Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 1 Measurements And Experimentation

Unit 1

Exercise 1

(A) Objective Questions

I. Multiple choice Questions. Select the correct option:

Question 1. Which of the following is not a fundamental unit?
(a) Second
(b) Ampere
(c) Candela
(d) Newton
Answer: (d) Newton
In simple words: Fundamental units are basic measurements like time or length that don't depend on others. Newton is a "derived" unit because it is made by combining mass, length, and time.

📝 Teacher's Note: Remind students that there are only seven SI fundamental units. Any unit that can be broken down into these seven is considered a derived unit.

🎯 Exam Tip: Memorize the seven fundamental units (m, kg, s, A, K, mol, cd). If a unit isn't on this list, it's likely a derived unit.

Question 2. Which of the following is a fundamental unit?
(a) \( \text{m/s}^2 \)
(b) Joule
(c) Newton
(d) metre
Answer: (d) metre
In simple words: A metre is a base unit used to measure length directly. The other options are combinations of different base units used for more complex measurements like speed or energy.

📝 Teacher's Note: Use the analogy of "Lego bricks" (fundamental units) and "Lego models" (derived units). Metre is a single brick.

🎯 Exam Tip: Look for single-word units that represent basic physical properties like length, mass, or time to find fundamental units.

Question 3. Which is not a unit of distance?
(a) metre
(b) millimetre
(c) Leap year
(d) kilometre
Answer: (c) Leap year
In simple words: Metres and kilometres measure how far apart two things are. A leap year measures a specific amount of time (366 days).

📝 Teacher's Note: Students often confuse "Light year" (distance) with "Leap year" (time). Clarify that "Light year" is the distance light travels in a year.

🎯 Exam Tip: Read the units carefully. Any unit ending in "year" or "day" without the prefix "light" is always a unit of time.

II Fill in the blanks

Question 1. The unit is which we measure the quantity is called constant quantity.
Answer: constant quantity
In simple words: A unit is a fixed standard that everyone agrees on so that measurements are the same no matter who takes them.

📝 Teacher's Note: Explain that for a measurement to be reliable, the standard must not change over time or across different locations.

🎯 Exam Tip: The definition of a unit often requires it to be "invariable" and "universally accepted."

Question 2. One light year is equal to \( 9.46 \times 10^{15} \text{ m} \).
Answer: \( 9.46 \times 10^{15} \text{ m} \)
In simple words: This is a massive number that shows how incredibly far light can travel in one full year.

📝 Teacher's Note: Show the calculation (\( \text{Speed of Light} \times \text{Seconds in a Year} \)) to help students understand where the value comes from.

🎯 Exam Tip: Be careful with the power of ten. It is positive 15, representing a very large distance.

Question 3. One mean solar day = 86400 sec
Answer: 86400 sec
In simple words: If you multiply 24 hours by 60 minutes and then by 60 seconds, you get the total number of seconds in one average day.

📝 Teacher's Note: Have students perform the multiplication \( 24 \times 60 \times 60 \) once to memorize this standard value.

🎯 Exam Tip: This number is frequently used in numerical problems involving time conversion. Knowing it by heart saves time.

Question 4. One year = \( 3.1536 \times 10^7 \text{ sec} \)
Answer: \( 3.1536 \times 10^7 \text{ sec} \)
In simple words: This is scientific notation for the total number of seconds that pass during 365 days.

📝 Teacher's Note: Explain that scientific notation makes it easier to work with very large numbers used in physics.

🎯 Exam Tip: Remember the exponent 7 for "year to seconds" and 15 for "light year to metres."

Question 5. One micrometre = \( 10^{-6} \text{ m} \).
Answer: \( 10^{-6} \text{ m} \)
In simple words: A micrometre is one-millionth of a metre. It is used to measure tiny things like the width of a human hair.

📝 Teacher's Note: Introduce the Greek symbol \( \mu \) (mu) for micro. Use a metric prefix chart to show the relationship between milli, micro, and nano.

🎯 Exam Tip: A negative exponent like \( -6 \) means the quantity is much smaller than one, whereas a positive exponent means it is larger.

(B) Subjective Questions

Question 1. What do you understand by the term measurement?
Answer: “Measurement implies comparison of a physical quantity with a standard unit to find out how many times the given standard is contained in the physical quantity.” Physics, like other branches of science requires experimental study which involves measurement.
In simple words: Measuring is just comparing something unknown (like the length of a desk) to something known (like a 1-metre ruler) to see how big it is.

📝 Teacher's Note: Use a simple object and a ruler in class to demonstrate that measurement is essentially counting "units."

🎯 Exam Tip: Use the keywords "comparison" and "standard unit" to get full marks for this definition.

Question 2. What do you understand by the terms
1. unit
2. magnitude, as applied to a physical quantity?
Answer:
(i) Unit : Unit “is a standard quantity of the same kind with which a physical quantity is compared for measuring it. ” In order to measure a physical quantity, a standard is needed (which is acceptable internationally). The standard should be some convenient, definite and easily reproducible quantity of the same kind in terms of which the physical quantity as a whole is expressed. This standard is called a unit
(ii) Magnitude of a physical quantity : The number of times a standard quantity is present in a given physical quantity is called magnitude of physical quantity.
Physical quantity = Magnitude × Unit
In simple words: If you say a pencil is 10 centimetres long, "centimetre" is the unit (the tool we use) and "10" is the magnitude (how many units we have).

📝 Teacher's Note: Emphasize that a physical quantity is meaningless without both parts. "10" is just a number; "10 cm" is a measurement.

🎯 Exam Tip: Remember the formula: \( \text{Physical Quantity} = N \times u \) where \( N \) is magnitude and \( u \) is unit.

Question 3. A body measures 25 m. State the unit and the magnitude of unit in the statement.
Answer: Here S.I. unit of length i.e. metre (m) has been used. Magnitude of the given quantity = 25
Metre : It is defined as 1,650,763.73 times the wavelength of specified orange red spectral line a emission spectrum of Krypton-86 or 1,553,164.1 times the wavelength of the red line in emission spectrum of cadmium. or one metre is defined as the distance travelled by the light in 1/299,792,458 of a second in air/vacuum.
In simple words: The unit is "metre" and the magnitude (the number) is "25". The rest of the definition describes exactly how long "one metre" is scientifically.

📝 Teacher's Note: Mention that scientific definitions change as technology improves. We now use the speed of light because it's the most constant thing in the universe.

🎯 Exam Tip: For this type of question, clearly label "Magnitude = 25" and "Unit = metre" before providing any definitions.

Question 4. State four characteristics of a standard unit.
Answer: Characteristics of standard unit :
1. It should be of convenient size.
2. It should not change with respect to place and time.
3. It should be well defined.
4. It should be easily reproduced.
In simple words: A good unit should be easy to use, stay exactly the same everywhere, have a very clear meaning, and be easy for other scientists to copy.

📝 Teacher's Note: Discuss why using "foot-spans" was a bad idea—it changes from person to person (not constant or well-defined).

🎯 Exam Tip: These points are standard. Try to use words like "invariable" (doesn't change) for the second point to sound more scientific.

Question 5. Define the term fundamental unit. Name fundamental units of mass; length; time; current and temperature.
Answer: Fundamental unit : A fundamental or basic unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit. e.g. units of mass, length, time and temperature.

In simple words: Fundamental units are the "original" units. They don't need any other units to define them. Like a second just measures time—it isn't made of anything else.

📝 Teacher's Note: Ensure students understand that Kelvin is the SI unit for temperature, even though we use Celsius more often in daily life.

🎯 Exam Tip: When asked to list units, always include both the full name (kilogramme) and the correct symbol (kg).

Question 6. What do you understand by the term derived unit? Give three examples.
Answer: Derived units. “Derived units are those which can be expressed in terms of fundamental units.”
Example.
1. Force = mass × acceleration
\( = \text{mass} \times \frac{\text{velocity}}{\text{time}} \)
\( = \text{mass} \times \frac{\frac{\text{distance}}{\text{time}}}{\text{time}} \)
\( = \frac{\text{mass} \times \text{distance}}{(\text{time})^2} \)

\( \therefore \text{Force} = \frac{\text{mass} \times \text{length}}{(\text{time})^2} \)

2. S.I. unit of area i.e. \( \text{m}^2 \) is a derived unit.
Area = length × breadth
Now metre is unit of length and breadth, so S.I. unit of area is obtained by multiplying the fundamental unit ‘m’ with itself. So, \( \text{m}^2 \) is the derived unit of area.

3. Density = Mass/volume
S.I. unit of density i.e. \( \text{kg/m}^3 \) is the derived unit of density because it can be obtained by combining two fundamental units kilogram and metre.
In simple words: Derived units are "combo" units. For example, Area is just Length multiplied by itself. It isn't a new base unit; it's just metres squared.

📝 Teacher's Note: Show the derivation steps on the board. Start with the formula for the quantity and replace each part with its fundamental unit.

🎯 Exam Tip: If you forget an example, think of any formula you know (like \( \text{speed} = d/t \)) and see how the units combine (\( \text{m/s} \)).

Question 7. (a) Define metre according to old definition.
(b) Define metre in terms of wavelength of light.
(c) Why is the metre length in terms of wavelength of light considered more accurate?
Answer:
(a) Metre : One metre is defined as the one ten millionth part of distance from the pole to the equator.
(b) Metre : One metre is defined as 1,650, 763.73 times the wavelength of specified orange red spectral line in emission spectrum of Krypton = 86.
OR
One metre is defined as 1,553,164.1 times the wavelength of the red line in emission spectrum of cadmium.
(c) Metre length in terms of wavelength of light is considered more accurate because
1. The wavelength of light does not change with time, temperature, pressure etc.
2. It can be reproduced anywhere at any time because Krypton is available every where.
In simple words: The old way of measuring a metre was by looking at the size of the Earth. The new way uses light waves because they never change and can be measured perfectly in any lab.

📝 Teacher's Note: Explain that physical bars of metal can expand or contract with temperature, but light wavelengths are "nature's constant ruler."

🎯 Exam Tip: Focus on the point that light wavelengths are unaffected by external factors like pressure or temperature.

Question 8. Name the convenient unit you will use to measure :
(a) length of a hall
(b) width of a book
(c) diameter of hair
(d) distance between two cities.
Answer:
(a) Foot (Ft)
(b) Centimetre (cm)
(c) Micrometre (µm)
(d) Kilometre (km)
In simple words: We choose units that match the size of the object—kilometres for long distances and micrometres for tiny hairs.

📝 Teacher's Note: Point out that while "Metre" is the standard SI unit, "convenience" means choosing a multiple or sub-multiple that doesn't result in too many zeros.

🎯 Exam Tip: "Convenient" units are often multiples of SI units, like km, or sub-multiples, like cm.

