Frank Brothers Solutions for ICSE Class 9 Physics Chapter 1 Measurement

Page No: 15

Question 1. What is measurement?
Answer: Measurement is an act or the result of comparison of a quantity whose magnitude is unknown with a predefined standard.
In simple words: Measurement is like comparing an unknown size (like the length of a table) with a fixed tool (like a ruler) to find out how big it is.

📝 Teacher's Note: Use a physical ruler in class to demonstrate this. Show how the 'predefined standard' (the ruler) helps define the 'unknown magnitude' (the object's length).

🎯 Exam Tip: Remember the two key parts of measurement: the unknown magnitude and the predefined standard. Include both for a complete definition.

Question 2. What are fundamental quantities?
Answer: The physical quantities like mass, length and time which do not depend on each other are known as fundamental quantities.
In simple words: These are the "base" measurements that don't need any other measurements to be calculated—they are independent.

📝 Teacher's Note: Explain that these are the building blocks of all other measurements in physics, just like primary colors are the base for all other colors.

🎯 Exam Tip: Always list mass, length, and time as the primary examples of fundamental quantities to score full marks.

Question 3. Name the three fundamental quantities.
Answer: Length, mass, time are the three fundamental quantities.
In simple words: The three basic things we measure in the physical world are how long something is, how heavy it is, and how long it takes.

📝 Teacher's Note: Quiz students on these three until they are ingrained, as they form the basis of the MKS, CGS, and FPS systems.

🎯 Exam Tip: This is a very common short-answer question. Make sure to list all three together.

Question 4. Define a unit.
Answer: Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.
In simple words: A unit is a fixed "piece" of measurement (like a centimeter or a kilogram) that we use as a yardstick to measure everything else.

📝 Teacher's Note: Use the example of "paws" vs. "centimeters" to show why we need standard units—everyone's hand size is different, but a centimeter is always the same.

🎯 Exam Tip: Keywords here are "standard quantity" and "comparison." Use these to show technical understanding.

Question 5. How is a standard metre defined?
Answer: A standard metre is equal to \( 1650763.73 \) wavelengths in vacuum, of the radiation from krypton isotope of mass \( 86 \).
In simple words: Scientists defined a meter very precisely by counting a specific number of tiny light waves coming from a specific element called Krypton.

📝 Teacher's Note: Explain that this definition was created so that any laboratory in the world could reproduce a perfect meter without needing to see the original "master" ruler kept in France.

🎯 Exam Tip: Memorize the specific number \( 1650763.73 \) and the element "Krypton-86" for high-precision marks.

Question 6. Name and describe the three systems of units.
Answer: Three systems of unit are:

• C.G.S system – fundamental unit of length is centimetre (cm), of mass is gram (gm), of time is second (s).
• F.P.S system – fundamental unit of length is foot (ft), of mass is pound (lb), of time is second (s).
• M.K.S system – fundamental unit of length is metre (m), of mass is kilogram (kg), of time is second (s).

In simple words: Different parts of the world used different "sets" of units. Some measured in centimeters/grams, some in feet/pounds, and the scientific standard uses meters/kilograms.

📝 Teacher's Note: Create a table on the board to compare these. Show students that "seconds" is the only unit that is the same in every single system.

🎯 Exam Tip: Use the abbreviations to remember the units: C-G-S is Centimeter-Gram-Second. This prevents mixing up the systems.

Question 7. What is the SI unit of mass and how is it defined?
Answer: The SI unit of mass is Kilogram. One standard kilogram is equal to the mass of a cylinder of nearly same height and diameter and made up of platinum and iridium alloy.
In simple words: A kilogram is defined by the weight of a special metal cylinder made of platinum and iridium that is kept safely in a lab.

📝 Teacher's Note: Mention that platinum and iridium are used because they don't rust or change weight easily over time.

🎯 Exam Tip: Be sure to mention the metal alloy "platinum and iridium" as it is a specific detail examiners look for.

Question 8. Name three units of length greater than a metre.
Answer: Three units of length greater than a metre are:

• Decameter = \( 10 \text{ metre} \)
• Hectometer = \( 100 \text{ metre} \)
• Kilometer = \( 1000 \text{ metre} \)

In simple words: If you need to measure longer distances, you group meters into tens (decameter), hundreds (hectometer), or thousands (kilometer).

📝 Teacher's Note: Use the "stairs" analogy for units: every step up is \( 10 \) times bigger, and every step down is \( 10 \) times smaller.

🎯 Exam Tip: Ensure you provide the numerical relationship (e.g., \( 1 \text{ km} = 1000 \text{ m} \)) to make your answer complete.

Question 9. List the fundamental units in the SI system.
Answer:

In simple words: This table shows the official world-standard units for measuring everything from heat and light to electric power and angles.

📝 Teacher's Note: Focus on the first seven units as they are the most common. Mention that 'Radian' and 'Steradian' are supplementary units.

🎯 Exam Tip: Note that symbols like 'K' and 'A' are capitalized because they are named after scientists (Lord Kelvin and Andre-Marie Ampere), while 'm' and 'kg' are lowercase.

Question 10. Define a light year.
Answer: Light year is defined as the distance travelled by light in vacuum in one year.
In simple words: Even though it has the word "year" in it, a light year measures distance, not time. It's how far a beam of light travels in a single year.

📝 Teacher's Note: This is a classic "trick" question. Emphasize to the class that it is a unit of distance, not time, used to measure space between stars.

🎯 Exam Tip: Always specify "in vacuum" in your definition, as light's speed can change if it's passing through air or glass.

Question 11. Name two units of length smaller than a metre.
Answer: Two units of length smaller than a metre are:

• Decimeter = \( 0.1 \text{ metre} \)
• Centimeter = \( 0.01 \text{ metre} \)

In simple words: For measuring small things, we divide a meter into \( 10 \) equal parts (decimeters) or \( 100 \) equal parts (centimeters).

📝 Teacher's Note: Show these units on a standard school ruler so students can visualize the scale.

🎯 Exam Tip: You can also use millimeter (\( 10^{-3} \text{ m} \)) as a valid answer for this question.

Question 12. Why is a leap year mentioned in a measurement context?

Answer: Leap year because it is a unit of time.

In simple words: A leap year is a specific way we measure a large amount of time, specifically every four years when we add an extra day.

📝 Teacher's Note: This often appears in "Odd One Out" questions alongside units of length. Remind students to categorize units by the physical quantity they measure.

🎯 Exam Tip: Ensure you clearly state that a leap year is a unit of time, not distance or mass.

Question 13. Define the order of magnitude of a physical quantity with an example.

Answer: Order of magnitude of a physical quantity is defined as its magnitude in powers of ten when that physical quantity is expressed in powers of ten with one digit towards the left decimal.
For example, volume = \( 52.37 \text{ m}^3 \) then the order of magnitude is \( 10^2 \text{ m}^3 \).

In simple words: Order of magnitude is like a "rounded" version of a number using only powers of ten to show roughly how big or small it is.

📝 Teacher's Note: Teach students the rule for rounding: if the number is \( \geq 5 \), the power of ten increases by one. If it is \( < 5 \), the power stays the same.

🎯 Exam Tip: Always convert the number to scientific notation (e.g., \( 5.237 \times 10^1 \)) before determining the order of magnitude.

Question 14. Is a micron the same as a millimeter? Explain.

Answer: No, micron is not same as millimeter because micron is equal to \( 10^{-6} \text{ metre} \) while a millimeter is equal to \( 10^{-3} \text{ metre} \).

In simple words: A micron is much smaller than a millimeter—in fact, there are one thousand microns in a single millimeter.

📝 Teacher's Note: Use a microscope analogy; millimeters can be seen with the naked eye, but microns are the size of cells seen under a microscope.

🎯 Exam Tip: Use the power of ten notation (\( 10^{-3} \) vs \( 10^{-6} \)) to provide a precise scientific comparison.

Question 15. State the following unit relationships:

Answer:
(a) \( 1 \text{ fermi} = 10^{-15} \text{ m} \)
(b) \( 1 \text{ m} = 10^{10} \text{ \AA} \)
(c) \( 1 \text{ parsec} = 3.26 \text{ ly} \)
(d) \( 1 \text{ light year} = 9.46 \times 10^{15} \text{ m} \)
(e) \( 1 \text{ m} = 10^{6} \mu \)
(f) \( 1 \text{ micron} = 10^{4} \text{ \AA} \)
(g) \( 1 \text{ u} = 1.655 \times 10^{-27} \text{ kg} \)

In simple words: These are standard conversion factors used to jump between very tiny units (like atoms) and very large units (like distances between stars).

📝 Teacher's Note: Group these into "Microscopic units" (fermi, Angstrom, micron) and "Astronomical units" (parsec, light year) to help students organize them in their minds.

🎯 Exam Tip: Pay close attention to the sign of the exponent; \( 10^{6} \) is a million times bigger, while \( 10^{-6} \) is a million times smaller.

Question 16. Name two units of mass smaller than a kilogram and state their values in kilograms.

Answer: Two units of mass smaller than a kilogram are:
(a) \( \text{Gram} = 10^{-3} \text{ kg} \)
(b) \( \text{Milligram} = 10^{-6} \text{ kg} \)

In simple words: We use grams for things like food ingredients and milligrams for tiny things like the medicine in a pill.

📝 Teacher's Note: Relate these to common objects—a paperclip is about 1 gram, and a grain of salt is closer to a milligram.

🎯 Exam Tip: Always provide the scientific notation (\( 10^{-3} \)) as it is expected in physics papers rather than just writing "1/1000".

Question 17. What is a leap year?

Answer: A leap year refers to a year in which February has 29 days and the total days in the year are 366 days.

In simple words: Every four years, we add one extra day to February to keep our calendar in sync with the Earth's orbit, making the year 366 days long.

📝 Teacher's Note: Explain that Earth takes about 365.25 days to orbit the sun, so those extra quarter-days add up to one full day every four years.

🎯 Exam Tip: Mention both the "29 days in February" and the "366 days total" for a complete answer.

Page No: 16

Question 18. Convert \( 5400 \text{ \AA} \) into nanometers (nm) and meters (m).

Answer:
\( 1 \text{ \AA} = 10^{-1} \text{ nm} \)
\( \implies \) So, \( 5400 \text{ \AA} = 5400 \times 10^{-1} = 540 \text{ nm} \)
\( 1 \text{ \AA} = 10^{-10} \text{ m} \)
\( \implies \) So, \( 5400 \text{ \AA} = 5400 \times 10^{-10} = 5.4 \times 10^{-7} \text{ m} \)

In simple words: This shows how a measurement in Angstroms (tiny units for atoms) looks when converted into slightly larger units like nanometers or meters.

📝 Teacher's Note: Practice moving the decimal point: \( 10^{-1} \) means moving it one place left, and scientific notation (\( 5.4 \times 10^{-7} \)) is the standard format for final answers.

🎯 Exam Tip: Show the step-by-step conversion starting from the unit relation to ensure partial marks even if you make a calculation error.

Question 19. If one bacterium is \( 2 \mu \) long, calculate the number of bacteria that can fit in a length of 3 m.

Answer:
\( 1 \mu = 10^{-6} \text{ m} \)
Length of one bacteria = \( 2 \mu = 2 \times 10^{-6} \text{ m} \)
Length given = 3 m
Number of bacteria = length given / length of one bacteria
\( \implies \) \( = \frac{3}{(2 \times 10^{-6})} \)
\( \implies \) \( = 1.5 \times 10^6 \)

In simple words: We find out how many bacteria fit by dividing the total length (3 meters) by the size of one bacterium. The result is 1.5 million bacteria!

📝 Teacher's Note: Remind students that they must convert all measurements to the same unit (meters) before doing the division.

🎯 Exam Tip: When a power of ten is in the denominator (bottom), it becomes positive when moved to the numerator (top). For example, \( 1/10^{-6} = 10^6 \).

Question 20. Fill in the following blanks with the correct numerical values.

Answer:
(i) quintal = \( 100 \text{ kg} \)
(ii) \( 1 \text{ mg} = 10^{-6} \text{ kg} \)
(iii) 1 mean solar day = \( 86400 \text{ s} \)
(iv) 1 nano second = \( 10^{-9} \text{ s} \)
(v) 1 year = \( 3.15 \times 10^7 \text{ s} \)

In simple words: These are common conversions for mass and time that scientists use to simplify large or very small calculations.

📝 Teacher's Note: Show the calculation for the mean solar day: \( 24 \text{ hours} \times 60 \text{ minutes} \times 60 \text{ seconds} = 86400 \text{ seconds} \).

🎯 Exam Tip: Memorizing the number of seconds in a day (86400) and a year (\( \approx \pi \times 10^7 \)) saves a lot of time during exams.

Page No: 28

Question 1. Define the "Pitch of the screw" and provide its formula.

Answer: When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = \( \frac{\text{distance travelled by screw in n rotations}}{\text{n rotations}} \)

In simple words: Pitch is the distance a screw travels in one full turn. It is also the gap between two "threads" or ridges on the screw.

📝 Teacher's Note: Use a real screw and nut to demonstrate this concept. One full rotation results in the screw advancing by the distance of exactly one thread gap.

