutes in daily life, scientists have chosen the "second" as the official standard.

📝 Teacher's Note: Remind students that S.I. stands for System International, the global standard for science.

🎯 Exam Tip: Don't confuse "common unit" with "S.I. unit."

Question 2M. Which of the following is not an S.I. unit?
(a) metre
(b) kilogramme
(c) second
(d) litre
Answer: (d) litre
In simple words: Metre, kilogram, and second are the official "base" units. Litre is used for volume, but the official S.I. unit for volume is the cubic metre.

📝 Teacher's Note: Explain that Litre is a "derived" or "accepted" metric unit, but not a fundamental S.I. unit.

🎯 Exam Tip: This is a classic trick question. Always look for "Litre" or "Celsius" as non-S.I. units.

Question 3M. A unit of time which is not a constant quantity is:
(a) leap year
(b) year
(c) lunar month
(d) second
Answer: (a) leap year
In simple words: Most years are 365 days, but a leap year has 366. Because it changes, it's not a "constant" unit.

📝 Teacher's Note: This highlights rule #3 from Question 3S: a good unit should not change with time.

🎯 Exam Tip: Think about the definition of "constant" (stays the same) to solve this.

Question 4M. \( 10^{-10} \text{ m} \) is equal to:
(a) 1 nm
(b) 1 \(\mu\)m
(c) 1 \(\text{\AA}\)
(d) 1 mm
Answer: (c) 1 \(\text{\AA}\)
In simple words: Ten-billionths of a metre is called one Angstrom.

📝 Teacher's Note: This is a memory-based conversion question.

🎯 Exam Tip: Memorize the powers: nano (9), micro (6), Angstrom (10).

Question 5M. The unit micron is the unit of:
(a) mass
(b) time
(c) length
(d) force
Answer: (c) length
In simple words: "Micron" is just a short name for a micrometer, which measures very small distances.

📝 Teacher's Note: Microns are often used in biology to measure the size of cells.

🎯 Exam Tip: Micron always refers to length.

Numerical Problems

Question 1N. Wavelength of light of particular colour is 5800 \(\text{\AA}\). (a) Express it in (i) nm and (ii) m. (b) What is the order of its magnitude in metre?
Answer:
(a) (i) 1 \( \text{\AA} \) = \( 10^{-1} \text{ nm} \)
\( 5800 \text{ \AA} = 5800 \times 10^{-1} \text{ nm} = 580 \text{ nm} \)

(ii) 1 \( \text{\AA} \) = \( 10^{-10} \text{ m} \)
\( 5800 \text{ \AA} = 5800 \times 10^{-10} \text{ m} = 5.8 \times 10^{-7} \text{ m} \)

(b) The order of its magnitude in metre is \( 10^{-6} \text{ m} \) because the numerical value of 5.8 is more than 3.2.
In simple words: We converted the tiny light waves from Angstroms to nanometers and meters. Because 5.8 is quite large for its decimal place, we round the order of magnitude up to the next power of ten.

📝 Teacher's Note: Rule for Order of Magnitude: If the number is greater than 3.2, add 1 to the power of ten.

🎯 Exam Tip: Always show the step-by-step conversion of powers of ten to get full marks.

Question 2N. The size of a bacteria is 1 \(\mu\). How many bacteria will be in a 1 m length?
Answer:
Size of a bacteria = 1 \( \mu \)
Since 1 \( \mu = 10^{-6} \text{ m} \)
Number of bacteria = Total length / size of one bacteria
\[ \text{Number} = \frac{1 \text{ m}}{10^{-6} \text{ m}} = 10^6 \]
In simple words: Because bacteria are so tiny (one-millionth of a meter), you can fit exactly one million of them in a line that is only one meter long.

📝 Teacher's Note: This is a classic division problem involving scientific notation. Remind students that \( 1 / 10^{-6} \) becomes \( 10^6 \).

🎯 Exam Tip: Don't forget the final answer is a count, so it has no units.

Question 3N. The distance of a galaxy is \( 5.6 \times 10^{25} \text{ m} \). If the speed of light is \( 3 \times 10^8 \text{ m/s} \), (a) how much time does light take to travel? (b) What is the order of magnitude of time in seconds?
Answer:
(a) Time taken by light = Distance travelled / speed of light
\[ t = \frac{5.6 \times 10^{25}}{3 \times 10^8} \text{ s} \]
\( t = 1.87 \times 10^{17} \text{ s} \)

(b) Order of magnitude = \( 10^{17} \text{ s} \)
(This is because the numerical value of 1.87 is less than 3.2, so the power of ten stays as 17).
In simple words: The galaxy is so far away that even fast light takes over a hundred quadrillion seconds to reach us.

📝 Teacher's Note: Review the "3.2 rule" for order of magnitude. Since 1.87 is small, we keep the original exponent.

🎯 Exam Tip: Keep units throughout your calculation to avoid confusion (\( m / (m/s) = s \)).

Question 4N. The wavelength of a light is 589 nm. (a) Express it in metres. (b) Find the order of magnitude in metre.
Answer:
(a) Wavelength of light = 589 nm
\( = 589 \times 10^{-9} \text{ m} \)
\( = 5.89 \times 10^{-7} \text{ m} \)

(b) Order of magnitude = \( 10^1 \times 10^{-7} \text{ m} = 10^{-6} \text{ m} \)
(This is because the numerical value of 5.89 is more than the numerical value 3.2).
In simple words: We converted nanometers to standard meters. Since 5.89 is a "big" number for its power, we say the order of magnitude is \( 10^{-6} \).

📝 Teacher's Note: This reinforces the order of magnitude rule for numbers between 1 and 10.

🎯 Exam Tip: Always convert to standard scientific notation (one digit before the decimal) before deciding on the order of magnitude.

Question 5N. The mass of an oxygen atom is 16.00 u. Express this in kg and find the order of magnitude.
Answer:
Mass of an oxygen atom = 16.00 u
Now, 1 u = \( 1.66 \times 10^{-27} \text{ kg} \)
Hence, mass of oxygen in kg = \( 16 \times 1.66 \times 10^{-27} \text{ kg} \)
\( = 26.56 \times 10^{-27} \text{ kg} \)
Standard form = \( 2.656 \times 10^{-26} \text{ kg} \)
Because the numerical value of 2.656 is less than the numerical value of 3.2, the order of magnitude of mass of oxygen in kg
\( = 10^{-26} \text{ kg} \)
In simple words: One oxygen atom is incredibly light—it's roughly 26 zeros after a decimal point!

📝 Teacher's Note: The OCR has a slight logic error in the step \( 10^1 \times 10^{-27} \); the standard form calculation above is the correct way to reach \( 10^{-26} \).