Question 9. (a) Define mass.
(b) State the units in which mass is measured in (1) C.GS. system (2) S.I. system.
(c) Name the most convenient unit of mass you will use to measure :
1. Mass of small amount of a medicine.
2. The grain output of a state
3. The bag of sugar
4. Mass of a cricket ball.
Answer:
(a) Mass: The quantity of matter contained in a body is known as its mass.
(b) In C.GS. system, mass is measured in gram. In S.I. system, mass is measured in Kilogram.
(c) (i) Microgram (µg). (ii) Tonne (t) (iii) Quintal (µt) (iv) Gram (g)
Note : 1 microgram = \( 10^{-6} \text{ g} = 10^{-9} \text{ kg} \)
1 Tonne = 1000 kg ; 1 Quintal = 100 kg
In simple words: Mass is the amount of "stuff" in an object. We use tonnes for huge crops of grain and tiny micrograms for medicine.

📝 Teacher's Note: Differentiate between mass (amount of matter) and weight (force of gravity on that matter). Mass doesn't change on the moon!

🎯 Exam Tip: Note that "Tonne" is the standard unit for massive commercial quantities like grain production.

Question 10. (a) Define time.
(b) State or define the following terms :
1. Solar day
2. Mean solar day
3. An hour
4. Minute
5. Second
6. Year.
Answer:
(a) Time : It is defined as the time interval between two events
(b)
(i) Solar day : The time taken by the earth to complete one rotation about its own axis is called solar day.
(ii) Mean solar day : The average of the varying solar days, when the earth completes one revolution around the sun, is called mean solar day.
(iii) An hour : It is defined as the 1/24 th part of the mean solar day.
(iv) Minute : It is defined as the 1/1440 part of the mean solar day.
(v) Second : “A second is defined as 1/86400 th part of a mean solar day.”
OR
Second may also be defined “as to be equal to the duration of 9,192,631,770 vibrations corresponding to the transition between two hyperfme levels of caesium – 133 atom in the ground state.”
(vi) Year : One year is defined as the time in which earth completes one complete revolution around the sun.
In simple words: Time measures the gap between things happening. A solar day is one full spin of the Earth. Smaller units like hours and seconds are just slices of that day.

📝 Teacher's Note: Explain "Mean solar day" by noting that the actual length of a day changes slightly throughout the year, so we use the average (mean) for our clocks.

🎯 Exam Tip: The scientific definition of a "second" involving Caesium vibrations is a favorite high-mark descriptive question. Memorize the number roughly if possible.

Unit II

Practice Problems 1

Question 1. A student calculates experimentally the value of density of iron as \( 7.4 \text{ g cm}^{-3} \). If the actual density of iron is \( 7.6 \text{ g cm}^{-3} \), calculate the percentage error in experiment.
Answer:
Experimental value of density of iron = \( \rho_1 = 7.4 \text{ g cm}^{-3} \)
Actual value of density of iron = \( \rho_2 = 7.6 \text{ g cm}^{-3} \)
Absolute error = \( \rho_2 – \rho_1 = 7.6 – 7.4 = 0.2 \text{ g cm}^{-3} \)

\( \text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100 \)

\( = \frac{0.2}{7.6} \times 100 = \frac{100}{38} = 2.63\% \)
In simple words: First find how far off the measurement was (0.2). Then divide that mistake by the real value (7.6) and turn it into a percentage.

📝 Teacher's Note: Emphasize that we always divide by the "Actual Value" (the truth), not the student's experimental value, when finding error.

🎯 Exam Tip: Always show the formula for percentage error before doing the calculation to earn step-marks.

Question 2. A student finds that boiling point of water in a particular experiment is 97.8°C. If the actual boiling point of water is 99.4°C, calculate the percentage error.
Answer:
Experimental value of boiling point of water = B.P1 = 97.8°C
Actual value of boiling point of water = B.P2 = 99.4°C
Absolute error = B.P.2 – B.P1 = 99.4 – 97.8 = 1.6°C

\( \text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100 \)

\( = \frac{1.6}{99.4} \times 100 = 1.60\% \)
In simple words: The thermometer was off by 1.6 degrees. Compared to the real temperature, that is a 1.6% error.

📝 Teacher's Note: Percentage error tells us how accurate an experiment was. A lower percentage means the student did a better job!

🎯 Exam Tip: Round your final answer to two decimal places unless the question says otherwise.

Question 3. A pupil determines velocity of sound as \( 320 \text{ ms}^{-1} \). If actual velocity of sound is \( 332 \text{ ms}^{-1} \), calculate the percentage error.
Answer:
Velocity of sound determined by pupil = \( V_1 = 320 \text{ ms}^{-1} \)
Actual value of velocity of sound = \( V_2 = 332 \text{ ms}^{-1} \)
Absolute error = \( V_2 – V_1 = 332 – 320 = 12 \text{ ms}^{-1} \)

\( \text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100 \)

\( = \frac{12}{332} \times 100 = 3.61\% \)
In simple words: The measurement was 12 units away from the real speed. This represents about a 3.6% mistake.

📝 Teacher's Note: Remind students to include the units in their "Absolute error" step, but note that the final "Percentage error" has no units.

🎯 Exam Tip: Check your subtraction carefully. A small mistake in the absolute error will ruin the whole calculation.

Exercise 2

Question 1. (a) What do you understand by the term order of magnitude of a quantity?
(b) Why are physical quantities expressed in the order of magnitude? Support your answer by an example.
Answer:
(a) Order of a magnitude of a quantity : The exponent part of a particular measurement is called order of magnitude of a quantity.
OR
The order of magnitude of a given numerical quantity is the nearest power of ten to which its value can be written, (b) Measurement of certain physical quantities are either too large or too small that these cannot be expressed conveniently. It is difficult to write or remember them. So such quantities can be expressed in the order of magnitude.
For example : The diameter of the sun is 1,390,000,000 m. It is difficult to write or remember such a measurement. So it is expressed as \( 1.39 \times 10^9 \text{ m} \).
Here power of ten i.e. 9 (i.e. exponent part of the measurement) gives the order of magnitude of the given quantity.
So order of magnitude of diameter of the sun is \( 10^9 \text{ m} \).
In simple words: Order of magnitude is a "rough estimate" using powers of ten. Instead of writing out a billion zeros, we just say "\( 10^9 \)" to show how huge it is.

📝 Teacher's Note: Use the example of counting people in a stadium. You don't need the exact number to know it's "on the order of" tens of thousands (\( 10^4 \)).

🎯 Exam Tip: When writing the order of magnitude, the number multiplied by the power of 10 must be between 0.5 and 5 to be accurate.

Question 2. Express the order of magnitude of the following quantities :
1. 12578935 m
2. 222444888 kg
3. 0.000,000,127 s
4. 0.000,000,000,00027 m
Answer:
(i) \( 12578935 \text{ m} = 1.2578935 \times 10^7 \text{ m} \)
So order of magnitude = \( 10^7 \text{ m} \)
(ii) \( 222444888 \text{ kg} = 2.22444888 \times 10^8 \text{ kg} \)
Order of magnitude = \( 10^8 \text{ kg} \)
(iii) \( 0.000,000,127 \text{ s} = 1.27 \times 10^{-7} \text{ s} \)
Order of magnitude = \( 10^{-7} \text{ s} \)
(iv) \( 0.000,000,000,00027 \text{ m} = 2.7 \times 10^{-13} \text{ m} \)
Order of magnitude = \( 10^{-13} \text{ m} \)
In simple words: We move the decimal point until we have a simple number and then count how many "jumps" it took to find the power of ten.

📝 Teacher's Note: Practice scientific notation with students. Moving the decimal to the left gives a positive exponent; to the right gives a negative exponent.

🎯 Exam Tip: The order of magnitude is just the power of ten. Don't include the decimal number in your final answer for "order of magnitude."

Question 3. (a) What do you understand by the term degree of accuracy?
(b) Amongst the various physical measurements recorded in an experiment, which physical measurement determines the degree of accuracy?
Answer:
(a) Degree of accuracy : It means that we can measure a quantity, without any error of estimation.
In any experiment, all observations should be taken with same degree of accuracy.
(b) Amongst the various physical measurements recorded in an experiment, least accurate observation determines the degree of accuracy.
In simple words: Accuracy is how close we can get to the "truth." In a chain of measurements, the final answer is only as accurate as your "weakest link" (the least accurate tool used).

📝 Teacher's Note: Use the analogy of a chef: even if most ingredients are perfectly weighed, one "handful" of salt ruins the precision of the whole recipe.

🎯 Exam Tip: The phrase "least accurate observation" is the key to part (b). Remember it!

Question 4. (a) State the formula for calculating percentage error
(b) Is it possible to increase the degree of accuracy by mathematical manipulations? Support your answer by an example.
Answer:
(a) The percentage error can be calculated by the formula :

\( \text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100 \)

(b) It is not possible to increase the degree of accuracy by mathematical manipulations.
For examples : When a number of values are added or subtracted, the result cannot be more accurate than the least accurate value.
In simple words: You can't fix sloppy measuring just by doing extra math later. If you use a rough estimate in your math, the final answer will also be just a rough estimate.

📝 Teacher's Note: Demonstrate this with addition: \( 10.1 + 5.283 = 15.4 \). Even though the second number has 3 decimals, we can only trust one decimal because of the first number.

🎯 Exam Tip: Clearly state "No" for part (b) before giving your explanation and example.

If we add:
    5.283
+ 72.5
    2.0014
We get, 79.78314
In the above addition 72.5 has least accuracy. When we . say 72.5, it implies that value lies between 72.45 and 72.55 and 72.5 is the most probable value. Thus the error in 72.5 is +0.05. As the final result cannot be more accurate than least accurate observation, so the correct and most reliable answer in the above addition is 72.9.

Question 5. State the factors which determine number of significant figures for the calculation of final result of an experiment.
Answer: Factors which determine number of significant figures for the calculation of final result of an experiment are :
1. The nature of experiment.
2. The accuracy with which various measurements are made.
In simple words: Significant figures show which digits we are sure about. This depends on what we are doing and how good our measuring tools (like rulers or scales) were.

📝 Teacher's Note: Explain that significant figures represent the "precision" of an instrument. A school ruler can only give significant figures to the nearest millimetre.

🎯 Exam Tip: If an instrument has a least count of 0.1 cm, your measurements should have one decimal place to be significant.

Question 6. The final result of calculations in an experiment is 125,347,200. Express the number in terms of significant places when
1. accuracy is between 1 and 10
2. accuracy is between 1 and 100
3. accuracy is between 1 and 1000
Answer: Final result of calculations in an experiment = 125,347,200

1. When accuracy lies between 1 and 10, then final result may be written as \( 1.2 \times 10^8 \).
2. When accuracy lies between 1 and 100, then final result may be written as \( 1.25 \times 10^8 \).
3. When accuracy lies between 1 and 1000 than final result may be written as \( 1.253 \times 10^8 \).
In simple words: As we get more accurate, we can keep more digits to show exactly how much we know. Lower accuracy means we have to round the number off more.