🎯 Exam Tip: Include the formula in your answer even if not explicitly asked, as it shows a higher level of understanding of the measurement instrument.

Question 2. Is least count the same as pitch?
Answer: No, least count is not same as pitch because least count is found by dividing pitch by number of divisions on the circular scale.
In simple words: Pitch is the distance the screw moves in one full turn, while least count is the smallest possible measurement the tool can actually take.

📝 Teacher's Note: Use the example of a staircase: the height of one full flight is the "pitch," but the height of a single step is the "least count."

🎯 Exam Tip: Always remember that the least count is a calculated value involving the pitch and the divisions on the scale.

Question 3. What are two common uses of a Vernier caliper?
Answer: Two uses of vernier caliper are:
(i) Measuring the internal diameter of a tube or a cylinder.
(ii) Measuring the length of an object.
In simple words: It is a special tool used to measure how long things are and also to find out the width of the inside of a pipe or hole.

📝 Teacher's Note: Show students the two sets of jaws on a Vernier caliper—the large ones for outside and the small ones for inside measurements.

🎯 Exam Tip: Listing "internal diameter" and "external length" shows you understand the function of both sets of jaws.

Question 4. Mention two limitations of using a metre rule.
Answer: Two limitations of metre rule:
(i) There comes an error of parallax due to thickness of the metre rule.
(ii) We cannot use metre rule for measuring small thickness.
In simple words: It is hard to read correctly if you don't look straight down at the line, and it is not accurate enough to measure very thin things like a hair.

📝 Teacher's Note: Demonstrate parallax error by having students look at a ruler from different angles and observing how the measurement seems to change.

🎯 Exam Tip: "Parallax error" and "least count limitation" are the formal terms to use for these two points.

Question 5. Define pitch of a screw and least count of a measuring instrument.
Answer: When one complete rotation is given to the screw hand, it moves forward or backward by a distance called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = \( \frac{\text{distance travelled by screw in n rotations}}{\text{n rotations}} \)
Least count refers to the smallest reading that can be accurately measured while using an instrument. The least count is the value of one division on its scale.
In simple words: Pitch is how far a screw advances when you turn it once. Least count is the tiniest bit of distance any tool is capable of measuring.

📝 Teacher's Note: Remind students that "Least Count" determines the precision of the instrument. The smaller the least count, the more precise the tool.

🎯 Exam Tip: Make sure to provide both definitions clearly when a question asks for two separate terms.

Question 6. If the initial water level is 30 ml and it rises to 50 ml after dropping in a copper piece, what is the volume of the copper?
Answer: Initial level of water in cylinder = 30 ml
Level of water in cylinder after immersing piece of copper = 50 ml
Volume of copper piece = 50 - 30 = 20 ml
In simple words: You find the volume of an irregular object by seeing how much water it pushes out of the way when you drop it in a measuring jar.

📝 Teacher's Note: This is based on the Displacement Method. Ensure students subtract the initial volume from the final volume, not the other way around.

🎯 Exam Tip: Always include the unit "ml" (or \( cm^3 \)) in your final calculation to avoid losing easy marks.

Page No: 29

Question 7. Provide a labeled diagram of a Vernier caliper.
Answer: A well labeled diagram of Vernier caliper is shown in figure:
In simple words: This diagram shows the different parts of a Vernier caliper, including the two sets of jaws for measuring inside and outside, and the sliding scale that gives extra precision.

📝 Teacher's Note: Ask students to identify the "Inside Jaws" (at the top) and "Outside Jaws" (at the bottom) on the real instrument.

🎯 Exam Tip: When drawing this, ensure the Vernier scale is correctly placed on the sliding part, and don't forget the depth strip (Strip T) at the far end.

Question 8. What is the purpose of the ratchet in a screw gauge?
Answer: The ratchet is used in a screw gauge to hold the object under measurement gently between the studs.
In simple words: It is a knob that you turn until it clicks; this makes sure you don't squeeze the object too hard and get a wrong reading.

📝 Teacher's Note: Explain that tightening the screw too much can deform the object (like a thin wire), which leads to an inaccurate measurement.

🎯 Exam Tip: Use the word "gently" to describe how the ratchet holds the object; this is a key technical point.

Question 9. What is zero error in a screw gauge, and what are its two types?
Answer: If the zero of the circular scale does not coincide with the zero of the main scale (pitch scale), this is known as zero error. There are two types of zero error –
1. If the zero of the circular scale remains below the line of graduation then it is called positive zero error.
2. If the zero of the circular scale lies above the line of graduation then it is called negative zero error.
For positive zero error correction, the zero error should always be subtracted from the observed reading.
For negative zero error correction, the zero error must be added to the observed reading.
In simple words: Zero error is when the tool doesn't start exactly at zero. Depending on which way it is off, you either add or subtract that amount to get the real answer.

📝 Teacher's Note: Think of it like a weight scale that shows 1 kg before you even step on it. You would need to subtract that 1 kg from your final result.

🎯 Exam Tip: Remember: Positive error = Subtraction; Negative error = Addition. This "opposite" rule is vital for corrections.

Question 10. Name the two scales found in a screw gauge.
Answer: Two scales in a screw gauge are:
(i) A linear scale called the main scale graduated in half millimeters.
(ii) A circular scale divided into 50 or 100 equal parts.
In simple words: The tool has one straight scale like a regular ruler and one circular scale on a knob that you turn.

📝 Teacher's Note: Point out that the main scale measures larger increments (like 0.5mm), while the circular scale measures tiny fractions of those increments.

🎯 Exam Tip: Mention the specific divisions (like "50 or 100 parts") to show detailed knowledge of the instrument's design.

Question 11. What is backlash error and how is it avoided?
Answer: Due to constant use, there is space for the play of screw gauge but gradually this space increases with the use or wear and tear, so that when the screw is moved by rotating it in some direction, it slips in the nut and does not cover any linear distance for some rotation of the screw head. The error due to this is known as backlash error. It is avoided by turning the screw always in the same direction.
In simple words: Over time, the screw gets a bit loose. To keep the measurement accurate, you should always turn the knob in the same direction when making a measurement.

📝 Teacher's Note: Use the analogy of a loose steering wheel in an old car—you have to turn it a bit before the wheels actually move.

🎯 Exam Tip: The key to avoiding backlash error is "always turning the screw in the same direction." This is the phrase the examiner looks for.

Question 12. Describe the procedure to measure the diameter of a wire using a screw gauge.
Answer: Following procedure is used to measure the diameter of a wire:
(i) Calculate the least count and zero error of the screw gauge.
(ii) Place the wire in between the studs. Turn the ratchet clockwise so as to hold the wire gently in between the studs. Record the main scale reading.
(iii) Now record the division of circular scale that coincides with the base line of main scale. This circular scale division multiplied by least count will give circular scale reading.
(iv) The observed diameter is obtained by adding the circular scale reading to the main scale reading. Subtract the zero error if any, with its proper sign, from the observed diameter to get the true diameter.
In simple words: First, check the tool for mistakes. Then put the wire in, read the two scales, add them together, and fix any starting error.

📝 Teacher's Note: Emphasize the calculation step (Circular division \(\times\) Least Count) as students often forget this and just add the raw division number.

🎯 Exam Tip: Listing the steps in a numbered sequence (i, ii, iii...) makes your answer much clearer and easier to grade.

Question 13. Calculate the reading of a screw gauge where the pitch is 1mm and circular scale divisions are 100.
Answer: (i) Pitch of the screw = 1mm. Divisions on circular scale = 100.
Least count (L.C.) = pitch / divisions on circular scale.
\( \implies \) L.C. = 1/100 = 0.01 mm = 0.001 cm.
(ii) If the main scale reading = 2 mm = 0.2 cm, and the 97th division of the circular scale coincides with the base line:
Circular scale reading = 97 \(\times\) 0.001 cm = 0.097 cm.
Total reading = 0.2 + 0.097 = 0.297 cm.
In simple words: This shows how we combine the reading from the straight scale and the tiny fractions from the turning scale to get a very precise measurement.

📝 Teacher's Note: Practice several of these calculations in class. The shift from mm to cm (\( 0.01 \text{ mm} = 0.001 \text{ cm} \)) is where most errors happen.

🎯 Exam Tip: Always show your unit conversion (mm to cm) early in the problem to keep your final answer consistent.

Question 14. Why does a screw gauge provide more accurate measurements than a Vernier caliper?
Answer: Screw gauge measures a small length to a high accuracy because it has the lowest least count among the given three instruments. And low least count means high accuracy.
In simple words: A screw gauge can measure much smaller things than other tools, so its results are much closer to the real size.

📝 Teacher's Note: Compare the least counts: Metre rule (1mm) > Vernier (0.1mm) > Screw Gauge (0.01mm). The smaller the "ruler markings," the more accurate the tool.

🎯 Exam Tip: Use the term "least count" to explain accuracy; it shows scientific reasoning.

Question 15. Match the instruments with their standard least counts.
Answer: Instruments which have least count as follows:
(i) 0.1 mm = Vernier calipers
(ii) 1 mm = Metre scale
(iii) 0.01 mm = Screw gauge
In simple words: This shows which tools are meant for general measuring (metre scale) versus very tiny, precise measuring (screw gauge).

📝 Teacher's Note: Have students memorize these three values as they are the foundations for almost all measurement problems in this chapter.

🎯 Exam Tip: Be careful with units—sometimes the question might ask for these in cm (e.g., 0.01 cm for Vernier).

Question 16. Which instrument would you use to measure a needle's diameter or a paper's thickness?
Answer:
(i) The diameter of a needle = Screw Gauge
(ii) The thickness of a paper = Screw Gauge
(iii) The diameter of a pencil = Screw Gauge
Because screw gauge is more accurate and can measure the thicknesses which metre scale and vernier calliper cannot.
In simple words: For anything extremely thin, like a needle or a piece of paper, you need a screw gauge because it is the most precise tool.

📝 Teacher's Note: While a Vernier *could* measure a pencil, the Screw Gauge is much better for consistently round, thin objects.

🎯 Exam Tip: If the object is smaller than 1 mm, always suggest the "Screw Gauge" as the preferred instrument.

Question 17. Calculate the rod length if the main scale is 3.3 cm and the 6th vernier division coincides.
Answer: Least count of vernier calliper shown in the figure above = 0.01 cm.
Main scale reading = 3.3 cm.
Vernier scale reading = 6 \(\times\) 0.01 cm = 0.06 cm.
\( \implies \) Length of the rod = 3.3 + 0.06 = 3.36 cm.
In simple words: You take the starting point on the main ruler and add the tiny extra bit found by matching up the sliding scale lines.

📝 Teacher's Note: Point out that the "6th division" means the line numbered 6 on the sliding part of the tool.

🎯 Exam Tip: Always show the addition: (Main Scale Reading) + (Vernier Scale Reading) = Final Answer.

Question 18. What is positive zero error and how is it corrected?
Answer: If the zero of the circular scale remains below the line of graduation then it is called positive zero error. When there is positive zero error, then the instrument reads more than the actual reading. Therefore in order to get the correct reading, the zero error should always be subtracted from the observed reading.
In simple words: If the tool starts at a number higher than zero, your answer will be too big. To fix it, you subtract that extra starting amount.

📝 Teacher's Note: "Below the line" refers to the reference line on the main scale when the jaws are closed.

🎯 Exam Tip: The keyword here is "subtracted." Examiners specifically check for this correction method.

Question 19. Calculate the least count of a screw gauge with a pitch of 0.5mm and 100 divisions.
Answer: Pitch of the screw gauge = 0.5mm = 0.05 cm
Circular scale divisions = 100
Least Count (L.C.) = pitch / divisions
\( \implies \) L.C. = 0.05 / 100 = 0.0005 cm.
In simple words: By dividing the distance of one full turn by 100, we find the tiny distance the tool can measure.

📝 Teacher's Note: Notice the conversion to cm. Many students calculate 0.005 and leave it at that, which is incorrect if the question asks for cm.

🎯 Exam Tip: Always state your formula (\( L.C. = \text{Pitch} / \text{Divisions} \)) before plugging in the numbers.

Question 20. What is negative zero error and how is it corrected?
Answer: If the zero of the circular scale lies above the line of graduation then it is called negative zero error. When there is negative zero error, then the instrument reads less than the actual reading. Therefore in order to get the correct reading, the zero error should always be added to the observed reading.
In simple words: If the tool starts "below" zero, your measurement will be too small. You have to add that missing bit back to get the real size.

📝 Teacher's Note: Explain that negative error means the zero on the knob has already passed the reference line before the tool is even fully closed.

🎯 Exam Tip: Just as positive error is subtracted, negative error is always "added" to the observed value.

Question 21. Mark the following statements as True or False.
Answer:
(i) Accuracy is higher in Vernier calipers than screw gauge. — False, because accuracy is higher in screw gauge due to lower least count.
(ii) A screw gauge can measure thickness up to 0.01 mm. — True.
(iii) Length of 0.5 cm can only be measured with a ruler. — False, it can be measured more accurately with a screw gauge.
(iv) The ratchet is used to move the screw quickly. — False, it is used to hold the object gently.
(v) Volume is measured in cubic metres. — True.
In simple words: These statements check your knowledge of tool precision, the specific job of parts like the ratchet, and the correct units for volume.