🎯 Exam Tip: Don't forget that 16.00 u means multiplying by the value of 1 atomic mass unit.

Question 6N. Light takes 8 minutes to reach from the Sun to the Earth. If speed of light is \( 3 \times 10^8 \text{ m/s} \), find the distance in km and its order of magnitude.
Answer:
Time taken by light to reach from the Sun to the Earth = 8 min = 480 s.
Speed of light = \( 3 \times 10^8 \text{ m/s} \)
Distance from the Sun to the Earth = Speed \( \times \) time
\( = 3 \times 10^8 \times 480 \text{ m} \)
\( = 1440 \times 10^8 \text{ m} \)
\( = 1440 \times 10^8 \times 10^{-3} \text{ km} \)
\( = 1440 \times 10^5 \text{ km} \)
\( = 1.44 \times 10^8 \text{ km} \)

Because the numerical value of 1.44 is less than the numerical value of 3.2, the order of magnitude of distance from the Sun to the Earth in km = \( 10^8 \text{ km} \).
In simple words: The Sun is about 144 million kilometers away. We calculate this by seeing how far light travels in the 8 minutes it takes to reach us.

📝 Teacher's Note: This is a very common practical problem. Ensure students convert minutes to seconds first.

🎯 Exam Tip: The final answer asks for KM, so ensure you divide your final meters by 1000.

Question 7N. Explain the statement: "the distance of a star from the Earth is 8.33 light minutes." Express it in metres.
Answer: The statement ‘the distance of a star from the Earth is 8.33 light minutes’ means that the light from the star takes 8.33 minutes to reach Earth.
The speed of light is \( 3 \times 10^8 \text{ m/s} \).
Hence, the distance covered by it in one second is \( d = 3 \times 10^8 \text{ m} \).
In one minute there are 60 seconds.
Therefore, the distance of 8.33 light minutes is:
\( d = 3 \times 10^8 \times 60 \times 8.33 \text{ m} \)
\( d \approx 1.5 \times 10^{11} \text{ m} \)
\( d = 150,000,000,000 \text{ m} \)
In simple words: This means if that star exploded right now, we wouldn't know for over 8 minutes because the news (light) has to travel across a 150-billion-meter gap.

📝 Teacher's Note: 8.33 light minutes is the distance of our Sun. Use this as a fun fact for the class.

🎯 Exam Tip: "Light minute" is defined by the speed of light. Distance = \( 3 \times 10^8 \times \text{time in seconds} \).

Exercise 1(B)

Question 1S. What do you understand by the least count of an instrument? Give an example.
Answer: The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.
In simple words: Least count is the "tiniest step" your tool can measure. On a normal ruler, it's 1 mm.

📝 Teacher's Note: The smaller the least count, the more precise the instrument is.

🎯 Exam Tip: Always provide an example when defining least count to show practical understanding.

Question 2S. A scale has 100 divisions in 1 m length. Calculate its accuracy. How can we increase the accuracy of this scale?
Answer: Total length of the scale = 1 m = 100 cm
No. of divisions = 100
Length of each division = Total length/total no. of divisions
= 100 cm / 100 = 1 cm
Thus, this scale can measure with an accuracy of 1 cm.
To increase the accuracy, the total number of divisions on the scale must be increased.
In simple words: This ruler only has marks every centimeter, so it's not very accurate. To fix it, you'd need to add more marks (like millimeters) in between.

📝 Teacher's Note: Accuracy is directly linked to the number of divisions within a fixed length.

🎯 Exam Tip: Accuracy and Least Count are terms often used interchangeably at this level.

Question 3S. Why can't the length of a pencil be expressed as 2.60 cm using a normal metre rule?
Answer: The least count of a metre rule is 0.1 cm (or 1 mm). The length cannot be expressed as 2.60 cm because a metre scale measures length correctly only up to one decimal place of a centimeter. The digit '0' in the second decimal place implies a level of precision that the tool does not possess.
In simple words: You can't say it's exactly 2.60 if your ruler only has marks for 2.6 and 2.7. That zero is just a guess!

📝 Teacher's Note: Explain the concept of "Significant Figures." We can only report numbers we can actually see on the scale.

🎯 Exam Tip: Use the term "limit of resolution" or "least count" to explain why extra decimals are incorrect.

Question 4S. How is the least count of vernier callipers calculated?
Answer: The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.
Let n divisions on vernier callipers be of length equal to that of (n – 1) divisions on the main scale and the value of 1 main scale division be x. Then,
Value of n divisions on vernier = (n – 1) x
Alternatively, value of 1 division on vernier = \( \frac{(n - 1) x}{n} \)
Hence,
\[ \text{Least count} = x - \frac{(n - 1) x}{n} = \frac{x}{n} \]
L.C. = (Value of one main scale division) / (Total no. of divisions on vernier callipers)
Value of one main scale division = 1 mm
Total no. of divisions on vernier = 10
Therefore, L.C. = 1 mm / 10 = 0.1 mm = 0.01 cm
In simple words: We find the size of a gap on the main ruler and divide it by how many tiny lines are on the sliding part.

📝 Teacher's Note: The "Vernier Principle" relies on the slightly different spacing between the two scales.

🎯 Exam Tip: Memorize the simplified formula \( \text{L.C.} = x/n \) where x is 1 MSD and n is the number of VSD.

Question 5S. Define Vernier constant.
Answer: Vernier constant is equal to the difference between the values of one main scale division and one vernier scale division. It is the same as the least count of vernier callipers.
In simple words: It is just a fancy name for the smallest measurement a vernier calliper can take.

📝 Teacher's Note: "Vernier Constant" and "Least Count" are identical terms for this instrument.

🎯 Exam Tip: If the question asks for "Vernier constant," don't get confused—just calculate the least count.

Question 6S. When is a vernier calliper said to be free from zero error?
Answer: A vernier callipers is said to be free from zero error, if the zero mark of the vernier scale coincides with the zero mark of the main scale when the jaws are closed.
In simple words: It's like a scale that says "0" when nothing is on it. The two zeros must line up perfectly.

📝 Teacher's Note: Show the students a physical vernier calliper and have them close the jaws to check for zero error.

🎯 Exam Tip: The keyword is "coincides."

Question 7S. What is zero error? Explain its types.
Answer: Due to mechanical errors, sometimes the zero mark of the vernier scale does not coincide with the zero mark of the main scale; the vernier callipers is said to have zero error. It is determined by measuring the distance between the zero mark of the main scale and the zero mark of the vernier scale. The zero error is of two kinds:
1. Positive zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the right of the zero mark of the main scale, the error is said to be positive.
2. Negative zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the left of the zero mark of the main scale, then the error is said to be negative.
In simple words: Positive error is like a scale that starts at 1 gram; negative error is like one that starts at -1 gram. We have to fix this to get the true weight.