📝 Teacher's Note: This exercise helps students see the connection between the "error" (like 10, 100, or 1000) and how many digits we are allowed to keep in the final answer.

🎯 Exam Tip: Always use scientific notation (\( 10^n \)) for very large numbers to clearly show the number of significant figures.

Unit 3

Practice Problems 1

Question 1. The main scale of vernier callipers has 10 divisions in a centimetre and 10 vernier scale divisions coincide with 9 main scale divisions. Calculate
1. pitch
2. L.C. of vernier callipers.
Answer: Main scale divisions of vernier callipers in one centimetre = 10

\( \text{Pitch} = \frac{\text{Unit of main scale}}{\text{Number of divisions in the unit}} \)

\( = \frac{1}{10} \text{ cm} = 0.1 \text{ cm} \)

\( \text{Least count} = \frac{\text{Pitch}}{\text{No. of vernier scale divisions}} \)

\( = \frac{0.1}{10} \text{ cm} = 0.01 \text{ cm} \)
In simple words: Pitch is the size of one gap on the main ruler. The Least Count (L.C.) is how much smaller we can measure using the extra sliding part.

📝 Teacher's Note: Explain that "Pitch" is effectively the "Least Count" of the main scale by itself. The Vernier scale helps us divide that pitch further.

🎯 Exam Tip: Always check if the question asks for the answer in 'cm' or 'mm'. Standard Vernier L.C. is 0.1 mm or 0.01 cm.

Question 2. In a vernier callipers 19 main scale divisions coincide with 20 vernier scale divisions. If the main scale has 20 divisions in a centimetre, calculate
1. pitch
2. L.C. of vernier callipers.
Answer: Main scale divisions of vernier callipers in one centimetre = 20 Unit

\( \text{Pitch} = \frac{\text{Unit of main scale}}{\text{Number of divisions in the unit}} \)

\( = \frac{1}{20} \text{ cm} = 0.05 \text{ cm} \)

\( \text{Least count} = \frac{\text{Pitch}}{\text{No. of vernier scale divisions}} \)

\( = \frac{0.05}{20} = 0.0025 \text{ cm} \)
In simple words: This tool is even more precise because the gaps on the ruler are smaller (20 in one cm) and the sliding part has more lines (20 divisions).

📝 Teacher's Note: Use this problem to show how increasing the number of divisions on either scale makes the instrument more sensitive and precise.

🎯 Exam Tip: The denominator for L.C. is always the total number of divisions on the *Vernier* scale (here it is 20, not 19).

Practice Problems 2

Question 1. Figure shows the position of vernier scale, while measuring the external length of a wooden cylinder.
1. What is the length recorded by main scale?
2. Which reading of vernier scale coincides with main scale?
3. Calculate the length.
Answer: Main scale divisions of vernier callipers in one centimetre = 10

\( \text{Pitch} = \frac{\text{Unit of main scale}}{\text{Number of divisions in the unit}} \)

\( = \frac{1}{10} \text{ cm} = 0.1 \text{ cm} \)

\( \text{Least count} = \frac{\text{Pitch}}{\text{No. of divisions on the vernier scale}} \)

\( \text{L.C.} = \frac{0.1}{10} \text{ cm} = 0.01 \text{ cm} \)

(i) Length recorded by main scale = 10.2 cm
\( \implies \) Main scale reading is 10.2 cm
(ii) Reading of vernier scale coinciding with main scale = 7th
\( \implies \) Vernier scale division (V.S.D.) is 7th
(iii) Length recorded by vernier callipers
= Main scale reading + L.C. × V.S.D.
= 10.2 + 0.01 × 7 = 10.2 + 0.07 = 10.27 cm
In simple words: To find the total length, look at the main ruler first (10.2). Then see which tiny line on the slider matches perfectly with the ruler above (the 7th). Add them together!

📝 Teacher's Note: The main scale reading is the mark just to the left of the Vernier zero. The coinciding division is the one that forms a single straight line with a mark on the main scale.

🎯 Exam Tip: Always write the formula \( \text{Length} = \text{MSR} + (\text{VSR} \times \text{LC}) \) to ensure you get full credit for the calculation process.

Question 2. In figure for vernier callipers, calculate the length recorded.
Answer: Main scale divisions of vernier callipers in one centimetre = 10

\( \text{Pitch} = \frac{\text{Unit of main scale}}{\text{Number of divisions in the unit}} \)

\( = \frac{1}{10} \text{ cm} = 0.1 \text{ cm} \)

\( \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of divisions on the vernier scale}} \)

\( \text{L.C.} = \frac{0.1}{10} \text{ cm} = 0.01 \text{ cm} \)

Here main scale reading = 7.3 cm
Vernier scale reading (V.S.D.) coinciding with main scale = 5th
Length recorded = Main scale reading + L.C. × V.S.D.
= 7.3 + 0.01 × 5 = 7.3 + 0.05 = 7.35 cm
In simple words: The tool shows 7.3 on the big ruler. The 5th tiny line on the slider matches up, so we add 0.05 cm to get the final answer.

📝 Teacher's Note: If a student gets 7.3 or 7.4, explain that they missed the "fine adjustment" provided by the Vernier scale.

🎯 Exam Tip: Vernier calliper answers should almost always have two decimal places if measured in centimetres.

Practice Problems 3

Question 1. (a) A vernier scale has 10 divisions. It slides over a main scale, whose pitch is 1.0 mm. If the number of divisions on the left hand of zero of the vernier scale on the main scale is 56 and the 8th vernier scale division coincides with the main scale, calculate the length in centimetres.
(b) If the above instrument has a negative error of 0.07 cm, calculate corrected length.
Answer: No. of divisions on vernier scale = 10
Pitch = 1.0 mm

\( \text{Pitch} = \frac{\text{Pitch}}{\text{No. of divisions on the vernier scale}} \)

\( \text{L.C.} = \frac{1.0}{10} \text{ mm} = 0.1 \text{ mm} = 0.01 \text{ cm} \)

There are 56 number of main scale division on the left hand of zero of the vernier scale.
\( \implies \) Main scale reading = 56 mm = 5.6 cm
Vernier scale reading coinciding with main scale = 8th
(a) Length recorded = Main scale reading + L.C. × V.S.D.
= 5.6 + 0.01 × 8 = 5.6 + 0.08 = 5.68 cm
(b) Negative error = -0.07 cm
\( \implies \text{Correction} = -(-0.07) \text{ cm} = +0.07 \text{ cm} \)
\( \therefore \text{Corrected length} = \text{Observed reading} + \text{Correction} \)
= 5.68 + (0.07) = 5.75 cm
In simple words: The tool was wrong by being 0.07 cm "behind" zero. So, after measuring, we have to ADD that 0.07 back to get the real length.

📝 Teacher's Note: Negative error means the tool is "under-reporting" the size. Therefore, the correction must be positive.

🎯 Exam Tip: Remember: Correction = - (Zero Error). This simple rule prevents mistakes with signs (+ or -).

Question 2. (a) A vernier scale has 20 divisions. It slides over a main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of sphere.
Answer: No. of divisions on vernier scale = 20
Pitch 0.5 mm

\( \text{Least count} = \frac{\text{Pitch}}{\text{No. of divisions on the vernier scale}} \)

\( \text{L.C.} = \frac{0.5}{20} \text{ mm} = 0.025 \text{ mm} = 0.0025 \text{ cm} \)

(a) There are 38 number of main scale divisions on the left of the zero of vernier scale.
\( \implies \text{Main scale reading} = \frac{38}{2} \text{ mm} = 1.9 \text{ cm} \)
Vernier scale division coinciding with main scale = 18th
Diameter of sphere = Main scale reading + L.C. × V.S.D.
= 1.9 + 0.0025 × 18 = 1.9 + 0.0450 = 1.945 cm
(b) Negative error = -0.04 cm
\( \implies \text{Correction} = -(-0.04) \text{ cm} = +0.04 \text{ cm} \)
Corrected diameter of sphere = Observed reading + Correction
= 1.945 + (0.04) = 1.985 cm
In simple words: Each line on the main ruler is 0.5 mm. So 38 lines means 19 mm (or 1.9 cm). We then add the Vernier adjustment and fix the negative starting error.

📝 Teacher's Note: Since the pitch is 0.5 mm, students must multiply the division count (38) by 0.5 to get the main scale reading in millimetres.

🎯 Exam Tip: Watch out! The question asks for the corrected *diameter* in (a) but the final part of (b) mentions *radius* in the prompt. Always divide diameter by 2 if radius is required.

Practice Problems 4

Question 1. The least count of a vernier callipers is 0.0025 cm and it has an error of + 0.0125 cm. While measuring the length of a cylinder, the reading on main scale is 7.55 cm, and 12th vernier scale division coincides with main scale. Calculate the corrected length.
Answer: Least count (L.C.) = 0.0025 cm
Error = +0.0125 cm
Correction = – (Error) = – (+0.0125) = – 0.0125 cm
Main scale reading = 7.55 cm
Vernier scale division (V.S.D.) coinciding with main scale = 12th
Length recorded = Main scale reading + L.C. × V.S.D.
= 7.55 + 0.0025 × 12 = 7.55+ 0.0300 = 7.58 cm
Correct length = Length recorded + Correction
= 7.58+ (-0.0125) = 7.58 – 0.0125 = 7.5675 cm = 7.567 cm
In simple words: The tool was starting "too far ahead" by 0.0125 cm. To fix this, we subtract that extra amount from our final measurement.

📝 Teacher's Note: Positive error means the tool over-reads. A subtraction is necessary to get the "true" value.

🎯 Exam Tip: In your final answer, maintain the same number of decimal places as used in the precision of the tool.

Question 2. The least count of a vernier callipers is 0.01 cm and it has an error of + 0.07 cm. While measuring the radius of a sphere, the main scale reading is 2.90 cm and the 5th vernier scale division coincides with main scale. Calculate the correct radius.
Answer: Least count (L.C.) = 0.01 cm
Error = + 0.07 cm
Correction = (Error) = – (+ 0.07) = – 0.07 cm
Main scale reading = 2.90 cm
Vernier scale division (V.S.D.) coinciding with main scale = 5th
Observed diameter of sphere = Main scale reading + L.C. × V.S.D.
= 2.90 + 0.01 x 5 = 2.90 + 0.05 = 2.95 cm
Corrected diameter = Observed diameter + Correction
= 2.95 + (-0.07) = 2.95 – 0.07 = 2.88 cm
\( \therefore \text{Corrected radius} = \frac{2.88}{2} = 1.44 \text{ cm} \)
In simple words: First find the full width (diameter) of the ball, fix the 0.07 cm error, then divide by 2 to get the radius from the center to the edge.

📝 Teacher's Note: This is a multi-step logic problem. Students often forget the very last step: dividing the diameter by 2 to find the radius.