📝 Teacher's Note: Go through each "False" statement and ask the students to rewrite them correctly. This reinforces the actual facts.

🎯 Exam Tip: For True/False questions that ask for a reason, always provide the "because..." part to ensure full credit.

Question 22. Define volume and state its SI unit.
Answer: The space occupied by a body is known as its volume. SI unit of volume is cubic metre (\( m^3 \)).
In simple words: Volume is just a measure of how much room an object takes up in three dimensions.

📝 Teacher's Note: Contrast this with surface area (2D space). Volume is for 3D objects like boxes, balls, or liquid in a jug.

🎯 Exam Tip: The symbol for the SI unit is \( m^3 \). Do not confuse it with Litre, which is a common unit but not the official SI unit.

Solution 23. Define volume and state its SI unit.
Answer: The space occupied by a body is known as its volume. SI unit of volume is cubic metre \( (m^3) \).
In simple words: Volume is the measure of how much three-dimensional space an object takes up, like how much water a bottle can hold.

📝 Teacher's Note: Use a box or a ball to explain that volume is about the "inside" space, while area is just about the "flat" surface.

🎯 Exam Tip: Always write the unit as cubic metre or \( m^3 \). Avoid writing just 'metre' as that is for length.

Question 24. Express 1 litre in terms of cubic metres.
Answer: \( 1 \text{ m}^3 = 1000 \text{ litre} \)
\( \implies \) \( 1 \text{ litre} = 1/1000 \text{ m}^3 \)
\( \implies \) \( = 0.001 \text{ m}^3 \)
In simple words: A litre is a smaller unit than a cubic metre; it takes one thousand litres to fill up a space of one cubic metre.

📝 Teacher's Note: Tell students that a cubic metre is like a large water tank, while a litre is like a standard water bottle. This helps them visualize the scale.

🎯 Exam Tip: Remember the conversion factor \( 1000 \). If you are moving from a smaller unit (litre) to a larger one (\( m^3 \)), you must divide.

Exercise Page No: 30

Question 25. Describe the measurement of volume of an irregular solid lighter than water by displacement method.
Answer: Measurement of volume of an irregular solid lighter than water by displacement method
(a) Fill the graduated cylinder with water.
(b) Take a heavy solid (eg. piece of metal) insoluble in water. This metal piece is called sinker.
(c) Tie the sinker to a strong string and lower it into the water gently and record the lower meniscus of water and let the value be \( V_1 \).
(d) Now take out the sinker and along with it tie the solid whose volume is to be measured and lower them into the water.
(e) Note the reading carefully and let the value be \( V_2 \).
(f) Volume of the solid, \( V = V_2 - V_1 \)
In simple words: Since some things float (like cork), we tie them to a heavy "sinker" to pull them underwater. We see how much the water rises with just the sinker, then with both, and the difference is the volume of the floating object.

📝 Teacher's Note: Demonstrate this in the lab using a cork and a stone. Explain that the sinker must be heavy enough to pull the lighter object completely under the surface.

🎯 Exam Tip: Clearly define "sinker" in your answer and use the formula \( V = V_2 - V_1 \) to show the final step.

Question 26. State the SI unit of volume and its relationship with the litre.
Answer: SI unit of volume is cubic metre or \( \text{metre}^3 \) \( (m^3) \).
The relation between liter and \( \text{metre}^3 \)
\( 1 \text{ metre}^3 = 1000 \text{ liter} \)
In simple words: The official scientific unit for volume is the cubic metre, and it is exactly equal to one thousand litres.

📝 Teacher's Note: Point out that in many everyday situations we use litres, but in physics calculations, we almost always use cubic metres.

🎯 Exam Tip: Be careful with the power; it is \( \text{metre}^3 \), not \( \text{metre}^2 \) (which is area).

Question 27. A screw has a pitch of 0.5 mm and a least count of 0.001 mm. Calculate the number of divisions on its circular scale.
Answer: Pitch of the screw = \( 0.5 \text{ mm} \)
Least count = \( 0.001 \text{ mm} \)
Number of divisions = pitch/least count
\( \implies \) \( = 0.5/0.001 \)
\( \implies \) \( = 500 \)
In simple words: To find the number of markings on the turning knob, we divide the distance of one full turn by the smallest possible measurement the tool can make.

📝 Teacher's Note: This formula \( (\text{LC} = \text{Pitch} / \text{Divisions}) \) is very important. Ask students to rearrange it to find any of the three values.

🎯 Exam Tip: Ensure both units (mm) are the same before dividing. In this case, they are, so no conversion is needed.

Question 28. Determine the final reading from the provided screw gauge image.
Answer: Since the screw advances by 1 division on the main scale when the circular head is rotated once
So, pitch of the screw = \( 1 \text{ mm} \)
Number of divisions = 50
Least count = pitch/number of divisions
\( \implies \) \( = 1/50 = 0.02 \text{ mm} \)
Now for the reading shown in figure,
The main scale reading = \( 4 \text{ mm} \)
From the figure it is clear that base line coincides with the \( 47^{th} \) division of the circular scale.
So, circular scale reading = \( 47 \times 0.02 \text{ mm} = 0.94 \text{ cm} \)
Hence, reading shown in the figure = \( 4 + 0.94 = 4.94 \text{ mm} \)
In simple words: We look at the straight scale to see how many whole millimeters have passed (4), then look at the knob to see the tiny extra bit (0.94), and add them together.

📝 Teacher's Note: Show the students how to align their eye with the reference line of the main scale to find the matching division on the circular scale.

🎯 Exam Tip: Always show the calculation of the least count first, even if it seems simple, to ensure you get marks for the method.

Question 29. What precautions should be taken while measuring volume of a solid lighter than water using displacement method?
Answer: Precautions to be taken while measuring volume of a solid lighter than water using displacement method

• The sinker should be insoluble in water
• The sinker should have a high density than water.
• Lower meniscus should be read to note down the readings and error due to parallax should be avoided.

In simple words: Make sure the heavy weight doesn't dissolve, is heavy enough to sink, and always look straight at the water line so you don't get a wrong reading.

📝 Teacher's Note: Explain that the 'lower meniscus' is the bottom of the curve formed by the water in the tube. This is the only accurate point to read.

🎯 Exam Tip: "Parallax error" is a keyword. Mentioning it shows the examiner you understand the common mistakes in manual reading.

Question 30. How do you measure the volume of an irregular solid that is soluble in water?
Answer: Measurement of volume of an irregular solid soluble in water using a graduated cylinder.
In this case, kerosene or any liquid whose density is lighter than water and in which the solid is not soluble is used.
(a) Fill the graduated cylinder with the liquid.
(b) Record the lower meniscus of liquid and let the value be \( V_1 \).
(c) Tie the solid whose volume is to be measured to a strong string and lower it into the water gently.
(d) Note the reading carefully and let the value be \( V_2 \)
(e) Volume of the solid, \( V = V_2 - V_1 \)
In simple words: If an object dissolves in water (like salt), we use a different liquid like oil or kerosene that won't dissolve it, and then measure how much that liquid rises.

📝 Teacher's Note: Use the example of a sugar cube. You can't measure it in water because it disappears, so you use a liquid like alcohol where it stays solid.

🎯 Exam Tip: Mentioning the use of an "alternative liquid" or "kerosene" is the most important part of this answer.

Exercise Page No: 38

Question 1. What is a simple pendulum? Draw a diagram.
Answer: A simple pendulum consists a heavy point mass (called the bob) suspended by a mass less and inextensible or non-elastic thread from a fixed point or rigid support. In simple words: A pendulum is just a heavy ball (the bob) hanging from a string that can swing back and forth from a fixed spot.

📝 Teacher's Note: Use a real pendulum in class to show that the 'string' doesn't stretch and the 'bob' is very heavy compared to the string.

🎯 Exam Tip: In the diagram, make sure to label the 'Bob', the 'String', and the 'Rigid Support'. These are the three essential parts.

Question 2. Define a seconds pendulum and state its length on Earth.
Answer: A seconds pendulum is a pendulum which takes 2 seconds to complete one oscillation. The length of seconds pendulum, where \( g = 9.8 \text{ ms}^{-2} \), is nearly 1 m.
In simple words: It's a special pendulum that takes exactly two seconds to swing out and come back. To do this on Earth, the string needs to be about one meter long.

📝 Teacher's Note: Explain that it's called a "seconds pendulum" because each single "tick" (half-oscillation) takes exactly one second.

🎯 Exam Tip: Remember the time period is \( 2 \) seconds, not \( 1 \) second. This is a very common mistake.

Question 3. What is a stopwatch used for?
Answer: A stopwatch is used to measure short intervals of time.
In simple words: A stopwatch is a tool that lets us start and stop time exactly when we want to, so we can see how fast something happens.

📝 Teacher's Note: Mention stopwatches during a sports day or for timing the oscillations of a pendulum in the lab.

🎯 Exam Tip: Mention "short intervals" to show you know it's for timing events, not for checking the time of day.

Question 4. State the SI unit of frequency.
Answer: SI unit of frequency is hertz (Hz).
In simple words: Frequency measures how many times something repeats in one second, and we name that count "Hertz."

📝 Teacher's Note: Relate this to radio stations (e.g., 91.1 MHz) to show students where they might have heard the term 'Hertz' before.

🎯 Exam Tip: The symbol for Hertz is always a capital \( H \) and a lowercase \( z \).

Question 5. If a pendulum completes one oscillation in one second, what is its frequency?
Answer: When a pendulum completes one oscillation in one second, then the frequency is one hertz.
In simple words: If it finishes one full swing-and-return every single second, it has a frequency of 1 Hz.

📝 Teacher's Note: Use the formula \( f = 1/T \). If \( T = 1 \), then \( f = 1/1 = 1 \). This helps students see the math behind the concept.

🎯 Exam Tip: Frequency is the inverse of the time period. If time is \( 1 \), frequency is \( 1 \).

Question 6. How are the time period and frequency of oscillation related?
Answer: The time period, \( T \) and frequency of oscillation, \( f \) are related as,
\( T = 1/f \) or \( f = 1/T \)
In simple words: They are opposites: if you know the time for one swing, you can find the frequency, and if you know the frequency, you can find the time.

📝 Teacher's Note: This is an "inverse relationship." As the time period gets longer, the frequency gets smaller (slower swings).

🎯 Exam Tip: Writing both forms of the equation \( (T = 1/f \text{ and } f = 1/T) \) shows a complete understanding.

Question 7. Define oscillation and amplitude for a simple pendulum.
Answer: One complete to and fro motion of a pendulum about its mean position is known as oscillation. Amplitude is the magnitude of the maximum displacement of the bob from the mean position on either side when an oscillation takes place.
In simple words: An oscillation is one full "there and back" trip. Amplitude is simply how far away from the middle the ball travels at its widest point.

📝 Teacher's Note: Draw a line in the middle (mean position) and show that amplitude is the distance from that middle line to the furthest point the bob reaches.

🎯 Exam Tip: Always use the phrase "from the mean position" when defining amplitude to get full marks.

Question 8. What is the SI unit of amplitude?
Answer: SI unit of amplitude is metre (m).
In simple words: Since amplitude is just a distance (how far it swings), we measure it in meters.

📝 Teacher's Note: Remind students that amplitude is a length measurement, so it shares the same unit as any other distance.

🎯 Exam Tip: Don't get confused by "vibrations"—if it's a distance, the unit is meters.

Question 9. Define a seconds pendulum.
Answer: A seconds pendulum is a pendulum which takes 2 seconds to complete one oscillation. The length of seconds pendulum, where \( g = 9.8 \text{ ms}^{-2} \), is nearly 1 m.
In simple words: It is a pendulum where one full swing takes exactly two seconds.

📝 Teacher's Note: This is a repeat question from earlier in the text; use it to quiz the students on their memory of the definition.

🎯 Exam Tip: If the question asks for the "length," approximately \( 1 \) metre (or \( 99.4 \) cm) is the expected answer.

Question 10. Calculate the length of a seconds pendulum using the formula.
Answer: \( G = 9.8 \text{ ms}^{-2} \)
Time, \( T \) for seconds pendulum = \( 2 \text{ s} \)
Now, \( T = 2\pi \sqrt{l/g} \)
\( \implies \) \( T^2 = 4\pi^2 l/g \)
\( \implies \) \( L = T^2g/4\pi^2 \)
\( \implies \) \( = 2^2 \times 9.8 / 4 \times 3.14^2 \)
\( \implies \) \( = 0.994 \text{ m} \)
In simple words: By using the math formula for swings and gravity, we find that a pendulum needs to be \( 0.994 \) meters long to take exactly two seconds for a full swing.

📝 Teacher's Note: Walk students through the algebra of squaring both sides to remove the square root. This is a fundamental skill in physics problems.

🎯 Exam Tip: Show every step of the formula rearrangement to earn method marks even if your final decimal calculation is slightly off.