📝 Teacher's Note: Use the "number line" logic. Right of zero is positive, left of zero is negative.

🎯 Exam Tip: Always specify the position of the "Vernier Zero" relative to the "Main Scale Zero" when describing these errors.

Question 8S. How is the zero error calculated for both positive and negative cases?
Answer:
1. To find Positive zero error: We note the division of the vernier scale which coincides with any division of the main scale. The number of this vernier division when multiplied by the least count gives the zero error.
Example: if the 6th division coincides and L.C. is 0.01 cm, Zero error = \( +6 \times 0.01 \text{ cm} = +0.06 \text{ cm} \).

2. To find Negative zero error: We note the division of the vernier scale coinciding with any division of the main scale. The number of this vernier division is subtracted from the total number of divisions on the vernier scale and then the difference is multiplied by the least count.
Example: if the 6th division coincides, total divisions are 10, and L.C. is 0.01 cm, Zero error = \( -(10 - 6) \times 0.01 \text{ cm} = -0.04 \text{ cm} \).
In simple words: For positive, just count forward. For negative, you count how many steps you are "missing" from the end.

📝 Teacher's Note: The negative error calculation is a common point of confusion. Remind students to always subtract the coinciding division from the total (usually 10).

🎯 Exam Tip: Always include the (+) or (-) sign with the zero error.

Question 9S. Label the parts of a vernier calliper and state their functions.
Answer:
Main parts and their functions:

• Main scale: It is used to measure length correct up to 1 mm.
• Vernier scale: It helps to measure length correct up to 0.1 mm.
• Outside jaws: It helps to measure length of a rod, diameter of a sphere, external diameter of a hollow cylinder.
• Inside jaws: It helps to measure the internal diameter of a hollow cylinder or pipe.
• Strip: It helps to measure the depth of a beaker or a bottle.

In simple words: The big teeth measure the outside, the little teeth measure the inside, and the tail measures the depth.

📝 Teacher's Note: The "Strip" is often called the "Depth gauge."

🎯 Exam Tip: You may be asked to name which specific part is used for a specific job, like measuring the "internal diameter" (Inside Jaws).

Question 10S. State three uses of vernier callipers.
Answer: Three uses of vernier callipers are
1. Measuring the internal diameter of a tube or a cylinder.
2. Measuring the length of an object.
3. Measuring the depth of a beaker or a bottle.
In simple words: It is a 3-in-1 tool for measuring outside, inside, and depth very precisely.

📝 Teacher's Note: These three uses correspond to the three measuring parts (Jaws and Strip).

🎯 Exam Tip: This is a standard descriptive question.

Question 11S. Explain the scale of a vernier calliper. What is its least count?
Answer: Two scales of vernier calipers are:
1. Main scale
2. Vernier scale
The main scale is graduated to read up to 1 mm and on vernier scale, the length of 10 divisions is equal to the length of 9 divisions on the main scale.
Value of 1 division on the main scale = 1 mm
Total no. of divisions on the vernier scale = 10
Thus, L.C. = 1 mm / 10 = 0.1 mm = 0.01 cm.
Hence, a vernier callipers can measure length correct up to 0.01 cm.
In simple words: The main ruler gives you the millimeters, and the sliding scale lets you split those millimeters into 10 smaller bits.

📝 Teacher's Note: This spacing (10 VSD = 9 MSD) is the most standard configuration for school laboratories.

🎯 Exam Tip: Always specify that the vernier scale allows measurement "up to two decimal places of a centimeter."

Question 12S. How do you take a final reading using vernier callipers?
Answer: To measure length, the object is placed between the fixed end and the vernier scale. We note the position of the zero mark of the vernier scale on the main scale. If it is beyond 1.2 cm, the main scale reading is 1.2 cm. Then, we find which division of the vernier scale (\( p \)) coincides perfectly with any main scale line. The final reading is:
\[ \text{Total reading} = \text{main scale reading} + (\text{vernier scale reading} \times \text{L.C.}) \]
Example: if MSR = 1.2 cm and 4th VSD coincides with L.C. 0.01 cm:
Total reading = \( 1.2 + (4 \times 0.01) = 1.24 \text{ cm} \).
In simple words: You read the big number first, then find the "matching line" on the slider to get the tiny extra decimal.

📝 Teacher's Note: Use the "integer and decimal" analogy. MSR provides the integer/first decimal, and the Vernier scale provides the second decimal.

🎯 Exam Tip: The formula \( \text{Reading} = \text{MSR} + (\text{VSD} \times \text{LC}) \) is the most important one to remember.

Question 13S. Which part of the vernier callipers is used for the following?
(a) External diameter (b) Internal diameter (c) Depth
Answer:
(a) Outside jaws
(b) Inside jaws
(c) Strip
In simple words: Bottom teeth for outside, top teeth for inside, tail for how deep.

📝 Teacher's Note: Quick-fire questions like this help solidify the functions of the instrument parts.

🎯 Exam Tip: "Outer jaws" and "Outside jaws" are the same thing.

Question 14S. Define Pitch and Least Count of a screw gauge.
Answer:
(i) Pitch: The pitch of a screw gauge is the distance moved by the screw along its axis in one complete rotation of the thimble.
(ii) Least count (L.C.) of a screw gauge: The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.
\[ \text{L.C.} = \frac{\text{Pitch of the screw gauge}}{\text{Total no. of divisions on its circular scale}} \]
Example: If pitch = 1 mm and divisions = 100, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm.
In simple words: Pitch is how much it opens per full turn. Least count is how much it opens per tiny "click" or mark on the knob.

📝 Teacher's Note: A screw gauge is much more precise than a vernier calliper. It can measure up to a thousandth of a centimeter.

🎯 Exam Tip: Ensure you include the units in your examples (0.001 cm is the standard for a good screw gauge).

Question 15S. How can the least count of a screw gauge be increased?
Answer: The least count of a screw gauge can be increased by:
1. Decreasing the pitch of the screw.
2. Increasing the total number of divisions on the circular scale.
In simple words: To make it more precise, you can either use a screw with finer threads or put more tiny marks on the knob.

📝 Teacher's Note: Increasing the least count technically means making the instrument "more precise" (the value of the least count actually gets smaller).

🎯 Exam Tip: Memorize these two ways as they are frequently asked in "factors affecting precision" questions.