🎯 Exam Tip: Double-check the question for the word "radius" vs "diameter" before you finish your answer.

Exercise 3

Question 1. Who invented vernier callipers?
Answer: Vernier callipers was invented by Pierre Vernier.
In simple words: Pierre Vernier was a French mathematician who came up with this clever sliding scale in the 1600s.

📝 Teacher's Note: Knowing the inventor helps students appreciate that scientific tools are products of human ingenuity.

🎯 Exam Tip: This is a straightforward knowledge question. Just remember the name "Pierre Vernier."

Question 2. What is the need for measuring length with vernier callipers?
Answer: For measuring the exact length with greater accuracy, especially when we are measuring a very small length, we use an appliance vernier calliper. A vernier calliper can measure accurately upto 1/100 th part of a centimetre.
In simple words: A regular ruler can only measure to the nearest millimetre. A Vernier calliper allows us to measure things 10 times more precisely.

📝 Teacher's Note: Use the example of measuring the thickness of a glass slide to show why a normal ruler isn't good enough.

🎯 Exam Tip: The keyword here is "greater accuracy" and the specific value "1/100th of a cm."

Question 3. Up to how many decimal places can a common vernier callipers measure the length in cm?
Answer: A common vernier calliper can measure the length accurately upto two places of decimal when length is measured in centimetre i.e. upto 1/100 th part of a centimetre.
In simple words: It can measure things as small as 0.01 cm.

📝 Teacher's Note: This is why all Vernier calliper answers in CM should have exactly two digits after the decimal point (e.g., 5.40 cm, not 5.4 cm).

🎯 Exam Tip: If the question asks for MM, it's one decimal place (0.1 mm). If it asks for CM, it's two decimal places (0.01 cm).

Question 4. Define the terms :
1. pitch
2. least count as applied to a vernier callipers.
Answer:
1. Pitch : “The pitch of a screw is the distance moved by the screw in one complete rotation of its head.”
OR
Pitch may also be defined as “the distance between two consecutive threads of screw measured along the axis of screw.”

\( \text{Pitch} = \frac{\text{Distance moved by thimble on M.S.}}{\text{Number of rotations of thimble}} \)

2. Least Count of Vernier Calliper : Least count of a vernier callipers is the difference between one main scale division (M.S.D.) and one vernier scale division (V.S.D.)
In simple words: Pitch is the distance of one full turn of a screw. Least count is the smallest difference between the two different ruler marks on the tool.

📝 Teacher's Note: Although "Pitch" is technically a term for screw gauges, the text uses it here to describe the smallest main scale unit of a Vernier Calliper.

🎯 Exam Tip: For Least Count, the formula \( 1 \text{ MSD} - 1 \text{ VSD} \) is the most fundamental and accurate definition.

Question 5. State the formula for determining :
1. pitch
2. least count for a vernier callipers.
Answer:
(i) \( \text{Pitch} = \frac{\text{Unit of main scale}}{\text{No. of divisions in the unit}} \)

For example : If one centimetre length has ten divisions then

\( \text{Pitch} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} \)

(ii) \( \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of vernier scale division}} \)

For example : If pitch is 0.1 cm and there are ten vernier scale divisions.

Then \( \text{L.C.} = \frac{0.1}{10} = 0.01 \text{ cm} \)
In simple words: Pitch = smallest step on the big ruler. L.C. = that step divided by the number of lines on the small slider.

📝 Teacher's Note: Encourage students to count the lines on their physical lab tools to see these formulas in action.

🎯 Exam Tip: Always provide the unit (cm or mm) in your example to make the answer complete.

Question 6. State the formula for calculating length if :
1. Number of vernier scale division coinciding with main scale and number of division of main scale on left hand side of zero of vernier scale are known.
2. The reading of main scale is known and the number of vernier scale divisions coinciding with main scale are known.
Answer:
1. If we know the number of vernier scale divisions (V.S.D.) coinciding with main scale and number of main scale divisions (M.S.D.) on left hand side of zero of vernier scale then Length recorded = Main scale reading + L.C. × V.S.D.
2. Same as in part (i).
In simple words: Read the number on the big ruler just before the slider's zero. Then multiply the matching line number by the tool's Least Count and add it to the first number.

📝 Teacher's Note: The "Main scale reading" is essentially the "integer" part of the measurement, and the "LC x VSD" is the "decimal" part.

🎯 Exam Tip: This is a recurring formula. Memorize it: \( \text{Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) \).

Question 7. (a) What do you understand by the term zero error?
(b) When does a vernier callipers has
1. positive error
2. negative error?
(c) State the correction if
Answer:
(a) Zero Error : A vernier callipers is said to have a zero error when zero of the main scale does not coincide with zero of vernier scale.
(b) Positive Error : If the zero of the vernier scale is on right hand side of zero of the main scale, then error is said to be positive and correction is said to be negative.

Negative Error : If the zero of the vernier scale is on the left hand side of zero of the main scale, the error is said be negative and the correction is said to be positive.

(c) When positive error is 7 divisions and L.C. is 0.01 cm
Then correction = – (+ 7 × L.C.)
= -7 × 0.01 cm = – 0.07 cm
When negative error is 7 divisions and count (L.C.) is 0.01 cm
Then correction = – (- 7 × L.C.)
= – ( – 7 × 0.01) cm = + 0.07 cm
In simple words: Zero error is like a scale that doesn't say "0" when nothing is on it. If it says 7 grams, that's positive error. If it starts behind zero, that's negative error.

📝 Teacher's Note: Draw two lines on the board that don't match up to represent zero error visually. It's much easier for students to see "right" vs "left" of zero.

🎯 Exam Tip: Be very careful with signs. Positive Error = Negative Correction (Subtraction). Negative Error = Positive Correction (Addition).

Question 8. Which part of vernier callipers is used to measure
(a) external diameter of a cylinder
(b) internal diameter of a hollow cylinder
(c) internal length of a hollow cylinder?
Answer:
(a) External Jaws of a vernier callipers are used to measure the external diameter of cylinder.
(b) Internal Jaws are used to measure internal diameter of a hollow cylinder.
(c) Tail of vernier callipers is used to measure the internal length of a hollow cylinder.
In simple words: Bottom jaws for outside width, top jaws for inside holes, and the skinny tail for depth.

📝 Teacher's Note: Bring a physical calliper and a test tube to class. Show how the jaws and tail provide three different measurements for the same object.

🎯 Exam Tip: The "Tail" is often also called the "Depth gauge." Both terms are usually acceptable.

Unit 4

Practice Problems 1

Question 1. The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on sleeve, when given four complete rotations calculate
1. pitch
2. least count.
Answer: Number of circular scale divisions (C.S.D.) = 50
Distance moved by screw (spindle) on sleeve = 2 mm
Number of complete rotations given = 4

(i) \( \text{Pitch} = \frac{\text{Distance moved by screw on sleeve}}{\text{No. of complete rotations}} \)

\( = \frac{2 \text{ mm}}{4} = 0.5 \text{ mm} \)

(ii) Pitch = 0.05 cm

\( \text{Least count} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} \)

\( = \frac{0.05}{50} \text{ cm} = 0.001 \text{ cm} \)
In simple words: Pitch is how much the screw moves forward in just one turn (0.5 mm). The Least Count is that tiny step divided by the 50 marks on the knob.

📝 Teacher's Note: Explain that a screw gauge is 10 times more accurate than a Vernier calliper because its least count is 0.001 cm instead of 0.01 cm.

🎯 Exam Tip: Always convert your final answer into CM if the tool's standard is discussed in CM. 1 mm = 0.1 cm.

Question 2. The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate
1. pitch
2. least count of the screw gauge.
Answer: Number of circular scale divisions = 100
Distance moved by spindle (screw) = 2.5 mm
No. of complete rotations given = 5

(i) \( \text{Pitch} = \frac{\text{Distance moved by screw on sleeve}}{\text{No. of complete rotations}} \)

\( = \frac{2.5}{5} = 0.5 \text{ mm} = 0.05 \text{ cm} \)

(ii) \( \text{Least count} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} \)

\( = \frac{0.05}{100} \text{ cm} = 0.0005 \text{ cm} \)
In simple words: This tool moves half a millimetre per turn. Because it has 100 tiny lines on the knob, it can measure items as thin as 0.0005 cm.

📝 Teacher's Note: Notice how doubling the circular divisions (from 50 in the last problem to 100 here) makes the tool twice as precise.

🎯 Exam Tip: Write down all given values (Divisions, Distance, Turns) clearly before starting your calculation to avoid simple logic errors.

Practice Problems 2

Question 1. Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of wire.
Answer: No. of circular scale divisions = 200
Pitch = 1 mm

\( \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of circular divisions}} \)

\( = \frac{1 \text{ mm}}{200} = 0.005 \text{ mm} \)
L.C. = 0.0005 cm
Main scale reading = 5 mm = 0.5 cm
(C.S.D.) circular scale reading = 34 divisions
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.5 + 0.0005 × 34
= 0.5 + 0.0170 = 0.5170 cm

\( \text{Radius of wire} = \frac{\text{diameter}}{2} = \frac{0.5170}{2} = 0.2585 \text{ cm} \)
In simple words: The big ruler shows 5 mm. The matching mark on the knob is 34. After adding them and getting the total width (diameter), we cut it in half to find the radius.

📝 Teacher's Note: In the diagram, look where the horizontal reference line hits the circular scale. It falls between 30 and 40—you have to count the lines carefully to get 34.

🎯 Exam Tip: This is a common trap! Always check if the question asks for "Diameter" or "Radius."

Question 2. Figure shows a screw gauge in which circular scale has 100 divisions. Calculate the least count and the diameter of a wire.
Answer: No. of circular scale division = 100
Pitch =0.5 mm

\( \text{Least count} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm} \)

L.C. = 0.005 mm = 0.0005 cm
Main scale reading = 4.5 mm = 0.45 cm
(C.S.D.) circular scale reading = 73 division
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.45 + 0.0005 × 73
= 0.45 + 0.0365 = 0.4865 cm
In simple words: The tool shows 4.5 mm on the big ruler. We add the tiny reading from the knob (mark 73) to get the final width of the wire.

📝 Teacher's Note: On a 0.5 mm pitch gauge, the main scale will have marks both above and below the line. Remind students to check if the bottom 0.5 mark is visible.

🎯 Exam Tip: Ensure that all your values are converted into CM before doing the final addition to avoid unit mix-ups.

Practice Problems 3

Question 1. A micrometre screw gauge having a positive zero error of 5 divisions is used to measure diameter of wire, when reading on main scale is 3rd division and 48th circular scale division coincides with base line. If the micrometer has 10 divisions to a centimetre on main scale and 100 divisions on circular scale, calculate
1. Pitch of screw
2. Least count of screw
3. Observed diameter
4. Corrected diameter.
Answer:
(i) Micrometre has 10 divisions in one centimetre on main scale.