Question 11. Why does the time period of a swing decrease when a person stands up?
Answer: The time period of swing decreases because in the standing position, the centre of the mass of the person shifts upwards. Due to which the effective length of the pendulum, \( l \) decreases. As \( T \propto \sqrt{l} \), therefore time decreases.
In simple words: When you stand up on a swing, you move your weight higher. This makes the "swing length" shorter, and shorter swings happen faster.

📝 Teacher's Note: Use this as a practical application of the pendulum theory. Effective length is from the pivot to the center of weight (center of mass).

🎯 Exam Tip: Mention the relationship \( T \propto \sqrt{l} \) to provide a scientific reason for the time decrease.

Question 12. How does the time period change if a pendulum is taken to the moon?
Answer: When a pendulum is taken from earth to moon surface, its time period will increase because the acceleration due to gravity on moon is less than that on earth and the time period depends inversely on square root of acceleration due to gravity.
In simple words: Because the moon's gravity is weaker, it doesn't "pull" the pendulum back down as fast. This makes the swing take more time.

📝 Teacher's Note: Explain that on the moon, everything "moves in slow motion" including pendulums because gravity is only \( 1/6^{th} \) of Earth's.

🎯 Exam Tip: Use the term "inversely proportional" to describe the relationship between time and gravity.

Question 13. What happens if the time period of a pendulum becomes infinite?
Answer: If time period of a pendulum becomes infinite, the pendulum will not oscillate at all as pendulum will take infinite time to complete one oscillation.
In simple words: If the time for one swing is "forever," it means the pendulum has stopped moving entirely.

📝 Teacher's Note: This would happen in deep space where there is zero gravity to pull the bob back down.

🎯 Exam Tip: Clearly state that "it will not oscillate" for this conceptual question.

Question 14. Define the effective length of a simple pendulum.
Answer: Effective length of a simple pendulum is the distance of the point of oscillation (i.e. the centre of the gravity of bob) from the point of suspension.
In simple words: It is the distance from the top hook (where it hangs) all the way down to the exact middle of the heavy ball.

📝 Teacher's Note: Make sure students understand that they must add the radius of the bob to the length of the string to get the "effective length."

🎯 Exam Tip: Use the terms "point of suspension" and "centre of gravity" to give a professional definition.

Question 15. Show a graph of \( l \) against \( T^2 \) for a simple pendulum.
Answer: A graph of \( l \) against \( T^2 \) for a simple pendulum of length \( l \) and time period \( T \) is shown in the figure: In simple words: This graph is a straight line, which proves that if you make the string longer, the square of the time it takes to swing also grows in a perfectly steady way.

📝 Teacher's Note: This straight-line graph starting from the origin shows that length and the square of the time period are directly proportional.

🎯 Exam Tip: A straight-line graph through the origin always indicates a direct proportionality between the two quantities on the axes.

Question 16. State the laws of proportionality for the time period of a simple pendulum.
Answer: The time period of simple pendulum is directly proportional to the square root of the length of the pendulum.
The time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.
\( T = 2\pi \sqrt{l/g} \)
Where \( T \) is the time period,
\( l \) is length of the pendulum,
\( g \) is acceleration due to gravity.
In simple words: If you increase the length, the time goes up. If you increase gravity (like moving to a bigger planet), the time goes down.

📝 Teacher's Note: Focus on the words "directly" and "inversely" to help students understand the direction of change.

🎯 Exam Tip: Providing the formula \( T = 2\pi \sqrt{l/g} \) is the best way to summarize these laws in one step.

Question 17. Calculate the ratio of time periods for two pendulums with different gravity.
Answer: \( g_1 = 9.8 \text{ ms}^{-2} \)
\( g_2 = 4.36 \text{ ms}^{-2} \)
Now, time period of pendulum is, \( T = 2\pi \sqrt{l/g} \)
\( \implies \) \( T_1 = 2\pi \sqrt{l/g_1} \)
\( \implies \) \( T_1 = 2\pi \sqrt{l/9.8} = 2\pi \times 0.3194 \times 1 = 2.006 \text{ l} \)
Similarly,
\( \implies \) \( T_2 = 2\pi \sqrt{l/g_2} \)
\( \implies \) \( T_2 = 2\pi \sqrt{l/4.36} = 2\pi \times 0.4789 \times 1 = 3.007 \text{ l} \)
\( T_1 : T_2 = 2.0061 : 3.0071 \)
\( = 0.667 : 1 \)
In simple words: This math shows that if gravity is weaker (the second planet), the pendulum swings slower, making the ratio roughly \( 2 \) to \( 3 \).

📝 Teacher's Note: This problem assumes the length is the same for both. It shows students that even without changing the string, a pendulum's time changes if gravity changes.

🎯 Exam Tip: In ratio questions, simplify your final answer to the smallest possible whole numbers or a normalized decimal like \( X : 1 \).

Question 18. Does the mass of the bob affect the time period of a pendulum?
Answer: The time period of a pendulum is independent of mass of the bob.
In simple words: No matter if the ball is made of heavy lead or light plastic, as long as the string length is the same, it will take the same amount of time to swing.

📝 Teacher's Note: Students find this very surprising. Use two pendulums with the same length but different masses to prove it in class.

🎯 Exam Tip: "Independent of mass" is the key phrase to use here.

Question 19. If the ratio of time periods is 2:1, what is the ratio of their lengths?
Answer: \( T_1 : T_2 = 2 : 1 \)
But since \( T \propto \sqrt{l} \)
So
\( \implies \) \( \frac{T_1}{T_2} = \frac{\sqrt{l_1}}{\sqrt{l_2}} = \frac{2}{1} \)
\( \implies \) \( \frac{l_1}{l_2} = \frac{2^2}{1^2} \)
\( \implies \) \( = 4/1 \)
In simple words: To make a pendulum swing twice as slow, you don't just double the string—you have to make it four times longer!

📝 Teacher's Note: Since time is proportional to the square root of length, if you square the time ratio, you get the length ratio.

🎯 Exam Tip: Remember to square both sides of the ratio to find the length ratio from the time ratio.

Question 20. Define mass and how it is measured.
Answer: The quantity of matter contained in a body is called its mass. Mass of a body can be measured by using a beam balance. Mass is always constant for a given body.
In simple words: Mass is just how much "stuff" is inside an object. We measure it with a balance scale, and it stays the same even if you go to space.

📝 Teacher's Note: Differentiate mass from weight. Mass is "stuff," weight is "gravity's pull on stuff." This is a fundamental physics distinction.

🎯 Exam Tip: Use the words "quantity of matter" and "constant" for a high-scoring definition.

Question 21. State the principle of a beam balance.
Answer: A beam balance works on the principle of moments. According to the principle of moments, under equilibrium condition, the clockwise moment due to the body on one side of beam equals the anti clockwise moment due to standard weights on the other side of beam.
In simple words: A scale balances when the "turning force" on the left side is exactly the same as the "turning force" on the right side.

📝 Teacher's Note: Relate this to a see-saw. If both kids have the same weight and are at the same distance, the see-saw stays flat.

🎯 Exam Tip: Keywords are "principle of moments," "equilibrium," and "clockwise/anti-clockwise moment."

Question 22. List the precautions for measuring mass using a beam balance.
Answer: Precautions to be taken to measure the mass of a body using beam balance are:

• The beam must be gently lowered before adding or removing weights from the pan.
• The weights should not be carried with bare hands to avoid the change in weights due to moisture and dust particles from the surrounding.
• The lever should be turned gently, in order to prevent knife edges from chipping.
• Never keep the wet or hot objects on the pan.
• The weights should be placed into weight box after use.
• Whenever you are near the actual weight, you should carefully try the weights in the descending order.

In simple words: Be very careful with the scale! Don't touch weights with your fingers (the oil from your skin adds weight), keep the scale clean, and always start with the biggest weights first when testing.

📝 Teacher's Note: Show students a real weight box and the forceps used to pick up the weights. Explain that even a fingerprint can change a precise measurement.

🎯 Exam Tip: Mentioning the "forceps" or "descending order of weights" are specific technical points that show you know lab procedures.

Question 23. State the SI units of time and mass.
Answer: SI units of time and mass are second (s) and kilogram (kg) respectively.
In simple words: In science, we always try to measure time in seconds and weight (mass) in kilograms.

📝 Teacher's Note: These are the standard "building blocks" of the metric system (SI system).

🎯 Exam Tip: Use the standard symbols \( (s \text{ and } kg) \) correctly.

Question 24. What are the conditions for a beam balance to be "true"?
Answer: Conditions for a beam balance to be true are:

• Both the pans must be of equal weights.
• Both the arms must be of equal lengths.

In simple words: For a scale to be fair and accurate, both sides must be identical—the hanging pans must weigh the same, and the metal arms holding them must be the same length.

📝 Teacher's Note: If the arms are different lengths, even identical weights won't balance because of the "lever effect."

🎯 Exam Tip: "Equal weights" and "Equal lengths" are the two critical conditions for balance integrity.

Exercise Page No: 44

Question 1. Define the term "Least Count."
Answer: Least count of an instrument refers to the smallest reading that can be accurately measured while using the instrument. For an instrument provided with a scale the least count is the value of one division on its scale.
In simple words: Least count is simply the smallest "gap" or mark on your ruler or tool—it is the tiniest bit you can actually measure.

📝 Teacher's Note: On a standard school ruler, the least count is \( 1 \) mm. It's the "resolution" of the tool.

🎯 Exam Tip: Mention "smallest reading" and "one division" for a perfect definition.

Question 2. What is the maximum possible error for a standard ruler?
Answer: Maximum possible error is 0.1 cm.
In simple words: When you use a regular ruler, you might be off by up to one millimeter (\( 0.1 \) cm) just because of how the marks are spaced.

📝 Teacher's Note: The error of an instrument is typically equal to its least count. Since a ruler has \( 1 \) mm markings, the error is \( 1 \) mm (\( 0.1 \) cm).

🎯 Exam Tip: Use centimeters (\( 0.1 \) cm) if the question asks for standard units, or millimeters (\( 1 \) mm) if it asks for the smallest scale unit.

Question 3. What does the slope of a graph represent?
Answer: Slope of a graph identifies the proportional relationship between the quantities plotted.
In simple words: The slope tells you how fast the line is rising; it shows how one measurement changes when the other one changes.

📝 Teacher's Note: Explain that a steeper slope means a faster change. In physics, the slope often gives us a new value, like speed or gravity.

🎯 Exam Tip: Use the word "proportional relationship" to sound scientifically accurate.

Question 4. List the least count of common measuring instruments.
Answer: Least count of the following are:
(i) Metre scale = \( 0.1 \text{ cm} \)
(ii) Vernier calipers = \( 0.01 \text{ cm} \)
(iii) Screw gauge = \( 0.001 \text{ cm} \)
In simple words: This shows that a screw gauge is 10 times more precise than a vernier caliper, and 100 times more precise than a regular ruler!

📝 Teacher's Note: Note how each tool adds one more decimal place of accuracy. This is why we choose different tools for different jobs.

🎯 Exam Tip: Pay attention to the units—these are all in cm. If writing in mm, the values would be \( 1 \text{ mm} \), \( 0.1 \text{ mm} \), and \( 0.01 \text{ mm} \).

Question 5. State the least count of the following instruments:
(i) Stopwatch
(ii) Thermometer
(iii) Spring balance
Answer: Least count of the following are
(i) Stopwatch = \( 0.1 \) s
(ii) Thermometer = \( 0.1 \) °C
(iii) Spring balance = \( 1 \) gm wt
In simple words: The least count is the smallest value you can accurately read on these tools, like the tiny marks on a ruler.

Í Teacher's Note: Show students real laboratory instruments and ask them to count the divisions between two major marks to determine the least count themselves.

🎯 Exam Tip: Always include the correct units (s, °C, gm wt) when writing the least count of an instrument.

Question 6. Define Accuracy in measurement.
Answer: Accuracy is the extent to which a reported measurement approaches the true value of the quantity measured. This extent is usually described by the least count of the instrument and since the least count for a given instrument is limited hence, the accuracy of the instrument is limited.
In simple words: Accuracy tells us how close our measurement is to the real, perfect size of an object. It depends mostly on how good and precise our measuring tool is.

📝 Teacher's Note: Use the analogy of an archer hitting a bullseye to explain accuracy versus precision.

🎯 Exam Tip: Mention the "least count of the instrument" to explain why measurement accuracy has a limit.

Question 7. Describe the two types of error in a measurement.
Answer: Two types of error in a measurement are

• Random errors - these errors are due to various factors. In a number of observations we get different readings every time. These errors can be minimized by taking observations a large number of times and taking the arithmetic mean of the readings.
• Gross error - these errors are due to carelessness of the observer like parallax, improper setting of the instrument. These errors can be minimized only when the observer is careful in setting up of instrument and taking readings.

In simple words: Random errors happen by chance and are fixed by taking many measurements and averaging them. Gross errors happen because of human mistakes, like looking at a ruler from the wrong angle.

📝 Teacher's Note: Demonstrate parallax error by having students read a scale from the left, right, and center to see how the value "changes."