Question 16S. Label the parts of a screw gauge and state their functions.
Answer:
Main parts and their functions:

• Ratchet: It advances the screw by turning it until the object is gently held between the stud and spindle. It prevents over-tightening.
• Sleeve: It marks the main scale and base line.
• Thimble: It marks the circular scale.
• Main scale: It helps to read the length correct up to 1 mm.
• Circular scale: It helps to read length correct up to 0.01 mm.

In simple words: The thimble is the big knob you turn, and the ratchet is the little end bit that clicks when it's tight enough.

📝 Teacher's Note: The ratchet is the most important safety feature. If you don't use it, you might crush the object or damage the screw.

🎯 Exam Tip: If asked about preventing "excessive pressure," the answer is always the Ratchet.

Question 17S. What is the use of a screw gauge?
Answer: A screw gauge is used for measuring the diameter of circular objects, mostly thin wires, with a high accuracy of 0.001 cm.
In simple words: It is the best tool for measuring extremely thin things, like a single human hair or a thin copper wire.

📝 Teacher's Note: Compare this to the 0.01 cm accuracy of the vernier calliper.

🎯 Exam Tip: Mention the accuracy "0.001 cm" to provide a complete answer.

Question 18S. Explain how the ratchet works.
Answer: Ratchet helps to advance the screw by turning it until the object is gently held between the stud and spindle of the screw. Once the object is held, if the user continues to turn the ratchet, it simply slips and produces a clicking sound without moving the screw further.
In simple words: It's like a safety cap on a medicine bottle; it clicks when it's tight so you don't break anything.

📝 Teacher's Note: This clicking sound is a signal to the student to stop turning and take the reading.

🎯 Exam Tip: The ratchet ensures "uniform pressure" is applied every time.

Question 19S. Define zero error in a screw gauge. How is it corrected?
Answer: Due to mechanical errors, sometimes when the anvil and spindle end are brought in contact, the zero mark of the circular scale does not coincide with the base line of the main scale. It is either above or below the base line, in which case the screw gauge is said to have a zero error. It can be both positive and negative. It is corrected by subtracting the zero error (with its sign) from the observed reading.
\[ \text{Correct reading} = \text{Observed reading} - \text{zero error (with sign)} \]
In simple words: If the tool doesn't start at zero, you just subtract that "starting mistake" from your final answer.

📝 Teacher's Note: Explain that "subtracting a negative number" means you actually end up adding the value. This is a common mathematical hurdle for students.

🎯 Exam Tip: Always remember the formula: Correct = Observed - Error.

Question 20S. Draw a diagram of a positive zero error of +0.007 cm in a screw gauge.
Answer:
In simple words: In a positive error, the "0" on the knob hasn't quite reached the center line yet when the tool is closed.

📝 Teacher's Note: Positive error occurs when the zero of the circular scale is below the reference line.

🎯 Exam Tip: In diagrams, look at where the circular zero sits relative to the horizontal reference line.

Question 21S. What is backlash error? State its reason and how to avoid it.
Answer: Backlash error: If by reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction immediately but remains stationary for a part of rotation; it is called backlash error.
Reason: It happens due to wear and tear of the screw threads.
To avoid: While taking measurements, the screw should be rotated in one direction only. If the direction of rotation needs to be changed, it should be stopped for a while and then rotated in the reverse direction.
In simple words: It's like a loose steering wheel that you have to turn a little bit before the car actually starts turning. It happens because the parts are worn out.

📝 Teacher's Note: This is a mechanical limitation of all screw-based tools. High-quality instruments have less backlash.

🎯 Exam Tip: The main way to avoid this error is to "rotate in one direction only" when reaching the measurement point.

Question 22S. How do you measure the diameter of a wire using a screw gauge?
Answer: The wire is placed between the anvil and spindle end. The thimble is rotated until the wire is held gently. The ratchet is used to apply uniform pressure. The reading is taken as follows:
Main scale reading (MSR) = 2.5 mm
Circular scale reading (CSR) = 46th division coincides with the base line.
L.C. of screw gauge = 0.01 mm
Total reading = MSR + (CSR \( \times \) L.C.)
Total reading = \( 2.5 + (46 \times 0.01) = 2.5 + 0.46 = 2.96 \text{ mm} \).
In simple words: You look at the ruler to get the big number, then look at the knob to find the tiny extra bits, then add them together.

📝 Teacher's Note: Encourage students to count the divisions on the circular scale carefully. If the 46th line matches the center line, that's your CSR.

🎯 Exam Tip: Always state your units. If MSR is in mm, your total will be in mm unless you convert it.

Question 23S. Which instrument would you use to measure the following?
(a) Diameter of a wire
(b) Thickness of a glass plate
(c) Internal diameter of a test tube
(d) Radius of a thin wire
Answer:
(a) Screw gauge
(b) Screw gauge
(c) Vernier callipers
(d) Screw gauge
In simple words: Use the screw gauge for anything very thin and round, and vernier callipers for holes and tubes.

📝 Teacher's Note: Remind students that screw gauges are superior for very fine objects like wires.

🎯 Exam Tip: Internal diameters are the specialty of Vernier callipers.

Question 24S. What is the main advantage of a screw gauge over a vernier calliper?
Answer: A screw gauge measures length to a much higher level of accuracy (0.001 cm) compared to a vernier calliper (0.01 cm).
In simple words: A screw gauge is ten times more precise than a vernier calliper.

📝 Teacher's Note: Accuracy is inversely proportional to the least count.

🎯 Exam Tip: Use the term "high accuracy" or "greater precision."

Question 25S. Arrange the following in increasing order of their accuracy: (a) Vernier callipers (b) Metre scale (c) Screw gauge.
Answer: (b) Metre scale < (a) Vernier callipers < (c) Screw gauge.
In simple words: A normal ruler is the least accurate, then the vernier slider, and the screw gauge is the champion of precision.

📝 Teacher's Note: Increasing order means starting from the least accurate (Metre scale).

🎯 Exam Tip: You can also list them by least count: 0.1 cm, 0.01 cm, 0.001 cm.

Numerical Problems (Set 2)

Question 1N. A stopwatch has a range of 5s and 10 divisions. Calculate its least count.
Answer: Range of the stop watch = 5s
Total number of divisions = 10
L.C. = 5/10 = 0.5 s
In simple words: If a five-second gap is split into ten steps, each step is half a second.

📝 Teacher's Note: This is a very simple division problem to practice the L.C. formula: Range/Divisions.

🎯 Exam Tip: Always include the unit 's' for seconds.