\( \therefore \text{Pitch} = \frac{\text{Unit}}{\text{No. of divisions in unit}} \)

\( \text{Pitch} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} \)

(ii) No. of circular scale divisions = 100

\( \therefore \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of circular divisions}} \)

\( = \frac{0.1 \text{ cm}}{100} = 0.001 \text{ cm} \)

(iii) Reading on main scale is 3rd division.
\( \implies \text{Main scale reading} = 3 \text{ mm} = 0.3 \text{ cm} \).
Circular scale reading = 48 div.
Observed diameter = M.S. reading + L.C. × C.S. reading
= 0.3 + 0.001 × 48 = 0.3 + 0.048
= 0.348 cm

(iv) Positive zero error = +5 divisions
\( \therefore \text{Correction} = -(\text{Error} \times \text{L.C.}) \)
= -(5 × 0.001) = -0.005 cm
\( \therefore \text{Corrected diameter} = \text{Observed diameter} + \text{Correction} \)
= 0.348 – 0.005
= 0.343 cm
In simple words: The tool was wrong by 5 marks at the start (it said 0.005 cm even when closed). We subtract this extra amount from our final answer to be accurate.

📝 Teacher's Note: The "3rd division" on a scale with 10 div/cm is 3 mm. Since 1 cm = 10 mm, each division is 1 mm.

🎯 Exam Tip: "Positive error" means the tool reads *too high*. Always subtract positive errors from your observed reading.

Question 2. A micrometre screw gauge has a positive zero error of 7 divisions, such that its main scale is marked in 1/2 mm and the circular scale has 100 divisions. The spindle of the screw advances by 1 division complete rotation. If this screw gauge reading is 9 divisions on main scale and 67 divisions on circular scale for the diameter of a thin wire, calculate
1. Pitch
2. L.C.
3. Observed diameter
4. Corrected diameter.
Answer:
(i) Pitch = \( \frac{1}{2} \text{ mm} = 0.5 \text{ mm} = 0.05 \text{ cm} \)

(ii) No. of circular scale divisions = 100

\( \therefore \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} \)

\( \text{L.C.} = \frac{0.05 \text{ cm}}{100} = 0.0005 \text{ cm} \)

(iii) Main scale reading = 9 divisions = \( 9 \times \frac{1}{2} \text{ mm} \).
= 4.5 mm = 0.45 cm
Circular scale reading = 67 div.
Observed diameter = M.S. reading + L.C. × C.S. reading
= 0.45 + 0.0005 × 67 = 0.45 + 0.0335
= 0.4835 cm

(iv) Positive zero error = 7 divisions
Correction = – (Error × L.C.)
= -(7 × 0.0005) cm = -0.0035 cm
\( \therefore \text{Corrected diameter} = \text{Observed diameter} + \text{Correction} \)
= 0.4835 + (-0.0035)
= 0.4835 – 0.0035
= 0.4800 cm
In simple words: Each turn moves the tool 0.5 mm. Nine divisions on the main scale means 4.5 mm. We add the matching lines from the knob and then subtract the starting error of 7 divisions.

📝 Teacher's Note: The phrase "main scale is marked in 1/2 mm" means that each division on the main scale is 0.5 mm, which is common in high-precision screw gauges.

🎯 Exam Tip: When the zero error is given in "divisions," you must multiply it by the Least Count to find the actual value in CM before subtracting.

Question 3. The thimble of a screw gauge has 50 divisions for one rotation. The spindle advances 1 mm when the screw is turned through two rotations.
1. What is the pitch of screw?
2. What is the least count of screw gauge?
3. When the screw gauge is used to measure the diameter of wire the reading on sleeve is found to be 0.5 mm and reading on thimble is found 27 divisions. What is the diameter of wire in centimetres?
Answer: Pitch of screw gauge is the distance moved by spindle in one revolution = 1/2 = 0.5 mm
(i) \( \therefore \text{Pitch} = 0.5 \text{ mm} \) OR 0.05 cm
(ii) The thimble of a screw gauge has 50 divisions for one rotation.
\( \implies \text{No. of circular scale divisions} = 50 \)

\( \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} \)

\( = \frac{0.05}{50} \text{ cm} = 0.001 \text{ cm} \)

(iii) Main scale reading = 0.5 mm = 0.05 cm
Circular scale reading = 27 div.
Diameter of wire = M.S. reading + L.C. × C.S. reading
= 0.05 + 0.001 × 27 = 0.05 + 0.027
Diameter of wire = 0.077 cm
In simple words: Since two turns move the tool 1 mm, one turn (the pitch) is 0.5 mm. We use this to find the total width by combining the main ruler and the knob reading.

📝 Teacher's Note: Remind students that "Sleeve" is just another word for the main scale, and "Thimble" is the circular scale.

🎯 Exam Tip: Always pay attention to whether the movement is given for one rotation or multiple rotations. Pitch is ALWAYS for exactly ONE rotation.

Practice Problems 4

Question 1. A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on main scale is 3 divisions and 24th circular scale division coincides with base line. If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate
1. pitch
2. observed diameter.
3. least count
4. corrected diameter.
Answer:
(i) The number of divisions on the main scale are 20 to a centimetre

\( \implies \text{Pitch} = \frac{\text{Unit}}{\text{No. of divisions in unit}} = \frac{1 \text{ cm}}{20} = 0.05 \text{ cm} \)

(ii) No. of circular scale divisions = 50

\( \therefore \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} \)

= 0.05 / 50 cm
L.C. = 0.001 cm
(iii) Main scale reading = 3 division
\( \implies \text{Main scale reading} = 3 \times \text{Pitch} = 3 \times 0.05 = 0.15 \text{ cm} \).
Circular scale reading = 24 division
\( \therefore \text{Observed diameter} = \text{M.S. reading} + \text{L.C.} \times \text{C.S. reading} \)
= 0.15 + 0.001 × 24 = 0.15 + 0.024 = 0.174 cm
(iv) Negative zero error = 8 division
Correction = – (-8 × L.C.)
= – (-8 × 0.001) cm = +0.008 cm
Correct diameter = Observed diameter + Correction
= 0.174 + 0.008 = 0.182 cm
In simple words: The tool was starting "behind zero," so it missed the first 0.008 cm. We find the measurement normally and then add that missing part back.

📝 Teacher's Note: Negative zero error is like a stopwatch that starts at -1 second. You have to add that second back to get the real time.

🎯 Exam Tip: "Negative error" always means you have to ADD the value back. Use the formula: \( \text{Correct} = \text{Observed} - (\text{Zero Error}) \). Since error is negative, it becomes \( \text{Observed} + |\text{Error}| \).

Question 2. A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on main scale is 2 divisions and 79th circular scale division coincides with base line. If the number of divisions on main scale is 10 to a centimetre and circular scale has 100 divisions, calculate
1. pitch
2. observed diameter
3. least count
4. corrected diameter.
Answer:
(i) The number of divisions on the main scale are 10 to a centimetre

\( \implies \text{Pitch} = \frac{\text{Unit}}{\text{No. of divisions in unit}} = \frac{1 \text{ cm}}{10} = 0.1 \text{ cm} \)

(ii) No. of circular scale divisions = 100

\( \therefore \text{Least count (L.C.)} = \frac{\text{Pitch}}{\text{No. of circular scale divisions}} \)

\( = \frac{0.1}{100} \text{ cm} = 0.001 \text{ cm} \)

(iii) Main scale reading = 2 division
\( \implies \text{Main scale reading} = 2 \times \text{Pitch} = 0.2 \text{ cm} \).
Circular scale reading = 79 division
\( \therefore \text{Observed diameter} = \text{M.S. reading} + \text{L.C.} \times \text{C.S. reading} \)
= 0.2 + 0.001 × 79 = 0.2 + 0.079 = 0.279 cm
(iv) Negative zero error = 7 division
\( \therefore \text{Correct} = – (-7 \times \text{L.C.}) = – (-7 \times 0.001) = +0.007 \text{ cm} \)
Corrected diameter = Observed diameter + Correction
= 0.279 + 0.007 = 0.286 cm
In simple words: We find the width (0.279 cm) and then add the 0.007 cm that the tool "missed" because it started behind zero.

📝 Teacher's Note: This is a great problem to test if students understand that "MSR" is not just the division number (2), but the division multiplied by the pitch (0.2 cm).

🎯 Exam Tip: Always show the multiplication of the zero error division by the L.C. to earn partial credit even if you make a final addition error.

Exercise 4

Question 1. For what range of measurement is micrometre screw gauge used?
Answer: Micrometre screw gauge is used to measure upto the accuracy of 0.001 cm.
In simple words: It is used for extremely thin things, like the thickness of a sheet of paper or a single hair.

📝 Teacher's Note: Compare this with the Vernier Calliper (0.01 cm accuracy) to show why we have different tools for different jobs.

🎯 Exam Tip: The key value to remember for a screw gauge is 0.001 cm or 0.01 mm.

Question 2. What do you understand by the following terms as applied to micrometre screw gauge?
1. Sleeve cylinder
2. Sleeve scale
3. Thimble
4. Thimble scale
5. Base line.
Answer:
1. Sleeve cylinder : A hollow cylinder attached to a nut of the screw gauge is known as sleeve cylinder. The spindle of the screw passes through sleeve cylinder
2. Sleeve scale : It is also known as main scale. A reference line or base line graduated in mm, drawn on the sleeve cylinder, parallel to axis of nut is known as sleeve scale.
3. Thimble : A hollow circular cylinder connected to the screw, which rotates along with nut on turning, is called thimble.
4. Thimble scale : It is also known as circular scale. A scale marked on tapered end of a hollow cylinder, which can move over the sleeve cylinder, is known as thimble scale.
5. Base line : A reference line drawn on the sleeve cylinder parallel to the axis of nut is known as base line.
In simple words: The sleeve is the fixed part with the ruler. The thimble is the part you twist. The base line is the reference mark we use to read the numbers.

📝 Teacher's Note: Use a real screw gauge to point out these parts. Seeing the tapered end of the thimble helps explain why it's easy to read the lines on the sleeve.

🎯 Exam Tip: When defining these, associate "Sleeve" with "Main Scale" and "Thimble" with "Circular Scale" to keep them clear.

Question 3. What is the function of ratchet in screw gauge?
Answer: When the flattened end of the screw comes in contact with stud, ratchet becomes free and makes a rattling noise. It indicates that screw should not be further pushed towards the stud.
In simple words: The ratchet is a safety knob. It clicks when the tool is tight enough, preventing you from crushing the object you are trying to measure.

📝 Teacher's Note: Explain that tightening too much can damage the delicate threads of the screw, ruining the instrument's accuracy forever.

🎯 Exam Tip: The keyword is "uniform pressure" or "preventing over-tightening."