🎯 Exam Tip: To fix random errors, use the keyword "arithmetic mean." For gross errors, emphasize "careful observation."

Question 8. Why is 3000g considered a more accurate measurement than others in a specific set?
Answer: 3000g is the most accurate measurement because it has maximum number of significant figures = 4.
In simple words: Measurements with more digits (significant figures) are more precise because they show that the tool used was capable of measuring very tiny differences.

📝 Teacher's Note: Explain that the trailing zeros in "3000g" are significant if they come from a measurement process, showing precision to the nearest gram.

🎯 Exam Tip: Higher number of significant figures always implies higher precision and accuracy in a scientific context.

Question 9. Compare the accuracy of three measurements based on significant figures.
Answer: Basically there is no difference between the quantity being measured but there is a difference of significant figures in the measurement.
1. Number of significant figures = 3
2. Number of significant figures = 4
3. Number of significant figures = 5
Since (3) part has maximum number of significant figures = 5, therefore it is most accurate among the given three.
In simple words: Even if the amount is the same, writing it with more decimal places shows that a more sensitive tool was used, making it more accurate.

📝 Teacher's Note: Provide examples like 2.1 m, 2.10 m, and 2.100 m to show how significant figures change without changing the total length.

🎯 Exam Tip: When asked to find the "most accurate" measurement, look for the one with the highest number of significant digits.

Question 1. Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.
Answer: Unit is a standard quantity of the same kind with which a physical quantity is compared for measuring it.
In simple words: A unit is like a fixed "yardstick" (like a centimeter) that we use to see how many of them fit into the object we are measuring.

📝 Teacher's Note: Use the example of measuring a desk with "hand-spans" versus a "meter rule" to show why we need standard units.

🎯 Exam Tip: Use the words "standard quantity" and "compared" to form a technically correct definition.

Question 2. Define fundamental units.
Answer: The units which can neither be derived from one another, nor can they be further resolved into other units are known as fundamental units.
In simple words: Fundamental units are the "base ingredients" of measurement, like Mass (kg), Length (m), and Time (s), that don't depend on anything else.

📝 Teacher's Note: Compare fundamental units to primary colors; you can't make red from other colors, but you can use red to make many other shades.

🎯 Exam Tip: Remember that fundamental units are independent and cannot be broken down further.

Question 3. Define derived units.
Answer: The units which can be expressed in terms of fundamental units of mass, length and time are known as derived units.
In simple words: Derived units are "recipes" made by combining base units—for example, Speed is made by dividing distance (Length) by Time.

📝 Teacher's Note: Show the derivation of Area (\( \text{m}^2 \)) or Velocity (\( \text{m/s} \)) on the board to illustrate how fundamental units combine.

🎯 Exam Tip: Provide an example like "Newton" or "Watt" to show you understand how they are built from kg, m, and s.

Question 4. Define a standard metre.
Answer: A standard metre is equal to \( 1650763.31 \) wavelengths in vacuum, of the radiation from krypton isotope of mass \( 86 \).
In simple words: Scientists defined a meter very precisely by counting a huge number of light waves from a specific element called Krypton.

📝 Teacher's Note: Explain that this definition was chosen because the speed of light and wavelengths are constant and don't change like a physical metal bar might.

🎯 Exam Tip: Memorize the isotope "Krypton-86" as it is a specific detail often required in exams.

Question 5. Define a standard kilogram.
Answer: One standard kilogram is equal to the mass of a cylinder of nearly same height and diameter and made up of platinum and iridium alloy.
In simple words: A kilogram is defined by the exact weight of a special metal cylinder kept in a laboratory in France.

📝 Teacher's Note: Platinum and Iridium are used because they are very stable and do not corrode or react with the environment easily.

🎯 Exam Tip: Mention the "Platinum-Iridium alloy" to show a complete scientific understanding of the standard kilogram.

Question 6. State the SI unit of electric current.
Answer: SI unit of electric current is Ampere (A).
In simple words: An Ampere measures how much electricity is flowing through a wire, similar to how we measure how much water flows through a pipe.

📝 Teacher's Note: Remind students that the unit symbol "A" is capitalized because it is named after the scientist André-Marie Ampère.

🎯 Exam Tip: Always use the capital letter 'A' for the symbol, even if you write 'ampere' in lowercase.

Question 7. Define a light year.
Answer: Light year is defined as the distance travelled by light in vacuum in one year.
In simple words: Even though it has the word "year," a light year measures a huge distance—how far light can travel in a whole year.

📝 Teacher's Note: This is a common "trick" question. Emphasize that light year is a unit of distance, not time.

🎯 Exam Tip: Include "in vacuum" in your definition to ensure full marks.

Question 8. Which is bigger: 1 Parsec or 1 Light Year?
Answer: 1 Parsec is bigger because 1 Parsec is \( 3.26 \) times a light year.
In simple words: A Parsec is about three times longer than a light year, used to measure even greater distances in space.

📝 Teacher's Note: Use a simple diagram to show that while light years are huge, astronomers need even larger units like parsecs for measuring distance to other galaxies.

🎯 Exam Tip: Use the numerical ratio "\( 3.26 \) times" to justify why Parsec is larger.

Question 9. Which is smaller: 1 Fermi or 1 Micron?
Answer: 1 Fermi is smaller because 1 Fermi is \( 10^{-15} \text{ m} \) while 1 micron is \( 10^{-6} \text{ m} \).
In simple words: A Fermi is trillions of times smaller than a micron, used to measure the tiny parts inside an atom.

📝 Teacher's Note: Explain negative exponents: \( 10^{-15} \) has much more zeros after the decimal than \( 10^{-6} \), making it significantly smaller.

🎯 Exam Tip: Writing the power of 10 values for both (\( 10^{-15} \) vs \( 10^{-6} \)) is the most effective way to answer comparison questions.

Question 10. Define Parsec and its relation to Astronomical Unit (A.U.).
Answer: Parsec refers to the distance at which an arc of length equal to 1 astronomical unit subtends an angle of one second at a point. No, parsec is not same as astronomical unit (A.U.). 1 Parsec = \( 2 \times 10^5 \) A.U.
In simple words: A parsec is a huge distance used in space math. It is equal to about \( 200,000 \) times the distance between the Earth and the Sun (AU).

📝 Teacher's Note: "Subtends an angle of one second" refers to a very tiny angle (1/3600 of a degree), showing how far away you'd have to be for the Earth-Sun distance to look that small.

🎯 Exam Tip: Clearly state that they are not the same and give the approximate conversion value.

Question 11. State the least count of a vernier caliper used in laboratory.
Answer: Least count of a vernier caliper used in laboratory is \( 0.1 \text{ mm} = 0.01 \text{ cm} \).
In simple words: A laboratory vernier caliper can measure things as tiny as one-tenth of a millimeter.

📝 Teacher's Note: Help students convert between mm and cm correctly, as this is a common source of calculation errors in labs.

🎯 Exam Tip: Providing the value in both mm and cm (\( 0.1 \text{ mm} \) and \( 0.01 \text{ cm} \)) shows a thorough understanding.

Question 12. Why is a vernier caliper important for measuring small lengths?
Answer: Vernier caliper is an instrument used for measuring small lengths of solid objects where an ordinary scale cannot be applied. We can measure the length accurately up to the order of \( 10^{-2} \text{ cm} \), \( 10^{-3} \text{ cm} \) depending upon the vernier used. Therefore a vernier caliper is important to measure the fraction of a smallest division of a measuring scale which otherwise could not be done by the judgment of the eye.
In simple words: It lets us measure tiny gaps that are smaller than the lines on a regular ruler, which we normally couldn't see accurately with just our eyes.

📝 Teacher's Note: Explain that the "Vernier scale" slides along the "Main scale" to reveal these hidden fractions of a millimeter.

🎯 Exam Tip: Focus on the phrase "measure the fraction of a smallest division" as this is the primary purpose of the Vernier scale.

Question 13. Define the least count of an instrument.
Answer: Least count of an instrument refers to the smallest reading that can be accurately measured while using the instrument. For an instrument provided with a scale the least count is the value of one division on its scale.
In simple words: Least count is the smallest "step" a tool can measure—like 1mm on a standard ruler.

📝 Teacher's Note: On a standard 15cm ruler, the smallest lines are 1mm apart; that is its least count.

🎯 Exam Tip: Use the term "value of one division" to define least count for scaled instruments.

Question 14. Can we measure the thickness of a paper with a vernier caliper? Explain.
Answer: No, we cannot measure the thickness of a paper with vernier caliper as its least count is only \( 0.1 \text{ mm} \). We should use screw gauge instead as its least count is \( 0.01 \text{ mm} \) as the thickness of the paper is in the range of \( 10^{-2} \text{ mm} \).
In simple words: A sheet of paper is too thin for a vernier caliper to catch. We need a screw gauge because it can measure much thinner things.

📝 Teacher's Note: Show students that if they try to measure paper with a vernier, the jaws appear closed, whereas a screw gauge will show a clear reading.

🎯 Exam Tip: Always compare the "least count" of the two instruments to justify your choice of tool.

Question 15. What is zero error and what are its types?
Answer: If the zero of the one scale (vernier scale or circular scale of screw gauge) does not coincide with the zero of the main scale, this is known as zero scale, zero error arises. There are two types of zero error –

• If the zero of the scale remains below the line of graduation of the main scale then it is called positive zero error
• If the zero of the scale lies above the line of graduation of the main scale then it is called negative zero error

In simple words: Zero error is when a tool doesn't start exactly at 0 when it's closed. It can be positive (starts too high) or negative (starts too low).

📝 Teacher's Note: Think of it like a weight scale; if it shows "2 kg" when no one is standing on it, that's a positive zero error.

🎯 Exam Tip: Remember: Positive error is subtracted from the reading, and negative error is added to it.

Question 16. Why is a screw gauge called by that name?
Answer: Screw gauge consists essentially of a screw with a uniform pitch which moves in a nut, thus it is named as screw gauge because the major working part is a screw.
In simple words: It gets its name because its main part is a long screw that you turn to take measurements.

📝 Teacher's Note: Unscrew the thimble of a screw gauge to show students the internal screw threads and how they relate to the pitch.

🎯 Exam Tip: Mention the "screw and nut" mechanism to explain the name and working principle.

Question 17. Define the time period of a simple pendulum and state its mathematical expression.
Answer:
The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum. \[ T = 2\pi \sqrt{\frac{l}{g}} \] Where \( T \) is the time period,
\( l \) is length of the pendulum,
\( g \) is acceleration due to gravity
In simple words: The time period is simply the amount of time (in seconds) it takes for a pendulum to swing all the way forward and back to its starting point once.

📝 Teacher's Note: Use a classroom demonstration to show that only the length of the string affects the time period. Students often incorrectly assume that a heavier weight (bob) will make it swing faster.

🎯 Exam Tip: Always define the variables used in your formula. Forgetting to state that 'l' is the length and 'g' is acceleration due to gravity can result in a loss of marks.

Question 18. Define the time period of a simple pendulum and state its mathematical expression.
Answer:
The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum. \[ T = 2\pi \sqrt{\frac{l}{g}} \] Where \( T \) is the time period,
\( l \) is length of the pendulum,
\( g \) is acceleration due to gravity
In simple words: Imagine a swing at the park; the time it takes for one full trip out and back is its time period. A longer swing takes more time for each trip!

📝 Teacher's Note: Explain that "one oscillation" means the complete cycle from the mean position to one extreme, then to the other extreme, and back to the mean.

🎯 Exam Tip: Remember the units! Time period \( T \) is measured in seconds (s), length \( l \) in meters (m), and \( g \) in \( \text{m/s}^2 \). Using standard SI units prevents calculation errors.

Question 19. Why is a screw gauge made of stainless steel?
Answer: Material used for making screw gauge is stainless steel to avoid expansion and contraction due to change in weather as stainless steel absorbs a little heat.
In simple words: We use stainless steel because it doesn't change its size much when the weather gets hot or cold, keeping the measurements accurate.

📝 Teacher's Note: Discuss thermal expansion. If the tool grew in summer, its marks would be in the wrong place!

🎯 Exam Tip: Use the keyword "thermal expansion" or "change in weather" to explain why durable materials are used for precision tools.

Question 20. Define pitch of a screw and provide its formula.
Answer: When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.

\( \implies \) Pitch of the screw = \( \frac{\text{distance traveled by screw in n rotations}}{\text{n rotations}} \)
In simple words: Pitch is how far the screw moves in or out when you turn the handle exactly one full circle.

📝 Teacher's Note: On the board, draw a screw and highlight the gap between two ridges; this physical distance is the pitch.

🎯 Exam Tip: Be sure to write the formula. If distance is given for 5 rotations, divide by 5 to find the pitch.

Question 21. How is a negative zero error corrected?
Answer: If the zero of the circular scale lies above the line of graduation then it is called negative zero error. When there is negative zero error, then the instrument reads less than the actual reading. Therefore in order to get the correct reading, the zero error should always be added from the observed reading.
In simple words: If the tool starts "below zero," it will give you a result that is too small. You have to add that missing amount back to get the real answer.

📝 Teacher's Note: Remind students of the "Opposite Rule": Negative error? Addition. Positive error? Subtraction.