Question 2N. 10 vernier divisions = 9 m.s.d. and 1 m.s.d. = 1 mm. Find the least count.
Answer: Value of 1 m.s.d. = 1 mm
L.C. = Value of 1 m.s.d. / number of divisions on vernier scale
= 1 mm / 10 = 0.1 mm or 0.01 cm
In simple words: This is the most common vernier calliper in the world. Its tiny step is one-tenth of a millimeter.

📝 Teacher's Note: This configuration (n VSD = n-1 MSD) is the standard for almost all introductory physics labs.

🎯 Exam Tip: 0.01 cm is the standard answer for most "find the least count" questions about verniers.

Question 3N. A microscope has a main scale with 20 divisions in 1 cm and a vernier scale with 25 divisions. Calculate its least count.
Answer: Value of 1 m.s.d. (x) = 1/20 cm = 0.05 cm
No. of divisions on the vernier scale (n) = 25
Hence, the L.C. of the microscope = x/n = (0.05 / 25) cm = 0.002 cm
In simple words: This microscope is very precise; it can measure things as tiny as two-thousandths of a centimeter.

📝 Teacher's Note: Help students with the math here: \( 0.05 / 25 = 0.002 \).

🎯 Exam Tip: Don't forget that 20 divisions in 1 cm means each division is 0.05 cm.

Question 4N. A pencil thickness is observed as 1.4 mm. The instrument has a positive zero error of 0.02 cm. Calculate the corrected reading.
Answer: Observed reading = 1.4 mm
Zero error = + 0.02 cm = + 0.2 mm
Correct reading = observed reading – zero error (with sign)
= 1.4 mm – 0.2 mm = 1.2 mm
In simple words: The ruler was wrong by 0.2 mm (it was starting too far ahead), so we have to subtract that "extra" bit to get the real size.

📝 Teacher's Note: Ensure students convert cm to mm before subtracting. Mixing units is a fatal error.

🎯 Exam Tip: For positive error, you always subtract. For negative error, you always add.

Question 5N. (i) 10 vernier divisions = 9 m.s.d. and 1 m.s.d. = 1 mm. Find L.C. (ii) If the jaws are closed and the 3rd vernier division coincides with a main scale division, find the zero error.
Answer:
(i) Value of 1 m.s.d. = 1 mm
L.C. = 1 mm / 10 = 0.1 mm or 0.01 cm
(ii) Coinciding division = 3
Zero error = +3 \( \times \) L.C. = +3 \( \times \) 0.01 cm = +0.03 cm
In simple words: The tool is starting 3 "tiny steps" ahead of zero.

📝 Teacher's Note: This is a two-step logic problem. First find the L.C., then use it to find the value of the zero error.

🎯 Exam Tip: Explicitly mention that the error is "positive" because the vernier zero is ahead of the main scale zero.

Question 6N. (i) 20 vernier divisions = 19 m.s.d. and 1 m.s.d. = 1 mm. Find L.C. (ii) If the main scale reading is 35 mm and 4th vernier division coincides, find total reading and cylinder radius.
Answer:
(i) Value of 1 m.s.d = 1 mm = 0.1 cm
L.C. = (0.1/20) cm = 0.005 cm
(ii) Main scale reading = 35 mm = 3.5 cm
Coinciding division (p) = 4.
Vernier scale reading = 4 \( \times \) 0.005 cm = 0.02 cm
Total reading = 3.5 + 0.02 = 3.52 cm
Radius of the cylinder = Diameter / 2 = 3.52/2 = 1.76 cm
In simple words: We find the full width (diameter) of the cylinder first, then cut it in half to find the radius.

📝 Teacher's Note: Radius vs Diameter is a classic "read the question carefully" trap.

🎯 Exam Tip: Always highlight the word "Radius" in your question paper so you don't forget the final division by 2.

Question 7N. (a) L.C. is 0.01 cm and MSR is 1.8 cm. If 4th division coincides, find total reading. (b) If zero error is -0.02 cm, find corrected reading.
Answer:
(a) Total reading = 1.8 + (4 \( \times \) 0.01) = 1.84 cm
(b) Observed reading = 1.84 cm
Zero error = -0.02 cm
Correct reading = Observed reading – Zero error (with sign)
= [1.84 – (-0.02)] cm = 1.86 cm
In simple words: The tool was starting behind zero, so we have to add that "missing" bit back to the measurement.

📝 Teacher's Note: This is a perfect example of why we say "subtract with sign." Subtracting -0.02 is adding 0.02.

🎯 Exam Tip: Negative error correction is addition. Positive error correction is subtraction.

Question 8N. A vernier calliper has L.C. 0.01 cm. In a reading, MSR is 3.3 mm and 6th VSD coincides. Find total reading.
Answer: L.C. = 0.01 cm
Main scale reading = 3.3 mm = 0.33 cm
Vernier scale reading = 6 \( \times \) 0.01 cm = 0.06 cm
Total reading = 0.33 + 0.06 = 0.39 cm
(Note: If MSR is 3.3 cm, reading is 3.36 cm).
In simple words: You just add the main part and the sliding part together.

📝 Teacher's Note: Watch out for unit consistency. 3.3 mm + 0.06 cm is a dangerous calculation if you don't convert first.

🎯 Exam Tip: Convert everything to cm before adding. It is the safest way to avoid mistakes.

Question 9N. A screw gauge has a pitch of 0.5 mm and 100 circular divisions. Calculate L.C.
Answer: Pitch = 0.5 mm
No. of divisions = 100
L.C. = 0.5/100 mm = 0.005 mm or 0.0005 cm
In simple words: This tool is incredibly sensitive, measuring as small as five-ten-thousandths of a centimeter.

📝 Teacher's Note: 0.5 mm is half a millimeter. Dividing that into 100 parts makes each part 0.005 mm.

🎯 Exam Tip: If the question asks for L.C. in cm, provide both mm and cm versions to be safe.

Question 10N. A screw advances 1 mm in 2 rotations. If there are 50 divisions on the circular scale, find (i) Pitch and (ii) L.C.
Answer:
(i) Pitch = Distance moved / number of rotations = 1 mm / 2 = 0.5 mm
(ii) L.C. = Pitch / No. of divisions = 0.5 / 50 = 0.01 mm
In simple words: If two turns move it 1 mm, one turn (pitch) is 0.5 mm. We then divide that turn into 50 tiny steps.

📝 Teacher's Note: This is a standard trick question. Students often use 1 mm as the pitch by mistake.

🎯 Exam Tip: Pitch is ALWAYS for one rotation.