Question 4. What do you understand by the terms
(a) pitch of screw
(b) least count of screw?
Answer:
(a) Pitch of screw : The pitch of screw is defined as the distance between two consecutive threads he screw, measured along the axis of the screw.
(b) Least count of the screw : Least distance of the screw is defined as the smallest distance moved by its tip when the screw turn through one division marked on it.
In simple words: Pitch is how much the tool opens or closes in one full turn. Least count is the tiniest amount it can measure (just one mark on the knob).

📝 Teacher's Note: Pitch is the distance light travels through the threads. If you look closely at a screw, it's the gap between the spiral lines.

🎯 Exam Tip: For pitch, use the phrase "linear distance moved in one rotation." For least count, use "minimum length measured."

Question 5. State the formula for calculating
1. pitch of screw
2. least count of screw.
Answer:
(i) Pitch of screw

\( = \frac{\text{Distance moved by thimble on main scale}}{\text{Number of rotations of thimble}} \)

E.g. If 5 mm is the distance moved by the thimble on the main scale in 5 rotations then,

\( \text{Pitch} = \frac{5 \text{ mm}}{5} = 1 \text{ mm} \)

(ii) Least count of screw

\( = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} \)

E.g. If pitch of screw is 1 mm and there are 100 divisions on the circular scalen, then

\( \text{Least count} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} \)

L.C. = 0.01 mm or 0.001 cm
In simple words: Pitch = Total distance / Total turns. Least Count = Pitch / Total marks on the knob.

📝 Teacher's Note: Remind students to always use the same units for Pitch and Circular Scale Divisions to get a valid Least Count.

🎯 Exam Tip: Always state the formula before plugging in numbers from the example given in the question.

Question 6. What do you understand by the following terms as applied to screw gauge?
(a) Zero error
(b) Positive zero error
(c) Negative zero error.
Answer: (a) Zero error : If the zero of the main scale does not coincide with zero of circular scale on bringing the screw end in contact with the stud, the screw gauge is said to have zero error.
(b) Positive zero error : If the zero of the circular scale is below the reference line of the main scale, then screw gauge is said to have positive zero error and the correction is negative.
(c) Negative zero error : If the zero of the circular scale is above the reference line of the main scale, then screw gauge is said to have negative zero error and correction is positive.
In simple words: Zero error means the tool is miscalibrated. If the "0" on the knob is too low, the tool over-reads (positive). If it's too high, it under-reads (negative).

📝 Teacher's Note: Use the diagrams to explain that the "Reference line" is the benchmark. If the circular "0" hasn't reached it yet when closed, it's a positive error.

🎯 Exam Tip: "Below the reference line" = Positive. "Above the reference line" = Negative. This is a very common diagram-based question.

Question 7. How do you account for (a) positive zero error (b) negative zero error, for calculating correct diameter of wires?
Answer:
(a) Positive zero error : If the zero line, marked on circular scale, is below the reference line of the main scale, then there is a positive zero error and the correction is negative. In the figure 5th circular scale division is coinciding with reference line.
\( \therefore \text{Correction} = – \text{Coinciding division of C.S.} \times \text{L.C.} \)
= – 5 × 0.001 cm = -0.005 cm
If the observed diameter is 0.557 cm, then:
Corrected diameter = Observed diameter + Correction
= 0.557 cm – 0.005 cm = 0.552 cm

(b) Negative zero error : If the zero line marked on circular scale, is above the reference line of the main scale, then there is a negative error and the correction is positive.
In the figure, there is 96th division on the circular scale which coincides with reference line.
\( \therefore \text{Correction} – + [n - \text{coinciding division of C.S.} \times \text{L.C.}] \)
where n is the total number of circular scale divisions.
\( \therefore \text{Correction} = + [100 – 96] \times 0.001 \text{ cm} = 0.004 \text{ cm} \)
If observed diameter is 0.557 cm, then :
Corrected diameter = Observed diameter + Correction
= 0. 557 cm + 0.004 cm = 0.561 cm
In simple words: For positive error, we just multiply the line number by LC and subtract it. For negative error, we first subtract the line number from the total divisions (usually 100), then multiply by LC and ADD it.

📝 Teacher's Note: Be careful with the logic for negative error. We don't add the value for "96"; we add the value for the "4" lines it moved past zero.

🎯 Exam Tip: For negative zero error, always use the formula \( (n - \text{division}) \times \text{LC} \). Using the division number directly (like 96) is the most common mistake students make.

Unit 5

Exercise 5

Question 1. (a) What do you understand by the term volume of substance?
(b) State the unit of volume in SI system.
Answer:
(a) Volume : The space occupied a substance (solid, liquid or gas) is called volume.
(b) SI unit of volume is Cubic metre (\( \text{m}^3 \)).
One cubic metre : Is the volume occupied by a cube whose each side is equal to 1 m.
In simple words: Volume is how much room something takes up. A big box that is one metre wide, tall, and deep has a volume of one cubic metre.

📝 Teacher's Note: Remind students that volume applies to gases too, even though we can't always "see" them filling space.

🎯 Exam Tip: Don't forget the power of 3 in the symbol. Writing \( \text{m}^2 \) (area) instead of \( \text{m}^3 \) (volume) will lose you the mark.

Question 2. How is SI system of unit of volume is related to 1 litre ? Explain.
Answer: We know one cubic metre is the SI unit of volume and \( 1 \text{ m}^3 = 1 \text{ m} \times 1 \text{ m} \times 1 \text{ m} \)
= 100 cm × 100 cm × 100 cm
\( 1 \text{ m}^3 = 10^6 \text{ cm}^3 \)
Also, 1 litre = 1000 mL = 1000 \( \text{cm}^3 \) = \( 10^3 \text{ cm}^3 \)
[ \( \because 1 \text{ mL} = 1 \text{ cm}^3 \) ]

\( 1 \text{ litre} = 10^3 \text{ cm}^3 \)

Multiply both sides by \( 10^3 \)
\( 10^3 \text{ litre} = 10^3 \times 10^3 \text{ cm}^3 = 10^6 \text{ cm}^3 \)
1000 litre = \( 1 \text{ m}^3 \)
[ \( \because 1 \text{ m}^3 = 10^6 \text{ cm}^3 \) ]
So, \( 1 \text{ m}^3 = 1000 \text{ litre} \)
In simple words: One cubic metre (a big 1m box) can hold exactly 1000 litres of water.

📝 Teacher's Note: This derivation is important. Remind students that \( 1 \text{ cm}^3 \) is the same as \( 1 \text{ mL} \). This link between length units and volume units is vital.

🎯 Exam Tip: Memorize the final result: \( 1 \text{ m}^3 = 1000 \text{ L} \). This is a standard conversion used throughout physics.

Question 3. In which unit, volume of liquid is measured? How is this unit is related to S.I. unit of volume?
Answer: The volume of liquid is measured in litre of its sub-multiple millilitre (mL).
\( 1 \text{ m}^3 = 1000 \text{ litre} \)
and 1 litre = 1000 mL
\( \implies 1 \text{ m}^3 = 1000 \text{ litre} = 1000 \times 1000 \text{ mL} = 10^6 \text{ mL} \)
In simple words: We usually use litres for drinks. One big cubic metre holds a million tiny millilitres.

📝 Teacher's Note: Explain that "Litre" is not actually an SI unit, but it is "accepted for use with SI" because it is very common.

🎯 Exam Tip: Note that \( 1 \text{ mL} = 1 \text{ cm}^3 \). This allows you to convert from length measurements directly to liquid volume.

Question 4. Explain the method in steps to find the volume of an irregular solid with the help of measuring cylinder.
Answer: Volume of an irregular solid
1. Take a measuring cylinder and fill water up to certain level. Note down the level of water in measuring cylinder. Let it be \( V_1 \).
2. Tie the irregular solid body with a thin and strong thread and lower the body gently so that the solid body is completely immersed in the water. The level of water rises. Solid body displaces water of its own volume. Note down the new level of water. Let it be \( V_2 \).
3. Take the difference of two level of water, i.e., \( (V_2 – V_1) \). This will give the volume of irregular solid body.
In simple words: Drop the object in a tube of water. The water level goes up because the object takes up space. The amount the water rises is exactly the volume of the object.

📝 Teacher's Note: This is called the "Displacement Method." Remind students that the object must be completely under the water and must not touch the bottom or sides of the tube.

🎯 Exam Tip: Always mention that the object must be "completely immersed" to get the full volume.

Question 5. Amongst the units of volume (i) \( \text{cm}^3 \) (ii) \( \text{m}^3 \) (iii) litre (iv) millilitre, which is most suitable for measuring:
(a) Volume of a swimming tank
(b) Volume of a glass filled with milk
(c) Volume of an exercise book
(d) Volume of air in the room.
Answer:
(a) litre
(b) \( \text{cm}^3 \)
(c) millilitre
(d) \( \text{m}^3 \)
In simple words: We use huge units (\( \text{m}^3 \)) for rooms and small units (\( \text{cm}^3 \) or mL) for things we can hold in our hands.

📝 Teacher's Note: Explain that "suitability" means using a unit that doesn't make the number too big (like a million mL for a room) or too small (like 0.00001 \( \text{m}^3 \) for a glass of milk).

🎯 Exam Tip: For liquid containers, Litre and Millilitre are preferred over cubic metres.

Question 6. Find the volume of a book of length 25 cm, breadth 18 cm and height 2 cm in \( \text{m}^3 \).
Answer: Length of book = \( l = 25 \text{ cm} \)
Breadth of book = \( b = 18 \text{ cm} \)
Height of book = \( h = 2 \text{ cm} \)
Volume of biook = \( l \times b \times h \)
\( V = 25 \times 18 \times 2 \)
\( V = 900 \text{ cm}^3 \)

\( V = \frac{900}{10^6} \text{ m}^3 ; V = \frac{9}{10000} \text{ m}^3 \)

V = 0.0009 \( \text{m}^3 \)
In simple words: Multiply the three sides to get the volume in centimetres (900). Then divide by a million to turn it into cubic metres.

📝 Teacher's Note: Students often divide by 100 instead of \( 100^3 \) when converting from \( \text{cm}^3 \) to \( \text{m}^3 \). Emphasize that because it is cubic, you must convert each of the three dimensions.

🎯 Exam Tip: Always show the multiplication (\( L \times B \times H \)) before doing the conversion to \( \text{m}^3 \).

Question 7. The level of water in a measuring cylinder is 12.5 ml. When a stone is lowered in it, the volume is 21.0 ml. Find the volume of the stone.
Answer: Level of water in measuring cylinder = \( V_1 = 12.5 \text{ ml} \)
When stone is lowered, then level of water in measuring cylinder = \( V_2 = 21.0 \text{ ml} \)
Volume of stone = \( V_2 – V_1 \)
V = 21.0 – 12.5
V = 8.5 ml
In simple words: The water went up by 8.5 ml when the stone was put in, so the stone's size is exactly 8.5 ml.

📝 Teacher's Note: This is a direct application of Archimedes' principle. The volume of water displaced is equal to the volume of the submerged object.