🎯 Exam Tip: In problems, always calculate the zero error first, then determine the sign for the final correction.

Question 22. How is backlash error avoided?
Answer: Due to constant use, there is space for the play of screw gauge but gradually this space increases with the use or wear and tear, so that when the screw is moved by rotating it in some direction, it slips in the nut and does not cover any linear distance for some rotation of the screw head. The error due to this is known as backlash error. It is avoided by turning the screw always in the same direction.
In simple words: Over time, the screw gets a bit loose inside. To keep it accurate, always turn the knob in the same direction when making a measurement.

📝 Teacher's Note: Use the analogy of a loose steering wheel in an old car; you have to turn it slightly before the wheels actually move.

🎯 Exam Tip: The phrase "turning the screw always in the same direction" is the key to this answer.

Question 23. Why do screws twist while nails slide straight?
Answer: A screw are threaded to twist in, when turned with a screw driver while nails are smooth to slide in straight when pounded with hammer.
In simple words: Screws have ridges (threads) that pull them in as you turn them, while nails are smooth so they can be pushed straight in with a hit.

📝 Teacher's Note: This illustrates the mechanical advantage of the inclined plane (threads) versus simple force.

🎯 Exam Tip: This is a conceptual question about the physical structure—use the word "threaded" for screws and "smooth" for nails.

Question 24. What are the two motions of a screw in a screw gauge?
Answer: Screw has two types of motions: linear and circular motions.
In simple words: As you turn the handle (circular motion), the screw also moves forward or backward along a straight line (linear motion).

📝 Teacher's Note: This is why the tool has two scales: one for the circular turn and one for the linear distance moved.

🎯 Exam Tip: Mention both "linear" and "circular" to show the relationship between the two scales on the instrument.

Answer 25: Unit of Least count of an instrument is cm.

Question 26. Express 1 micron in meters.
Answer: \( 1 \text{ micron} = 10^{-6} \text{ m} \).
In simple words: A micron is a millionth of a meter—it's about the size of a single bacterium.

📝 Teacher's Note: Microns are commonly used in biology to measure cells. 1,000,000 microns make up 1 meter.

🎯 Exam Tip: Standard scientific notation (\( 10^{-6} \)) is better than writing decimals with many zeros.

Question 27. State the principle of moments as applied to a beam balance.
Answer: A physical balance works on the principle of moments. According to the principle of moments, under equilibrium condition, the clockwise moment due to the body on one side of beam equals the anti clockwise moment due to standard weights on the other side of beam.
In simple words: A scale balances when the "turning force" on the left is exactly the same as the "turning force" on the right.

📝 Teacher's Note: Think of a seesaw: it stays flat only when the weight and distance on both sides match up perfectly.

🎯 Exam Tip: Use the terms "equilibrium," "clockwise moment," and "anti-clockwise moment" for full marks.

Question 28. Define light year and give its value in meters.
Answer: \( 1 \text{ light year} = 9.46 \times 10^{15} \text{ m} \).
In simple words: A light year is about 9.5 trillion kilometers! It's a massive distance used only for space.

📝 Teacher's Note: Ask students to try and calculate this by multiplying the speed of light (\( 3 \times 10^8 \text{ m/s} \)) by the number of seconds in a year (\( 365 \times 24 \times 3600 \)).

🎯 Exam Tip: Memorize the value \( 9.46 \times 10^{15} \text{ m} \) as it is a common constant in physics problems.

Question 29. Sketch the graph of \( T^2 \) versus \( l \) for a simple pendulum.
Answer: Graph of \( T^2 \) versus \( l \) for a simple pendulum is a straight line passing through the origin. In simple words: This graph shows that as you make the string longer, the time it takes for a swing increases in a very predictable, straight-line way.

📝 Teacher's Note: This straight line through the origin proves that \( T^2 \) is directly proportional to length \( l \). The slope of this line can be used to find the value of gravity \( g \).

🎯 Exam Tip: Always label your axes with the quantity and its unit—\( T^2 \) in \( \text{s}^2 \) and \( l \) in \( \text{m} \).

Question 30. Sketch the graph of \( T \) versus \( l \) for a simple pendulum.
Answer: Graph of \( T \) versus \( l \) for a simple pendulum is a curve (parabola). In simple words: This graph is a curve, showing that doubling the length doesn't just double the time; the relationship follows a square-root rule.

📝 Teacher's Note: Contrast this with the \( T^2 \) vs \( l \) graph. This one is a curve because \( T \propto \sqrt{l} \), while the other is straight because \( T^2 \propto l \).

🎯 Exam Tip: Do not use a ruler to draw this line; it must be a smooth, freehand curve starting from the origin.

Question 31. Is vibration the same as oscillation?
Answer: Yes, the vibration is same as the oscillation.
In simple words: Both words mean a back-and-forth motion around a center point, like a guitar string or a swing.

📝 Teacher's Note: In higher physics, "vibration" is often used for fast, high-frequency motion, while "oscillation" is used for slower motion, but their basic definition is identical.

🎯 Exam Tip: A simple "Yes" with a brief description of back-and-forth motion is sufficient here.

Question 32. How are time period and frequency of oscillation related?
Answer: The time period, \( T \) and frequency of oscillation, \( f \) are related as, \( T = 1/f \) or \( f = 1/T \).
In simple words: They are opposites: if you know how many seconds one swing takes, you can calculate how many swings happen in one second.

📝 Teacher's Note: Use a simple example: If a swing takes 0.5 seconds (\( T \)), it will happen 2 times per second (\( f = 1/0.5 \)).

🎯 Exam Tip: Mention the inverse relationship to show you understand the mathematical link between the two.

Question 33. Describe an ideal pendulum.
Answer: An ideal pendulum is a simple pendulum consists a heavy mass (called the bob) considered as a point mass suspended by a thread which is considered to be mass less and inextensible or non-elastic, from a fixed point or rigid support and in which there is no friction between the support and the string.
In simple words: An ideal pendulum is a "perfect" version that only exists in theory, where the string has no weight, doesn't stretch, and there is no air resistance to slow it down.

📝 Teacher's Note: Explain that we use "ideal" models to make math easier, then add in "real world" factors like air friction later.

🎯 Exam Tip: Keywords are "point mass," "massless thread," "inextensible," and "no friction."

Question 34. Why does a pendulum clock run faster in winter?
Answer: Wall clock with a pendulum will run at a faster rate in winter as it pendulum rod get shorter and the pendulum will swing at a faster rate thus the clock would run faster in winters.
In simple words: Cold weather makes metal shrink. Since a shorter pendulum swings faster, the clock finishes its "seconds" too quickly and gets ahead of time.

📝 Teacher's Note: This is a real-world application of thermal contraction. In summer, the rod expands and the clock runs slow.

🎯 Exam Tip: Connect the change in temperature to the change in "effective length" to explain the change in time period.

Question 35. Why is measurement needed in the real world?
Answer: Measurement is needed for precise description of any phenomenon happening in the world. For example, if a body is freely falling down to the ground, to understand this phenomenon we must know its velocity, time it will take to reach the ground, etc and to get answer to all our questions we need measurement.
In simple words: We need measurement so we can describe things exactly with numbers rather than just guessing "fast" or "heavy."

📝 Teacher's Note: Ask students how they would buy clothes or cook a recipe without measurement to show its everyday importance.

🎯 Exam Tip: Use the phrase "precise description" to define the necessity of measurement in science.

Question 36. Distinguish between fundamental units and derived units.
Answer:

In simple words: Fundamental units are the "base ingredients" (like bricks), while derived units are what you build with them (like a wall).

📝 Teacher's Note: Help students practice breaking down a derived unit like Newton (\( \text{kg} \cdot \text{m/s}^2 \)) into its fundamental parts.

🎯 Exam Tip: Always provide examples for both categories (e.g., metre for fundamental and speed for derived) in a comparison table.

Question 37. List the fundamental SI units and their symbols.
Answer:

In simple words: These are the seven basic units used by scientists all over the world to make sure everyone is measuring exactly the same way.

📝 Teacher's Note: Note that Radian (angle) and Steradian (solid angle) are supplementary SI units often added to this list.

🎯 Exam Tip: Pay attention to capitalization—units named after people (Kelvin, Ampere) have capitalized symbols (K, A).

Question 38. Why is it essential to maintain standard units?
Answer: The maintenance of standard units is essential because any variation in these standards would lead to wrong measurements, misleading results and confusing generalizations. The standards are preserved in such a way that they do not undergo any change with the change in temperature, pressure, humidity and other environmental changes.
In simple words: We need standard units so a "kilogram" is the same in every country. This prevents confusion in trade and ensures science experiments work correctly everywhere.

📝 Teacher's Note: Explain that if every shopkeeper had their own version of a "meter," buying 5 meters of cloth would result in different sizes at every shop.

🎯 Exam Tip: Use the term "universal acceptance" or "lack of variation" to describe why standards are needed.

Question 39. State the main characteristics of a standard unit.
Answer: Main characteristics of a standard unit are as follows:

• It must be well defined.
• It must be of proper size. Very small or large size may cause inconvenience.
• It should be easily accessible.
• It must be reproducible at all places without any difficulty.
• It must be accurately defined and must not change with time, place and physical conditions such as pressure, humidity, etc.
• It must be widely acceptable all over the world.

In simple words: A good unit should be easy to use, easy to share, and never change its size no matter the weather or location.

📝 Teacher's Note: Ask students why we don't use "the length of the King's foot" as a standard anymore. (It changes when the King grows or a new King takes over!)

🎯 Exam Tip: "Widely acceptable" and "Reproducible" are the two most important characteristics of a scientific standard.

Question 40. Define fundamental units and provide examples.
Answer: The units which can neither be derived from one another, nor can they be further resolved into other units are known as fundamental units. Some of the fundamental units are metre (length), kilogram (mass), second (time), Kelvin (temperature), ampere (current), etc.
In simple words: They are the basic "blocks" of science. You can't break a "second" down into other measurements.

📝 Teacher's Note: Reinforce that there are only 7 fundamental quantities in the modern SI system.

🎯 Exam Tip: This is a common 2-mark question; define the term and list at least 3 examples to get full marks.

Question 41. Observe the provided diagrams for Vernier Calipers and determine the zero error and correction.
Answer:
(a) Here, zero of the vernier scale coincides with the zero of the main scale,
So, the zero error and zero error correction is nil.
(b) Here, zero of the vernier scale lies on the right of the zero of the main scale.
Also, \( 5^{th} \) vernier division is coinciding with some main scale division
Thus, zero error = \( 0 + 5 \times L.C. = 0.05 \text{ cm} \)
Hence zero correction = \( -0.05 \text{ cm} \)
(c) Here, zero of the vernier scale lies on the left of the zero of the main scale.
Also, \( 5^{th} \) vernier division is coinciding with some main scale division
Thus, zero error = \( 0 - 5 \times L.C. = -0.05 \text{ cm} \)
Hence zero correction = \( 0.05 \text{ cm} \)

In simple words: Zero error happens when the "zero" marks on the two scales don't line up perfectly when the tool is closed. If the vernier zero is to the right, it's a positive error; if it's to the left, it's a negative error.

📝 Teacher's Note: Use a physical Vernier caliper to show students how the zero marks might drift over time due to wear and tear. Explain that "correction" is just the opposite of "error" to fix the final reading.

🎯 Exam Tip: Always remember: Final Reading = Observed Reading - Zero Error (with its sign). A negative zero error will actually be added because subtracting a negative makes a positive.

Question 42. Calculate the least count of a screw gauge given its pitch and divisions.
Answer:
Pitch of the screw = \( 0.5 \text{ mm} \)
No of divisions on circular scale = \( 50 \)
\ \text{Least count} = \frac{\text{pitch of the screw}}{\text{no of divisions on the circular scale}} \
So, \( L.C. = \frac{0.5}{50} = 0.01 \text{ mm} \)

In simple words: The least count is the smallest distance a tool can measure. For a screw gauge, you find it by dividing the distance moved in one full turn (pitch) by the number of marks on its rotating handle.

📝 Teacher's Note: Demonstrate how one full rotation of the thimble moves the screw forward by the pitch distance. This visual help students understand the ratio formula.

🎯 Exam Tip: Always state the formula before substituting values to ensure step-marks even if your calculation goes wrong.

Question 43. Calculate the least count of a vernier caliper given the MSD and VSD.
Answer:
\( 1 \text{ MSD} = 1 \text{ mm} \)
\( 10 \text{ VSD} = 9 \text{ mm} \)
\( 1 \text{ VSD} = \frac{9}{10} \text{ mm} \)
Least count = \( 1 \text{ MSD} - 1 \text{ VSD} \)
\( = \frac{1}{10} \text{ MSD} \)
\( = \frac{1}{10} \times 1 \text{ mm} = 0.1 \text{ mm} \)
\( = 0.01 \text{ cm} \)

In simple words: The least count is the tiny difference between one main scale division and one vernier scale division. It allows the tool to measure parts of a millimeter.

📝 Teacher's Note: Highlight that the Vernier scale is slightly "shorter" than the main scale (9 mm spread over 10 divisions). This difference is what we call the least count.