Question 11N. Pitch is 1mm and circular divisions are 100. (i) Find L.C. (ii) If MSR is 2mm and 45th CSR coincides, find total reading.
Answer:
(i) L.C. = 1/100 = 0.01 mm or 0.001 cm
(ii) Main scale reading = 2mm = 0.2 cm
CSR (p) = 45
Circular scale reading = 45 \( \times \) 0.001 = 0.045 cm
Total reading = 0.2 + 0.045 = 0.245 cm
In simple words: Combine the 2mm from the ruler with the 45 tiny clicks from the knob.

📝 Teacher's Note: 0.2 cm + 0.045 cm = 0.245 cm. Help students align the decimal places during addition.

🎯 Exam Tip: Show the multiplication by L.C. clearly as a separate step.

Question 12N. L.C. is 0.001 cm. MSR is 1 mm and 27th CSR coincides. (i) find diameter (ii) if zero error is 0.005 cm, find corrected reading.
Answer:
(i) MSR = 0.1 cm, CSR = 27
Diameter = 0.1 + (27 \( \times \) 0.001) = 0.127 cm
(ii) Zero error = + 0.005 cm
Correct reading = 0.127 – 0.005 = 0.122 cm
In simple words: Find the measurement first, then subtract the instrument's starting error.

📝 Teacher's Note: Positive error (0.005) means the instrument was "over-reading," so we subtract to get back to the truth.

🎯 Exam Tip: Always subtract positive errors from the observed reading.

Question 13N. A screw gauge moves 1 mm in 2 rotations and has 50 circular divisions. (i) Find pitch and L.C. (ii) If closed jaws show 4 divisions below the base line, find zero error.
Answer:
(i) Pitch = 1/2 = 0.5 mm, L.C. = 0.5/50 = 0.01 mm
(ii) Zero lies below the base line \( \implies \) positive error.
Zero error = + (4 \( \times \) L.C.) = + (4 \( \times \) 0.01) = + 0.04 mm
In simple words: "Below the line" means the zero hasn't made it to the start yet, so the tool is already counting 4 tiny steps.

📝 Teacher's Note: Use the "below = positive, above = negative" rule for screw gauge zero errors.

🎯 Exam Tip: Always include the sign and units in the zero error answer.

Question 14N. A screw moves 1 mm in 1 rotation. Divisions = 50. (i) Find pitch and L.C. (ii) If MSR is 4 mm and 47th CSR coincides, find total diameter.
Answer:
(i) Pitch = 1 mm, L.C. = 1/50 = 0.02 mm
(ii) MSR = 4 mm, CSR = 47
Circular reading = 47 \( \times \) 0.02 = 0.94 mm
Diameter = 4 + 0.94 = 4.94 mm
In simple words: One turn moves it 1 mm. We read 4 mm plus almost one more whole turn (47 out of 50 steps).

📝 Teacher's Note: This is a slightly different configuration where L.C. is 0.02 instead of the standard 0.01.

🎯 Exam Tip: Don't assume the L.C. is always 0.01. Calculate it every time based on the given pitch and divisions.

Question 15N. Pitch is 0.5 mm and L.C. is 0.001 mm. Find the number of circular scale divisions.
Answer: No. of divisions = Pitch / L.C.
= 0.5 / 0.001 = 500
In simple words: To reach such a tiny accuracy, the knob needs to have 500 little marks on it.

📝 Teacher's Note: This is a rearrangement of the L.C. formula: \( n = \text{Pitch} / \text{LC} \).

🎯 Exam Tip: This is a great test of your decimal division skills. 0.5/0.001 is 500.

Exercise 1(C)

Question 1S. What is a simple pendulum? Is the pendulum in a clock a true simple pendulum?
Answer: A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string. No, the pendulum used in a pendulum clock is not a simple pendulum because a simple pendulum is an ideal case. In real life, we cannot have a mass that is a single point, nor a string that weighs absolutely nothing.
In simple words: A simple pendulum is a physics "dream." Real clocks have strings with weight and bobs with size, which makes them slightly different from the perfect version in books.

📝 Teacher's Note: Use the word "Idealization" to explain how scientists simplify the world to understand rules like gravity better.

🎯 Exam Tip: Mention the two reasons why a real pendulum isn't "simple": string mass and bob size.

Question 2S. Define: (i) Oscillation (ii) Amplitude (iii) Frequency (iv) Time period.
Answer:
(i) Oscillation: One complete to and fro motion of the pendulum is called one oscillation.
(ii) Amplitude: The maximum displacement of the bob from its mean position on either side is called the amplitude of oscillation. It is measured in metres (m).
(iii) Frequency: It is the number of oscillations made in one second. Its unit is hertz (Hz).
(iv) Time period: This is the time taken to complete one oscillation. It is measured in second (s).
In simple words: Oscillation is a "round trip." Amplitude is how far it "swings out." Frequency is "how many trips per second," and Time period is "how long for one trip."

📝 Teacher's Note: Use a physical pendulum to demonstrate a single "to and fro" motion to define oscillation visually.

🎯 Exam Tip: Always remember the inverse relationship: Frequency (\( f \)) = 1 / Time Period (\( T \)).

Question 3S. Draw and label a diagram of a simple pendulum.
Answer:
In simple words: This is a map of a swinging weight. The center is where it wants to rest, and the amplitude is how far it pushes away from that center.

📝 Teacher's Note: The "effective length" is measured from the point of suspension to the center of gravity of the bob.

🎯 Exam Tip: Label the "Mean position" (center) and "Extreme positions" (sides) clearly.

Question 4S. Name two factors on which the time period of a simple pendulum depends.
Answer: Two factors on which the time period of a simple pendulum depends are:
1. Length of pendulum (\( l \))
2. Acceleration due to gravity (\( g \))
Formula: \( T = 2\pi\sqrt{\frac{l}{g}} \)
In simple words: Only the length of the string and how hard the planet pulls (gravity) decide how fast it swings.

📝 Teacher's Note: This was discovered by Galileo. It's why pendulums are so good for making clocks—they are very consistent.

🎯 Exam Tip: Length is directly proportional (under a root), and gravity is inversely proportional (under a root).

Question 5S. Name two factors on which the time period of a simple pendulum does not depend.
Answer: Two factors on which the time period of a simple pendulum does not depend are:
1. Material of the bob (mass of the bob).
2. Amplitude of oscillation (provided it is small).
In simple words: A heavy lead ball and a light wooden ball will swing at exactly the same speed! It also doesn't matter if you pull it back a little or a lot (if it's a small swing).

📝 Teacher's Note: This is often surprising to students. A common experiment is to time bobs of different weights to prove it.

🎯 Exam Tip: Mass and Amplitude are the "trap" answers in multiple-choice questions.