🎯 Exam Tip: Ensure you use the correct units (ml) in your final subtraction.

Question 8. A measuring cylinder is filled with water upto a level of 30 ml. A solid body is immersed in it so that the level of water rises to 37 ml. Now solid body is tied with a cork and then immersed in water so that the water level rises to 40 ml. Find the volume of solid body and the cork.
Answer: Level of water in measuring cylinder = \( V_1 = 30 \text{ ml} \)
Level of water in measuring cylinder when a solid body is immersed in it \( V_2 = 37 \text{ ml} \)
Level of water in measuring cylinder when a cork tied with the solid is immersed in water = \( V_3 = 40 \text{ ml} \)
Volume of solid body = \( V_2 – V_1 = 37 – 30 = 7 \text{ ml} \) or \( 7 \text{ cm}^3 \)
Volume or cork = \( V_3 – V_2 = 40 – 37 \)
= 3 ml or 3 \( \text{cm}^3 \) [ \( \because 1 \text{ ml} = 1 \text{ cm}^3 \) ]
In simple words: First we find the solid's volume (7 units). Then, since the cork floats, we tie it to the solid to push it under. The extra 3 units of water rise must be the cork's volume.

📝 Teacher's Note: Explain that "sinkers" (like the solid body) are needed to measure the volume of objects that float (like cork), otherwise they wouldn't displace their full volume of water.

🎯 Exam Tip: Remember to solve for the solid body first (\( V_2 - V_1 \)) and then use that result to find the second object (\( V_3 - V_2 \)).

Unit 6

Practice Problems 1

Question 1. Calculate the time period of simple pendulum of length 0. 84 m when \( g = 9.8 \text{ ms}^{-2} \).
Answer: Length of the pendulum = \( l = 0.84 \text{ m} \)
\( g = 9.8 \text{ ms}^{-2} \)

Time period (T) = \( 2\pi\sqrt{\frac{l}{g}} = 2 \times \frac{22}{7} \times \sqrt{\frac{0.84}{9.8}} \)

\( = \frac{44}{7} \times 0.2928 = 1.84 \text{ s} \)
In simple words: We plug the length of the string and the power of gravity into the pendulum formula to see how long one full swing takes.

📝 Teacher's Note: Remind students that \( \pi \) is approximately 22/7 or 3.14. Using 22/7 is usually easier for hand calculations in physics problems.

🎯 Exam Tip: Always write the base formula \( T = 2\pi\sqrt{l/g} \) first. Even if you make a calculation error, you will get marks for the formula.

Question 2. Calculate the time period of simple pendulum of length 1.44 m on the surface of moon. The acceleration due to gravity on the surface of moon is 1/6 the acceleration due to gravity on earth, [\( g = 9.8 \text{ ms}^{-2} \)]
Answer: Length of simple pendulum = l = 1.44 m
Time period (T) = ?
Acceleration due to gravity on the surface of moon
\( = g' = \frac{1}{6} \times g \)

\( g = \frac{9.8}{6} \)

T = \( 2\pi\sqrt{\frac{l}{g'}} \) : T = \( 2 \times \frac{22}{7} \times \sqrt{\frac{1.44 \times 6}{9.8}} \)

\( = \frac{44}{7} \times 0.9389 = 5.90 \text{ S} \)
In simple words: Gravity is much weaker on the moon, so the pendulum swings back and forth much more slowly than it does on Earth.

📝 Teacher's Note: Explain that since 'g' is in the bottom of the fraction, lower gravity means a BIGGER time period. The pendulum is sluggish on the moon.

🎯 Exam Tip: Note how the '6' from the gravity fraction flipped to the top in the square root calculation. This is a common math step in these problems.

Practice Problems 2

Question 1. Length of second’s pendulum is 100 cm. Find the length of another pendulum whose time period is 2.4 s.
Answer: We know time period of second’s pendulum is 2 s.
\( \therefore T_1 = 2 \text{ s} \); \( l_1 = 100 \text{ cm} \)
\( T_2 = 2.4 \text{ s} \); \( l_2 = ? \)

\( \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} \)

\( \frac{2}{2.4} = \sqrt{\frac{100}{l_2}} \)

\( \frac{1}{1.2} = \sqrt{\frac{100}{l_2}} \)

Squaring both sides,

\( \frac{1}{1.44} = \frac{100}{l_2} \)
\( l_2 = 100 \times 1.44 = 144 \text{ cm} \)
In simple words: A longer string makes for a slower swing. To slow the pendulum down from 2 seconds to 2.4 seconds, we need to lengthen the string to 144 cm.

📝 Teacher's Note: The phrase "seconds pendulum" is a secret way of saying the time period is exactly 2 seconds. Students must know this definition to solve the problem.

🎯 Exam Tip: When dealing with square roots in physics, "Squaring both sides" is the most efficient way to isolate the unknown variable.

Question 2. A pendulum of length 36 cm has time period 1.2 s. Find the time period of another pendulum, whose length is 81 cm.
Answer: \( l_1 = 36 \text{ cm} \); \( T_1 = 1.2 \text{ s} \)
\( l_2 = 81 \text{ cm} \); \( T_2 = ? \)

\( \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} \)

\( \frac{1.2}{T_2} = \sqrt{\frac{36}{81}} \)

\( \frac{1.2}{T_2} = \frac{6}{9} \)

\( T_2 = \frac{9 \times 1.2}{6} = 9 \times 0.2 = 1.8 \text{ s} \)
In simple words: We compare the two pendulums using a ratio. Since the second string is much longer, it takes longer (1.8 seconds) to complete a swing.

📝 Teacher's Note: Notice how the lengths 36 and 81 are perfect squares (6x6 and 9x9). In physics exams, numbers are often chosen like this to make the square root steps easier.

🎯 Exam Tip: Use the simplified ratio \( T \propto \sqrt{l} \) for comparison problems like this one.

Question 3. Calculate the length of second’s pendulum on the surface of moon when acceleration due to gravity on moon is \( 1.63 \text{ ms}^{-2} \).
Answer: Length of second’s pendulum = l = ?
Acceleration due to gravity on surface of moon
= \( g_m = 1.63 \text{ ms}^{-2} \)
Time period = T = 2 s

\( T = 2\pi\sqrt{\frac{l}{g}} \)

\( 2 = 2 \times \frac{22}{7} \times \sqrt{\frac{l}{1.63}} \)

\( \sqrt{\frac{l}{1.63}} = \frac{7}{22} \)

Squaring both sides

\( \frac{l}{1.63} = \frac{49}{484} \)

\( l = \frac{49}{484} \times 1.63 \)
l = 0.165 m
In simple words: On the moon, gravity is very weak. To make a pendulum that swings every 2 seconds, you need a very short string (only about 16.5 cm).

📝 Teacher's Note: On Earth, a seconds pendulum is about 1 metre long. This problem shows how much gravity changes that physical requirement.

🎯 Exam Tip: Ensure your final length is in metres if you used \( g \) in \( \text{ms}^{-2} \). You can convert it to cm at the very end if needed.

Practice Problems 3

Question 1. The length of two pendulum are 110 cm and 27.5 cm. Calculate the ratio of their time periods.
Answer: \( l_1 = 110 \text{ cm} \); \( l_2 = 27.5 \text{ cm} \)
Let \( T_1 \) and \( T_2 \) be the time period of two pendulums.

\( \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} \)

\( \frac{T_1}{T_2} = \sqrt{\frac{110}{27.5}} = \sqrt{\frac{4}{1}} \)

\( \frac{T_1}{T_2} = \frac{2}{1} \)
In simple words: The first pendulum is four times longer than the second. Because of the square root rule, its swing takes exactly twice as much time.

📝 Teacher's Note: Simplifying the fraction \( 110 / 27.5 \) to 4 is the key here. Always try to simplify before taking the square root.

🎯 Exam Tip: Ratios should be written in the form "2:1."

Question 2. A pendulum 100 cm and another pendulum 4 cm long are oscillating at the same time. Calculate the ratio of their time periods.
Answer: \( l_1 = 100 \text{ cm} \); \( l_2 = 4 \text{ cm} \)
Let \( T_1 \) and \( T_2 \) be the time period of two pendulums.

\( \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} \)

\( \frac{T_1}{T_2} = \sqrt{\frac{100}{4}} = \sqrt{\frac{25}{1}} \)

\( \frac{T_1}{T_2} = \frac{5}{1} \)
In simple words: The long string is 25 times longer than the short one. This means it takes 5 times as long to complete a swing.

📝 Teacher's Note: This extreme example helps students visualize that time doesn't increase at the same "speed" as length—length has to grow a lot to increase time just a little.

🎯 Exam Tip: "5:1" is the ratio. Remember to put the first length mentioned in the numerator.

Practice Problems 4

Question 1. The time periods of two pendulums are 1.44 s and 0.36 s respectively. Calculate the ratio of their lengths.
Answer: \( T_1 = 1.44 \text{ s} \); \( T_2 = 0.36 \text{ s} \)
Let \( l_1 \) and \( l_2 \) be the lengths of the two pendulums.

\( \sqrt{\frac{l_1}{l_2}} = \frac{T_1}{T_2} \)

\( \sqrt{\frac{l_1}{l_2}} = \frac{1.44}{0.36} = \frac{144}{36} = \frac{4}{1} \)

Squaring both sides,

\( (\sqrt{\frac{l_1}{l_2}})^2 = (\frac{4}{1})^2 \implies \frac{l_1}{l_2} = \frac{16}{1} \)
In simple words: One pendulum takes 4 times longer to swing. For that to happen, its string must be 16 times longer (4 x 4).

📝 Teacher's Note: This is the reverse of previous problems. Here we start with time and solve for length, which means we square the ratio instead of taking its root.

🎯 Exam Tip: Note that \( 1.44 / 0.36 \) is exactly 4. Simplify your numbers before squaring to keep the math easy.

Question 2. The time period of two pendulums are 2 s and 3 s respectively. Find the ratio of their lengths.
Answer: \( T_1 = 2 \text{ s} \); \( T_2 = 3 \text{ s} \)
Let \( l_1 \) and \( l_2 \) be the length of the two pendulums.

\( \sqrt{\frac{l_1}{l_2}} = \frac{T_1}{T_2} = \frac{2}{3} \)

Squaring both sides,

\( (\sqrt{\frac{l_1}{l_2}})^2 = (\frac{2}{3})^2 = \frac{4}{9} \)

\( \frac{l_1}{l_2} = \frac{4}{9} = 4 : 9 \)
In simple words: If you want one swing to take 2 seconds and another to take 3 seconds, their string lengths must be in a 4 to 9 ratio.

📝 Teacher's Note: Direct relationship is between \( T^2 \) and \( l \). If \( T \) doubles, \( l \) quadruples.

🎯 Exam Tip: Express the final ratio clearly as "4:9" for full marks.