🎯 Exam Tip: Be careful with units! Students often forget to convert from mm to cm. \( 0.1 \text{ mm} \) is the same as \( 0.01 \text{ cm} \).

Question 44. Define zero error in a screw gauge.
Answer:
If the zero of the circular scale does not coincide with the zero of the main scale (pitch scale) when the end of the movable screw is brought in contact with the fixed end then the screw gauge is said to have a zero error.

In simple words: A screw gauge has a zero error if it doesn't show exactly "0" when it is fully tightened and closed.

📝 Teacher's Note: Compare this to a bathroom scale that shows \( 2 \text{ kg} \) even when nobody is standing on it. That "starting mistake" is the zero error.

🎯 Exam Tip: In your definition, mention both the "circular scale zero" and "main scale zero" to be precise.

Question 45. Determine positive zero error in a screw gauge.
Answer:
In this case, the zero error is positive
Least count of screw gauge = \( 0.01 \text{ mm} \)
Thus, zero error = \( 0 + 4 \times L.C. = 0.04 \text{ mm} \)

In simple words: If the zero on the rotating scale is slightly below the reference line when closed, the error is positive. We multiply the division count by the smallest measurement possible (L.C.).

📝 Teacher's Note: Positive error means the tool is showing a value larger than zero when it should be zero. Therefore, we must subtract this from our final result.

🎯 Exam Tip: Use the plus sign explicitly for positive error to show the examiner you understood the direction of the shift.

Question 46. Determine negative zero error in a screw gauge.
Answer:
In this case, the zero error is negative
Least count of screw gauge = \( 0.01 \text{ mm} \)
Thus, zero error = \( (50-47) \times L.C. \)
\( = 3 \times 0.01 \)
\( = 0.03 \text{ mm} \)

In simple words: When the rotating zero goes past the reference line, the error is negative. We count how many divisions it went "over" from the total (like 50) and multiply by the least count.

📝 Teacher's Note: Note that in negative error calculations, you subtract the observed division from the total number of divisions on the circular scale.

🎯 Exam Tip: Remember that for negative zero error, the actual value used in the final formula is \( -0.03 \text{ mm} \).

Question 47. Can we measure the diameter of a wire by wrapping it around a pencil?
Answer:
No, we cannot measure the diameter of a wire by wrapping it around a pencil because it is not very accurate. We can use screw gauge for this purpose as it can measure the diameter correct up to \( 1/100 \) of millimeter or even less.

In simple words: Wrapping wire around a pencil is too "rough" and leaves gaps. A screw gauge is a specialized tool that can measure thickness thinner than a human hair.

📝 Teacher's Note: This is a good opportunity to discuss "sensitivity" of instruments. Different jobs require different levels of precision.

🎯 Exam Tip: Mention the specific precision of the screw gauge (\( 0.01 \text{ mm} \)) to make your answer scientifically strong.

Question 48. Calculate the thickness of the metal sheet from the given Vernier Caliper reading.
Answer:
\( 1 \text{ MSD} = 1 \text{ mm} \)
\( 10 \text{ VSD} = 9 \text{ mm} \)
\( 1 \text{ VSD} = \frac{9}{10} \text{ mm} \)
Least count = \( 1 \text{ MSD} - 1 \text{ VSD} \)
\( = \frac{1}{10} \text{ MSD} \)
\( = \frac{1}{10} \times 1 \text{ mm} = 0.1 \text{ mm} = 0.01 \text{ cm} \)
Thickness of the metal sheet:
Main scale reading = \( 3.9 \text{ cm} \)
Vernier scale division coinciding with main scale = \( 8 \)
Vernier scale reading = \( 8 \times 0.01 \text{ cm} = 0.08 \text{ cm} \)
Thickness of the metal sheet = \( 3.9 + 0.08 = 3.98 \text{ cm} \)

In simple words: To find the thickness, you look at where the zero of the sliding scale sits (Main scale) and then find which line on the sliding scale perfectly matches a line above it. Add them together.

📝 Teacher's Note: Guide students to look very closely at the lines. Only one line on the Vernier scale will truly "coincide" or line up perfectly with the Main scale.

🎯 Exam Tip: Always write: Total Reading = M.S.R. + (V.S.D. \( \times \) L.C.) to keep your working clear.

Question 49. Calculate the final reading of the screw gauge.
Answer:
When one complete rotation is given to the screw hand, it moves forward or backward by a distance is called pitch of the screw. It is distance between two consecutive threads of the screw.
Pitch of the screw = distance travelled by screw in n rotations / n rotations
Pitch of the screw = \( 1 \text{ mm} \)
No of divisions on circular scale = \( 100 \)
Least count = pitch / no of divisions =
\( = 0.01 \text{ mm} \)
Main scale reading = \( 3 \text{ mm} \)
Circular scale division coinciding with main scale = \( 42 \)
Circular scale reading = \( 0.01 \times 42 = 0.42 \)
Final reading = \( 3 + 0.42 = 3.42 \text{ mm} \)

In simple words: Just like a ruler, the main scale gives you the whole millimeters. The rotating scale gives you the tiny fractions. You add them to get the total thickness.

📝 Teacher's Note: Explain "Pitch" using a regular hardware screw. If you turn it once, it sinks in by exactly one thread-gap. That's the pitch.

🎯 Exam Tip: Pay attention to the baseline of the main scale. The number on the circular scale that aligns with this line is your CSR division.

Question 50. Calculate Least Count given threads and distance.
Answer:
Number of threads = \( 20 \)
Distance covered in 20 threads = \( 10 \text{ mm} \)
Pitch of the screw gauge = \( 10/20 = 0.5 \text{ mm} \)
No of divisions on circular scale = \( 50 \)
Least count = pitch / no of divisions = \( 0.01 \text{ mm} \)

In simple words: First find the pitch (distance of one thread) by dividing total distance by the number of threads. Then divide that pitch by circular scale markings to find the smallest measurable part.

📝 Teacher's Note: This problem tests the understanding of what "pitch" actually means. It's the distance per single thread or single rotation.

🎯 Exam Tip: Be careful with division order. It's distance divided by threads, not threads divided by distance.

Question 51. Define Oscillation, Time period, and Frequency.
Answer:

• Oscillation - One complete to and fro motion of a pendulum about its mean position is known as oscillation.
• Time period - The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum.
• Frequency - the number of oscillation made by the pendulum in one second is called frequency. Its SI unit is Hertz (Hz).

In simple words: Oscillation is one full "swing" and back. Time period is how long that swing takes. Frequency is how many swings happen every second.

📝 Teacher's Note: Use a playground swing as an analogy. One full trip from the center, to the front, to the back, and back to center is one oscillation.

🎯 Exam Tip: Don't forget to mention the SI unit for Frequency (Hertz) to get full marks for that definition.

Question 52. Calculate Time Period and Frequency of a pendulum.
Answer:
Time period from A to B = \( 0.5 \text{s} \)
For an ideal pendulum,
Time period from A to B = time period from B to C = time period from C to B = time period from B to A
Total time period, \( T = \text{time period from A to B} + \text{time period from B to C} + \text{time period from C to B} + \text{time period from B to A} \)
\( = 0.5 + 0.5 + 0.5 + 0.5 \)
\( = 2 \text{ s} \)
Frequency \( f = 1/T \)
\( = 1/2 = 0.5 \text{ Hz} \)

In simple words: If it takes half a second to go from the center to one side, it will take another half second to come back, then another half to go to the other side, and a final half to return. The whole cycle is 2 seconds.

📝 Teacher's Note: Emphasize that one oscillation consists of four "quarters" (mean to extreme, extreme to mean, mean to other extreme, other extreme to mean).

🎯 Exam Tip: Always remember that Frequency and Time Period are inverses of each other. If \( T = 2 \), \( f \) must be \( 1/2 \).

Question 53. Calculate the density of the metal.
Answer:
Mass of the metal = \( 540 \text{g} \)
Volume = \( 200 \text{ cm}^3 \)
Density = mass of metal / volume
\( = 540 / 200 = 2.70 \text{ g/cm}^3 \)

In simple words: Density tells you how "packed" something is. You find it by seeing how much mass is squeezed into a certain amount of space (volume).

📝 Teacher's Note: This specific result (\( 2.7 \text{ g/cm}^3 \)) is the density of Aluminium. It's a good way to help students relate numbers to real-world materials.

🎯 Exam Tip: Don't forget the units! Density is always "Mass units" per "Volume units", like \( \text{g/cm}^3 \) or \( \text{kg/m}^3 \).

Question 54. Calculate the mass of iron in an alloy.
Answer:
Mass of copper = \( 540 \text{ g} \)
Density of copper = \( 9 \text{ g/cm}^3 \)
Volume of copper used in the alloy = mass of copper / density
\( = 540 / 9 = 60 \text{ cm}^3 \)
Mass of iron = \( 240 \text{ g} \)
Density of iron = \( 8 \text{ g/cm}^3 \)
Volume of iron used in the alloy = mass of iron / density
\( = 240 / 8 = 30 \text{ cm}^3 \)
Total mass of the alloy = \( 540 + 240 = 780 \text{ g} \)
Total volume of the alloy = \( 60 + 30 = 90 \text{ cm}^3 \)
Density of the alloy = mass of the alloy / density of the alloy
\( = 780 / 90 = 8.67 \text{ g/cm}^3 \)

In simple words: When you mix two metals, you add their masses and their volumes separately. The new density is the total mass divided by the total volume.

📝 Teacher's Note: Students often try to average the densities (\( (9+8)/2 \)). Explain why this is wrong—it only works if you mix equal volumes, which isn't the case here.

🎯 Exam Tip: Break this problem into steps: find individual volumes first, then totals, then final density. It prevents silly mistakes.

Question 55. Take readings from the provided diagrams for measuring a rod, cavity, and depth.
Answer:
Least count of vernier caliper = \( 0.01 \text{ cm} \)
(i) Main scale reading = \( 5.5 \text{ cm} \)
Vernier scale division coinciding with main scale = \( 6 \)
Vernier scale reading = \( 6 \times 0.01 \text{ cm} = 0.06 \text{ cm} \)
Diameter of rod = \( 5.5 + 0.06 = 5.56 \text{ cm} \)
(ii) Main scale reading = \( 1.6 \text{ cm} \)
Vernier scale division coinciding with main scale = \( 5 \)
Vernier scale reading = \( 5 \times 0.01 \text{ cm} = 0.05 \text{ cm} \)
Internal diameter of a cavity = \( 1.6 + 0.05 = 1.65 \text{ cm} \)
(iii) Main scale reading = \( 3.1 \text{ cm} \)
Vernier scale division coinciding with main scale = \( 5 \)
Vernier scale reading = \( 5 \times 0.01 \text{ cm} = 0.05 \text{ cm} \)
Depth of a cavity = \( 3.1 + 0.05 = 3.15 \text{ cm} \)

In simple words: Vernier calipers are like 3 tools in one: the big bottom jaws measure outside width, the top small jaws measure inside width, and the stick at the back measures depth. All use the same scale logic.

📝 Teacher's Note: Point out the three different parts of the calipers in the diagram. This helps students realize that one instrument has multiple applications.

🎯 Exam Tip: In diagram (i), make sure to count the lines after the 5 cm mark carefully to identify \( 5.5 \text{ cm} \) as the start point.

Question 56. Define measurement and list SI units for basic quantities.
Answer:
Measurement is an act or the result of comparison of a quantity whose magnitude is unknown with a predefined standard. Fundamental units in a SI system:

Rules observed while writing unit of a physical quantity:
a. It must be of proper size. Very small or large size may cause inconvenience.
b. It should be easily accessible
c. It must be reproducible at all places without any difficulty.
d. It must be accurately defined and must not change with time, place and physical conditions such as pressure, humidity, etc.
e. It must be widely acceptable all over the world.

In simple words: Measuring is just comparing something we don't know (like how tall a tree is) to something we all agree on (like a meter stick). Standards like SI units make sure everyone's "meter" is the exact same length.

📝 Teacher's Note: Ask students why we use "Kelvin" for temperature instead of "Celsius" in science—this leads to a discussion about absolute zero.

🎯 Exam Tip: Memorize the table perfectly. Note that symbols like 'K' and 'A' are capitalized because they are named after scientists, while 'm' and 's' are lowercase.

Question 57. What steps are followed for measuring the length of an object using a vernier calipers?
Answer:
For measuring the length of an object using a vernier calipers, these steps are followed:

• First of all we find the least count and zero error of the vernier calipers.
• Place the object whose length is to be measured below the lower jaws and move the jaw till it touches the object. Record the main reading.
• Note the division on the vernier scale that coincides with some division of the main scale. Multiply this number of vernier division with least count. This is vernier scale reading.
• Record the observed length by adding the main scale reading and the vernier scale reading. Also, subtract zero error with its proper sign, if any, from the observed length to find the true length of the object.

In simple words: Check for errors, trap the object in the jaws, read the big numbers on the fixed part, find the matching line on the sliding part, and add them up.

📝 Teacher's Note: Emphasize that "zeroing" the instrument is the most important first step. Even the best measuring job is useless if the tool started wrong.