Question 6S. What happens to the time period if:
(a) Length is quadrupled (\( \times 4 \))?
(b) Gravity is reduced to one-fourth (\( \div 4 \))?
Answer:
(a) If length quadruples:
\( T = 2\pi\sqrt{\frac{l}{g}} \)
\( T' = 2\pi\sqrt{\frac{4l}{g}} = 2 \times T \)
Therefore, the time period is doubled.

(b) If gravity is reduced to one-fourth:
\( T' = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}} = 2 \times T \)
Therefore, the time period is doubled.
In simple words: Because of the square root sign in the formula, changing things by 4 only changes the time by 2 (since the square root of 4 is 2).

📝 Teacher's Note: This is a great way to practice the square root relationship in physics formulas.

🎯 Exam Tip: Always show the substitution into the square root to get full marks for the logic.

Question 7S. How is the frequency of a pendulum related to its time period?
Answer: Time period of a simple pendulum is inversely proportional to its frequency.
\[ f = \frac{1}{T} \]
In simple words: If it takes a long time for one swing (high T), you won't get many swings per second (low f). They are exact opposites.

📝 Teacher's Note: Reciprocals are the mathematical way to describe "opposites" in physics.

🎯 Exam Tip: If \( T = 2\text{s} \), then \( f = 0.5\text{ Hz} \).

Question 8S. Describe how to measure the time period of a simple pendulum in a lab.
Answer:
1. Displace the bob slightly from its mean position and release it to start the to-and-fro motion.
2. Use a stopwatch to measure the total time ‘t’ for 20 complete oscillations.
3. Find the time period ‘T’ by dividing the total time ‘t’ by 20 (\( T = t/20 \)).
In simple words: You don't just time one swing because it's too fast. You time 20 swings and then divide by 20 to be much more accurate.

📝 Teacher's Note: Timing multiple oscillations reduces the percentage error caused by the human reaction time when clicking the stopwatch.

🎯 Exam Tip: Mentioning "20 oscillations" and "dividing by 20" shows you know correct lab procedure.

Question 9S. How is the value of 'g' calculated from a simple pendulum graph?
Answer: The time period of a simple pendulum is directly proportional to the square root of its effective length (\( T \propto \sqrt{l} \)).
By finding the slope of the \( T^2 \) vs \( l \) graph, we can find \( g \). The slope is constant and equal to \( 4\pi^2 / g \).
\[ g = \frac{4\pi^2}{\text{Slope of } T^2 \text{ vs } l \text{ graph}} \]
In simple words: We draw a line showing how length and time-squared are linked. The steeper the line, the different the gravity.

📝 Teacher's Note: The graph is a straight line passing through the origin because \( T^2 = (\frac{4\pi^2}{g}) l \), which fits the \( y = mx \) pattern.

🎯 Exam Tip: Remember: \( g = 4\pi^2 \times (\text{slope of } l \text{ vs } T^2 \text{ graph}) \). Check which variable is on which axis!

Question 10S. Two pendulums have bobs of different weights. What is the ratio of their time periods?
Answer: The ratio of their time periods would be 1:1 because the time period does not depend on the weight (mass) of the bob.
In simple words: Weight doesn't change the swing speed, so two identical strings with different weights will swing perfectly together.

📝 Teacher's Note: This is a common logic test. If the lengths are the same, the ratio is always 1:1.

🎯 Exam Tip: Always justify the ratio by stating the independence of mass.

Question 11S. Pendulum A is twice as long as Pendulum B. Which one swings faster?
Answer: Pendulum B will swing faster (make more oscillations in the same time). This is because the time period (\( T \)) is directly proportional to the square root of length. Pendulum A, being longer, will have a greater time period (take more time for one swing) and thus complete fewer oscillations than B.
In simple words: Longer strings make for slow, lazy swings. Shorter strings zip back and forth much faster.

📝 Teacher's Note: A "faster" swing means a "shorter" time period.

🎯 Exam Tip: Use the proportionality \( T \propto \sqrt{l} \) to explain your reasoning.

Question 12S. Summarize the relationships affecting a simple pendulum.
Answer:
(a) Time period is directly proportional to the square root of the length.
(b) Time period does not depend on the mass of the bob.
(c) Time period does not depend on the amplitude of oscillations (for small swings).
(d) Time period is inversely proportional to the square root of acceleration due to gravity.
In simple words: Longer string = slower swing. Stronger gravity = faster swing. Mass and how hard you push it don't matter.

📝 Teacher's Note: These four points are the summary of the whole section on pendulums.

🎯 Exam Tip: Memorize this list—it can answer almost any conceptual question on this topic.

Question 13S. What is a seconds pendulum?
Answer: A pendulum with the time period of oscillation equal to exactly two seconds is known as a seconds pendulum.
In simple words: It is a pendulum that takes one second to swing one way and one second to swing back.

📝 Teacher's Note: The effective length of a seconds pendulum on Earth is approximately 1 metre (99.4 cm).

🎯 Exam Tip: "Two seconds" is the defining characteristic. Don't say "one second."

Question 14S. What is the frequency of a seconds pendulum? Does it change with amplitude?
Answer: The frequency of oscillation of a seconds’ pendulum is 0.5 \( s^{-1} \) (or 0.5 Hz). It does not depend on the amplitude of oscillation.
In simple words: Since it takes 2 seconds for one swing, it only finishes "half" a swing every single second.

📝 Teacher's Note: Frequency = 1 / 2 = 0.5. Simple math, but high exam value.

🎯 Exam Tip: Always include the units for frequency (Hz).

Question 1M. If the length of a pendulum is made four times, its time period becomes:
(a) double
(b) half
(c) four times
(d) remains same
Answer: (a) double
In simple words: The time period is linked to the square root of the length. Since the square root of 4 is 2, the time doubles.

📝 Teacher's Note: This is a direct test of the \( \sqrt{l} \) relationship.

🎯 Exam Tip: Remember: Length \( \times 4 \implies T \times 2 \); Length \( \times 9 \implies T \times 3 \).

Question 2M. The time period of a seconds pendulum is:
(a) 1 s
(b) 2 s
(c) 0.5 s
(d) 4 s
Answer: (b) 2 s
In simple words: Don't let the name "seconds" fool you; a full round trip takes two seconds.

📝 Teacher's Note: It's called a seconds pendulum because every "tick" is 1 second, but one oscillation is "tick-tock" (2 seconds).

🎯 Exam Tip: This is one of the most common "trap" questions in the syllabus.

Question 3M. Find the length of a seconds pendulum on Earth (g = 9.8).
(a) 0.5 m
(b) 1.0 m
(c) 2.0 m
(d) 1.5 m
Answer: (b) 1.0 m
In simple words: If you want a pendulum to take 2 seconds to swing, you need a string that is almost exactly one meter long.

📝 Teacher's Note: The exact calculation gives 0.994 m, which is rounded to 1.0 m for practical purposes.

🎯 Exam Tip: You can memorize "1 metre" as the standard length for a seconds pendulum on Earth.

Numerical Problems (Set 3)

Question 1N. A pendulum completes 40 oscillations in one minute. Find its frequency and time period.
Answer:
(a) Frequency = Oscillations per second
= (40/60) \( s^{-1} \) = 0.67 \( s^{-1} \)
(b) Time period = 1/frequency
= (1/0.67) s = 1.5 s
In simple words: It does 40 swings in 60 seconds, which is less than one swing per second. Each full swing takes 1.5 seconds.

📝 Teacher's Note: Always divide oscillations by 60 to get the per-second rate (frequency).

🎯 Exam Tip: Provide both frequency and time period as separate steps to ensure maximum marks.

Question 2N. Find the frequency of a pendulum with a time period of 2s. What is it called?
Answer: Time period = 2 s
Frequency = 1 / time period = (1/2) \( s^{-1} \) = 0.5 \( s^{-1} \)
Such a pendulum is called the seconds’ pendulum.
In simple words: This is a classic 2-second swing, which happens 0.5 times every second.

📝 Teacher's Note: This numerical reinforces the definition of a seconds pendulum.

🎯 Exam Tip: "Seconds pendulum" is the key identifying name here.

Question 3N. If the acceleration due to gravity falls to one-fourth, what happens to the time period?
Answer: Let the original time period be T and original gravity be g. Let the new gravity be g' = g/4.
\[ T = 2\pi\sqrt{\frac{l}{g}} \]
\[ T' = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}} = 2 \times T \]
Therefore, the time period will be doubled.
In simple words: If gravity gets weaker, the pendulum moves slower. On a planet with 1/4 gravity, the swing takes twice as long.

📝 Teacher's Note: This is the reciprocal logic of Question 6S(a). Weaker gravity = larger T.

🎯 Exam Tip: Always clarify that the time period increases because T is inversely proportional to \( \sqrt{g} \).

Question 4N. Find the length of a seconds pendulum if g = 10 \( \text{m/s}^2 \).
Answer: Given, g= 10 \( \text{m/s}^2 \) and time period (T) = 2s.
Let 'l' be the length of the seconds' pendulum.
\[ T = 2\pi\sqrt{\frac{l}{g}} \]
\[ 2 = 2\pi\sqrt{\frac{l}{10}} \]
\[ \frac{1}{\pi^2} = \frac{l}{10} \]
\[ l = \frac{10}{(3.14)^2} \approx 1.0142 \text{ m} \]
In simple words: Using a slightly higher gravity (10 instead of 9.8) means we need a slightly longer string (1.01m) to keep the 2-second swing.

📝 Teacher's Note: This is a standard calculation. Remind students that \( \pi^2 \approx 9.87 \), which is very close to 10.

🎯 Exam Tip: Always use the value of 'g' provided in the question, not the 9.8 default.

Question 5N. Compare the time periods of two pendulums of lengths 1m and 9m.
Answer: Let \( T_1 \) and \( T_2 \) be the time periods of the two pendulums.
\[ \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} \]
\[ \frac{T_1}{T_2} = \sqrt{\frac{1}{9}} = \frac{1}{3} \]
\( T_1 : T_2 = 1 : 3 \)
In simple words: The 9m pendulum is nine times longer, so its swing takes three times as much time as the 1m pendulum.

📝 Teacher's Note: Ratio problems are the best way to test proportionality without complex math.

🎯 Exam Tip: Express the final answer as a ratio (1:3).

Question 6N. If a pendulum takes 2.5s to complete one oscillation when g = 9.8, find its length.
Answer: Time period T = 2.5 s, g = 9.8.
\[ T = 2\pi\sqrt{\frac{l}{g}} \]
\[ 2.5 = 2(3.14)\sqrt{\frac{l}{9.8}} \]
\[ 0.398 = \sqrt{\frac{l}{9.8}} \]
Squaring both sides,
\[ l = (0.398)^2 \times (9.8) \]
\( l = 1.55 \text{ m} \)
In simple words: To get a 2.5 second swing on Earth, you'd need a string that is about one and a half meters long.

📝 Teacher's Note: Squaring decimals can be tricky. Help students with \( 0.4 \times 0.4 = 0.16 \) as a quick estimation guide.

🎯 Exam Tip: Keep at least 3 decimal places during the intermediate steps for better accuracy.

Question 7N. Compare the time periods of pendulums with lengths \( l_1 \) and \( l_2 \).
Answer: Let \( T_1 \) and \( T_2 \) be the time periods of the two pendulums of lengths \( l_1 \) and \( l_2 \), respectively.
Then, we know that the time period is directly proportional to the square root of the length of the pendulum.
\[ \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} \]
If \( l_1 = 4l_2 \), then \( T_1/T_2 = \sqrt{4}/1 = 2/1 = 2:1 \).
In simple words: This is the universal rule for comparing any two pendulums.

📝 Teacher's Note: This is the symbolic form of the relationship found in Question 11S.

🎯 Exam Tip: Using the ratio formula is the "best practice" for comparison questions.

Question 8N. If a bob takes 0.2s to move from the mean position to one extreme, find its time period.
Answer: Time period = Time taken to complete 1 full oscillation.
One full oscillation = \( 4 \times \) (time from mean to extreme)
= \( (4 \times 0.2) \text{ s} = 0.8 \text{ s} \)
In simple words: A full swing is 4 parts: out to the right, back to center, out to the left, and back to center. So we multiply the 0.2s part by four.

📝 Teacher's Note: Draw a pendulum and count the four segments of a full oscillation: \( M \rightarrow E_1 \rightarrow M \rightarrow E_2 \rightarrow M \).

🎯 Exam Tip: This is a very common conceptual numerical. Don't multiply by 2; multiply by 4.

Question 9N. How long does a seconds pendulum take to move from one extreme to the other?
Answer: Time period of a seconds’ pendulum = 2 s
Time taken to complete half oscillation, i.e. from one extreme to the other extreme = 1 s.
In simple words: Since a full trip takes 2 seconds, half the trip (one side to the other) takes exactly one second. This is why it's called a "seconds" pendulum.

📝 Teacher's Note: This explains the physical "tick" and "tock" of old grandfather clocks.

🎯 Exam Tip: Extreme-to-extreme is always half of the total time period (\( T/2 \)).