🎯 Exam Tip: When listing steps, use bullet points or numbered lists. Examiners find them much easier to grade than a long paragraph.

Question 58. Explain the types of zero error in Vernier Calipers and how to determine them.
Answer:
When two jaws are in contact, the zero of the vernier must coincide with the zero of main scale. If this is not so, the instrument has zero error. These are of two types:-
a. If the zero of the vernier scale lies to the right of the zero of main scale then it is called positive zero error.
b. If the zero of the vernier scale lies to the left of the zero of main scale then it is called negative zero error
To determine the zero error, note the vernier division coinciding with any main scale division when two jaws are completely closed. Suppose \( n^{th} \) division is coinciding with main scale. Multiply this with the least count of vernier calliper.
Positive zero error = \( n \times L.C. \)
Negative zero error = \( -(10 - n) \times L.C. \)
To get the correct reading, the zero error is usually subtracted from the observed reading with its proper sign
Correct reading = observed reading - zero error

In simple words: Positive error is like a runner starting ahead of the line. Negative error is like a runner starting behind the line. We must adjust the final score to make it fair.

📝 Teacher's Note: The formula for negative error \( -(10-n) \) often confuses students. Explain that since there are 10 total divisions, we are finding how many steps "back" the zero went.

🎯 Exam Tip: Always state "Correct reading = observed reading - zero error". This universal formula works for both positive and negative errors as long as you keep the signs correct.

Question 59. What is the procedure used to measure the diameter of a wire using a screw gauge?
Answer:
Following procedure is used to measure the diameter of a wire

• Calculate the least count and zero error of the screw gauge.
• Place the wire in between the studs. Turn the ratchet clockwise so as to hold the wire gently in between the studs. Record the main scale reading.
• Now record the division of circular scale that coincides with the base line of main scale. This circular scale division multiplied by least count will give circular scale reading.
• The observed diameter is obtained by adding the circular scale reading to the main scale reading. Subtract the zero error if any, with its proper sign, from the observed diameter to get the true diameter.

In simple words: Use the "clicking" knob (ratchet) to tighten the tool on the wire, read the hidden scale, read the spinning scale, and combine them while fixing any starting errors.

📝 Teacher's Note: Explain why we use the ratchet instead of the thimble—it prevents over-tightening and crushing the object, which would give a false reading.

🎯 Exam Tip: Use the term "ratchet" specifically in your answer; it's a key technical term that examiners look for.

Question 60. State the precautions to be taken when measuring length with a metre rule.
Answer:
In order to measure the length of an object using a metre rule, the metre rule must be placed with its marking close to the object, such that the zero marking on the scale coincides with one end of the object. Then the reading on the scale corresponding to the other end of the object will give the length of the object.
Precautions to be taken for measuring the length of the object, the eye must be kept vertically above the end of the object to avoid parallax and the corresponding marking along the line should be carefully read. The meter scale can measure up to an accuracy of 1mm or 0.1 cm

In simple words: Line up the zero mark perfectly. Most importantly, look straight down at the mark. If you look from the side, the lines will seem to shift—this is called parallax error.

📝 Teacher's Note: Demonstrate parallax error by holding a finger in front of a ruler and having students look from different angles. They will see the "reading" change.

🎯 Exam Tip: Mentioning "parallax error" and "vertical eye position" is crucial for scoring full marks on precaution questions.

Question 61. Explain the displacement method for measuring the volume of an irregular solid.
Answer:
For measurement of an irregular solid by the displacement method using a measuring cylinder.
a. Fill the graduated cylinder with the water.
b. Record the lower meniscus of water and let the value be \( V_1 \).
c. Tie the solid whose volume is to be measured to a strong string and lower it into the water gently.
d. When the object is completely immersed in water, the level of water rises.
e. Note the reading carefully and let the value be \( V_2 \)
f. Volume of the solid,

In simple words: Since you can't measure a lumpy rock with a ruler, you drop it in a cup of water. How much the water level goes up is exactly equal to the space (volume) the rock takes up.

📝 Teacher's Note: This is Archimedes' Principle in action. Remind students to use a thin string so the string's own volume doesn't change the reading too much.

🎯 Exam Tip: Use the term "meniscus" and specify that the object must be "completely immersed" to get an accurate total volume.

Question 62. Describe the steps to find the slope of a graph.
Answer:
To find a slope of a graph we take following steps
a. Take two points P and Q on the line, at a distance more than half the length of the line. The points should be so chosen that the value of x and y should be easily read on the axes.
b. Draw horizontal and vertical lines from these points on the axes, and read the values of x, say \( x_1 \) and the value of y, say \( y_1 \) for the point P. Similarly, the values of x, say \( x_2 \) and the value of y, say \( y_2 \) for the point Q.
c. Calculate \( \Delta y = y_2 - y_1 \) and \( \Delta x = x_2 - x_1 \). Calculate the slope of the graph \( m = \Delta y / \Delta x \)

In simple words: Slope tells you how steep the line is. You pick two points far apart, find how much the height changed (\( \Delta y \)), and divide it by how much the width changed (\( \Delta x \)).

📝 Teacher's Note: Encourage students to pick points that are "nice" numbers on the grid. This makes the math easier and reduces reading errors.

🎯 Exam Tip: Always show the triangle on your graph to indicate where you took your readings for the slope calculation.

Question 63. What conclusion can be drawn from the graph of \( T^2 \) versus \( l \) for a simple pendulum?
Answer:
The square of time period is directly proportional to the effective length of simple pendulum.
\( T^2 \propto l \)

In simple words: The graph is a straight line through the origin. This proves that if you make the string longer, the time for a swing increases in a very predictable, mathematical way.

📝 Teacher's Note: This linear relationship allows us to calculate the value of 'g' (acceleration due to gravity) very accurately in a school lab.

🎯 Exam Tip: If asked about the nature of the graph, always mention it is a "straight line passing through the origin," which indicates direct proportionality.

Question 64. Explain the importance of presenting data in a tabular form using the example of a Vernier caliper measurement, and state the rules for recording observations.
Answer: The presentation of the data in a tabular form helps in quick grasping and analyzing the recorded observations. For example to measure a diameter using a vernier caliper we record the following data in a tabular form as follows

Vernier calliper

The measurements corresponding to one physical quantity form one instrument must be recorded in one table. And the name of the instrument should be written on top. Observation must be recorded up to the limit of instrument and form the recorded individual observations, calculate the mean value and record it at the bottom of the column.
In simple words: Organizing your measurements into a table makes it much easier to see patterns and find the average. Always label the instrument name and the units you are using at the top.

📝 Teacher's Note: Teach students to draw the table before starting the experiment. This ensures they don't miss any critical data points like the Least Count or Zero Correction during the busy lab session.

🎯 Exam Tip: When presenting a table in an exam, always include a column for the observation number and ensure that every header has its correct unit (like cm or mm).

Question 65. Provide a sample table to record the time period of a simple pendulum for different effective lengths.
Answer:

Table to record time period of a simple pendulum for different effective lengths.

In simple words: This table is used to record how the time for one full swing changes when you change the length of the string. Usually, a longer string means a slower swing.

📝 Teacher's Note: Use this table format to help students plot a graph of \( T^2 \) vs \( L \). It visually demonstrates the direct proportionality between the square of the time period and the length.

🎯 Exam Tip: Ensure that the values recorded in the table have a consistent number of decimal places based on the precision of the stopwatch used.

Question 66. Define the following terms related to a simple pendulum: Oscillation, Amplitude, Frequency, and Time period.
Answer:

• Oscillation – One complete to and fro motion of a pendulum about its mean position is known as oscillation.
• Amplitude – Amplitude is the magnitude of the maximum deviation of the bob from the mean position on either side when an oscillation takes place.
• Frequency – the number of oscillation made by the pendulum in one second is called frequency. Its SI unit is Hertz (Hz).
• Time period – The time taken by a simple pendulum for an oscillation is known as the time period of a simple pendulum.

In simple words: An oscillation is a full swing and back. Amplitude is how far it swings to the side. Frequency is how fast it swings (how many per second), and Time Period is how long one full swing lasts.

📝 Teacher's Note: Use a physical pendulum to demonstrate that amplitude is measured from the center to the furthest side, not from one side to the other.

🎯 Exam Tip: Always mention 'Hertz (Hz)' as the SI unit for frequency to ensure you get full marks for the definition.

Question 67. Draw a diagram of a simple pendulum and illustrate what constitutes one complete oscillation.
Answer:

Diagram of simple pendulum

Figure below shows one oscillation of a simple pendulum
\( O \rightarrow A, A \rightarrow B, B \rightarrow O \) is one oscillation.
In simple words: A full oscillation is the journey of the pendulum starting from the middle, going to the left, then all the way to the right, and finally back to the middle.

📝 Teacher's Note: Clarify that an oscillation can also be measured from one extreme point to the other and back (\( A \to B \to A \)). The starting point doesn't change the total time period.

🎯 Exam Tip: In diagrams, clearly label the 'Mean position' (O) and 'Extreme positions' (A and B). These labels are standard in physics grading schemes.

Question 68. What is a simple pendulum? State the mathematical expression for its time period and define the variables involved.
Answer: A simple pendulum consists a heavy point mass (called the bob) suspended by a mass less and an inextensible or non-elastic thread from a fixed point or rigid support. The time period of simple pendulum is directly proportional to the square root of the length of the pendulum The time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity \[ T = 2\pi \sqrt{\frac{l}{g}} \] Where \( T \) is the time period, \( l \) is length of the pendulum, \( g \) is acceleration due to gravity
In simple words: A pendulum is just a heavy weight on a string that doesn't stretch. Its swing time depends on how long the string is and how strong gravity is.

📝 Teacher's Note: Remind students that "mass-less" and "inextensible" are ideal conditions. In real life, the string has some mass and might stretch slightly, but we ignore these for basic physics problems.

🎯 Exam Tip: Memorize the proportionality: \( T \propto \sqrt{l} \). If the length is quadrupled, the time period doubles. This is a very common objective question.

Question 69. Describe the procedure to measure the mass of a body using a physical balance. What are the precautions and the conditions for a beam balance to be true?
Answer: To measure mass of a body using a physical balance 1. Before starting, bring the plumb line just above the pointed projection by adjusting the leveling screws at the base. The beam is then gently raised using the lever. And it should be ensured that the pointer swings equally on both sides of the zero mark of the scale. 2. Now lower the beam gently and given body is kept on left pan. 3. Next, place some weight on the right pan form the weight box using the forceps. 4. Now the lever is turned towards right so that the beam rises and the power begins to swing to pointer swing on either side. It must be carefully noted that the side to which the pointer moves more, denotes lesser mass on that side. 5. Go on adjusting the standard weights till the pointer swings equally on both sides of the zero mark. 6. At this stage, the total mass of weights on the right pan gives the mass of the body. Three precautions to be taken to measure the mass of a body using beam balance are

• The beam must be gently lowered before adding or removing weights from the pan.
• The weights should not be carried with bare hands to avoid the change in weights due to moisture and dust particles from the surrounding
• Whenever you are near the actual weight, you should carefully try the weights in the descending order.

Conditions for a beam balance to be true are 7. Both the pans must be of equal weights. 8. Both the arms must be of equal lengths.
In simple words: You put the object on the left and standard weights on the right until the pointer swings the same amount on both sides. Use tweezers for the weights so oils from your hands don't make them heavy

📝 Teacher's Note: Emphasize the "descending order" rule for weights. It's the most efficient way to balance the pans without going back and forth too many times.

🎯 Exam Tip: The "forceps" and "plumb line" are key terms to include. Also, remember that the object goes in the LEFT pan and weights in the RIGHT pan.

Question 70. Given the time periods for 20 oscillations of a simple pendulum as 40s, 41s, 42s, 39s, and 38s, calculate the mean time period and the absolute error for each observation.
Answer: Time period of 20 oscillations of a simple pendulum are \( T_1 = 40 \text{ s} \quad T_2 = 41 \text{ s} \quad T_3 = 42 \text{ s} \quad T_4 = 39 \text{ s} \quad T_5 = 38 \text{ s} \) Mean time period of pendulum = \( \frac{40+41+42+39+38}{5} = 40 \text{ s} \) Absolute error \( \Delta T_1 = 40 - 40 = 0 \) Absolute error \( \Delta T_2 = 40 - 41 = -1 \) Absolute error \( \Delta T_3 = 40 - 42 = -2 \) Absolute error \( \Delta T_3 = 40 - 39 = 1 \) Absolute error \( \Delta T_1 = 40 - 38 = 2 \)
In simple words: The average time for the oscillations is 40 seconds. The absolute error shows how much each individual measurement was "off" from this average—some were exact, some were faster, and some were slower.

📝 Teacher's Note: Explain that "Absolute Error" can be negative. It represents the deviation from the mean value. The labels in the source have some typos (like repeating \( \Delta T_3 \)), but the math represents the sequence of observations.

🎯 Exam Tip: When calculating absolute error, follow the formula \( \text{Mean Value} - \text{Measured Value} \). Don't ignore the minus signs; they are mathematically important.