Exercise 6

Question 1. (a) Define simple pendulum.
(b) State two factors which determine time period of a simple pendulum.
(c) Write an expression for the time period of a simple pendulum.
Answer:
(a) Simple Pendulum : A simple pendulum consists of a heavy point mass (called bob) suspended from a rigid support by a massless, inextensible string.
(b) Factors on which time period of a simple pendulum depends :
\( T \propto \sqrt{l} \) or \( T^2 \propto l \)
1. i.e., if length increases, time period increases. That is why in summer pendulum of clock goes slow. \( T \propto \frac{1}{\sqrt{g}} \).
2. That is why when clock is taken to a mountain where ‘g’ decreases with altitude, time period increases and pendulum takes more time to complete an oscillation and hence the clock goes slow.
3. Mass or material of bob : Time period of simple pendulum is independent of mass.
4. Amplitude : Time period of simple pendulum is independent of amplitude. So long as swing is not too large.
(c) Expression for time period of a simple pendulum is given as :

\( T = 2\pi\sqrt{\frac{l}{g}} \)

Where l = length of the pendulum
g = acceleration due to gravity.
In simple words: A simple pendulum is just a heavy weight on a string. Its swing speed depends only on how long the string is and how strong gravity is. It doesn't matter how heavy the weight is!

📝 Teacher's Note: The "massless" string and "point mass" are ideal conditions. In a real lab, we use a very thin string and a small metal ball to get close to this ideal.

🎯 Exam Tip: "Length" and "Gravity" are the two factors. Mentioning that mass and amplitude *do not* affect it is also important for conceptual questions.

Question 2. Define the following in connection with a simple pendulum.
(a) Time period
(b) Oscillation
(c) Amplitude
(d) Effective length.
Answer:
(a) Time period (T) : “is the time taken to complete one oscillation.” Its unit is second (s) and time period is denoted by ‘T”
(b) Oscillation : “One complete to and fro motion of the pendulum” is called an oscillation. i.e., motion of bob from B to C and then C to B is one oscillation.
(c) Amplitude : The maximum displacement of bob from mean position on either side is called amplitude. Amplitude = AB or AC. It is denoted by ‘a’.
(d) Effective length : The length between the point of suspension and centre of gravity of bob of a pendulum is called effective length.
In simple words: Oscillation is one full trip (out and back). Time period is the time for that trip. Amplitude is how far it swings to one side. Length is measured from the top pivot down to the center of the heavy ball.

📝 Teacher's Note: Use the diagram to show "Mean Position" (A) and "Extreme Positions" (B and C). Amplitude is the distance from the center to the edge.

🎯 Exam Tip: For "Effective length," make sure to mention it's to the "centre of gravity" of the bob, not just its top surface.

Question 3. (a) What is a second’s pendulum?
(b) A second’s pendulum is taken on the surface of moon where acceleration due to gravity is l/6th of that of earth. Will the time period of pendulum remain same or increase or decrease? Give a reason.
Answer:
(a) Seconds’ pendulum : “A pendulum which has time period of two seconds” is called seconds’ pendulum.
OR
Seconds’ pendulum may also be defined as “a pendulum which completes one oscillation in two seconds.”
(b) We know time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.
\( T \propto \frac{1}{\sqrt{g}} \)
As acceleration due to gravity on the surface of moon decreases as compared to that of the earth.
\( \because g_{\text{moon}} < g_{\text{earth}} \implies T_{\text{moon}} > T_{\text{earth}} \)
\( \implies \) The time period of second’s pendulum increases when it is taken to the surface of the moon.
In simple words: A seconds pendulum takes 2 seconds for a swing. Because gravity is weaker on the moon, it doesn't "pull" the pendulum back as fast, so the swing takes much longer.

📝 Teacher's Note: It's called a "seconds pendulum" because it takes exactly *one* second to swing from one side to the other (half an oscillation).

🎯 Exam Tip: Use the keyword "inversely proportional" to explain the relationship between gravity and time period.

Question 4. Which of the following do not affect the time period of a simple pendulum?
(a) mass of bob
(b) size of bob
(c) effective length of pendulum
(d) acceleration due to gravity
(e) amplitude.
Answer: (a) Mass of the bob, and (b) Size of the bob, do not affect the time period of a pendulum. Also time period of pendulum is independent of the amplitude provided this is not too great.
In simple words: A heavy weight and a light weight will swing at the same speed. A big ball and a small ball will also swing at the same speed. Only the string length and gravity matter.

📝 Teacher's Note: This is a fundamental concept in physics—all objects fall/swing at the same rate regardless of their mass (Galileo's discovery).

🎯 Exam Tip: Make sure to add the condition "provided the amplitude is not too great," as very large swings *do* slightly affect the timing.

Question 5. A simple pendulum is hollow from within and its time period is T. How is the time period of pendulum affected when :
(a) 1/4 of bob is filled with mercury
(b) 3/4 of bob is filled with mercury
(c) The bob is completely filled with mercury?
Answer: We know that time period of a simple pendulum is independent of its mass. So in all the above said cases, time period of simple pendulumn remains same.
In simple words: Even if you make the pendulum heavier by filling it up, the swing time stays the same because mass doesn't change the time period.

📝 Teacher's Note: This is a trick question. Students often think "heavier = faster" or "heavier = slower," but in ideal physics, mass cancels out of the equations.

🎯 Exam Tip: "Remains same" is the standard answer for any variation in the mass of the bob.

Question 6. Two simple pendulums, A and B have equal lengths but their bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? What is the reason for your answer?
Answer: We know that time period of simple pendulum at a place is given by \( T = 2\pi\sqrt{\frac{l}{g}} \). This expression does not contain weight of bob i.e. is independent of the weight of bob.
\( \therefore \text{Time period of both pendulums will be same.} \)
\( \therefore \text{Ratio of their time periods = 1 : 1} \)
In simple words: Since the strings are the same length, the pendulums will swing at the same speed, no matter how much the weights weigh.

📝 Teacher's Note: This reinforces the concept from Question 5. The ratio will always be 1:1 if length and gravity are constant.

🎯 Exam Tip: Always justify the 1:1 ratio by stating that "Time period is independent of the mass of the bob."

Question 7. State the numerical value of the frequency of oscillation of a second’s pendulum. Does it depend on the amplitude of oscillation?
Answer: \( \text{Frequency} = \frac{1}{T} \). And T for seconds' pendulum is 2 seconds.
\( \therefore \text{Frequency} = \frac{1}{2} = 0.5 \text{ s}^{-1} \).
Oscillation of pendulum does not depend on amplitude.
In simple words: Frequency is how many full swings happen in one second. Since one swing takes 2 seconds, you only get half a swing per second (0.5).

📝 Teacher's Note: Frequency and Time Period are reciprocals. If one is large, the other is small.

🎯 Exam Tip: Frequency unit is Hertz (Hz) or \( \text{s}^{-1} \). Use either in your answer.

Question 8. (a) Name the two factors on which time period of a simple pendulum depends.
(b) Name the devices commonly used to measure
(i) mass and
(ii) weight of a body.
Answer:
(a) Factors on which time period of a simple pendulum depends :
1. i.e., if length increases, time period increases. That is why in summer pendulum of clock goes slow.
2. That is why when clock is taken to a mountain where ‘g’ decreases with altitude, time period increases and pendulum takes more time to complete an oscillation and hence the clock goes slow.
(b)
1. Mass of measured by physical balance.
2. Weight of a body is measured by spring balance.
In simple words: Swing time depends on the string's length and how strong gravity is. We use a beam balance for mass and a spring scale for weight.

📝 Teacher's Note: Use the "summer clock" example to show that metal strings expand in heat, making the clock run slow.

🎯 Exam Tip: Remember the distinction: Physical Balance = Mass; Spring Balance = Weight. This is a very common distinction question.

Question 9. Draw a graph of l, the length of simple pendulum against \( T^2 \), the square of its time period.
Answer: Nature : The graph of length (l) of simple pendulum against square of its time period (\( T^2 \)) is a straight line inclined to time axis.
In simple words: Since length and the square of time are directly related, their graph is a perfectly straight line starting from the zero corner.

📝 Teacher's Note: This straight-line relationship is used in labs to calculate the value of 'g' by finding the slope of the graph.

🎯 Exam Tip: Always label your axes: "Length (l)" on the Y-axis and "\( T^2 \)" on the X-axis.

Question 10. What do you understand by (a) amplitude and (b) frequency of oscillations of simple pendulum?
Answer:
(a) Amplitude : The maximum displacement of bob from mean position on either side is called amplitude. Amplitude = AB or AC. It is denoted by ‘a’.
(b) Frequency: “It is the number of vibrations or oscillations made in one second.” It is denoted by f or n and its unit is Hertz (Hz) or per second (\( \text{s}^{-1} \)).
In simple words: Amplitude is how far the swing goes. Frequency is how many full "out-and-back" swings happen in one second.

📝 Teacher's Note: Demonstrate frequency by shortening a pendulum—it swings faster, meaning more oscillations per second, hence a higher frequency.

🎯 Exam Tip: Frequency unit Hertz is named after Heinrich Hertz. Always capitalize the 'H' in Hz.

Unit 7

Exercise 7

Question 1. (a) What do you understand by the term graph?
(b) What do you understand by the terms (i) independent variable, (ii) dependent variable?
(c) Amongst the independent variable and dependent variable, which is plotted on X-axis?
Answer:
(a) Graph : A pictorial representation of two physical variables, recorded by ah experimenter is called graph.
(b)
1. Independent variable : A variable whose variation does not depend on that of another is known as independent variable.
2. dependent variable : A variable whose variation depends upon another variable is known as dependent variable.
(c) The independent variable is always plotted on x – axis.
In simple words: A graph is a picture of your data. The variable you control (like time) is the "independent" one on the bottom. The one that changes (like a plant's height) is the "dependent" one on the side.

📝 Teacher's Note: Use the example of studying: Hours spent studying (independent) affects the Grade (dependent). You decide the hours, the grade depends on it.

🎯 Exam Tip: In science experiments, Time is almost always the independent variable plotted on the X-axis.

Question 2. (a) State how will you choose a scale for the graph.
(b) State the two ratios of a scale, which are suitable for plotting points.
(c) State the two ratios of a scale, which are not suitable for plotting points.
Answer:
(a) We can choose any convenient scale to represent a given variable on a given axis, such that the whole range of variations are well spread out on the whole graph paper, to give the graph line a suitable size. For this a round number, nearest to or slightly less than minimum value should be taken as origin and a round number nearest to or slightly more than maximum value should be taken at the far end of the respective axis for a given variable.
(b) Two ratios of a scale suitable for ploting points are 1 : 2 and 1 : 4.
(c) Two ratios of a scale not suitable for plotting points are 1 : 3 and 1 : 7. Because such scales are impractical and pose difficulty in plotting intermediate points.
In simple words: Pick a scale that makes your graph big and easy to read. Ratios like 1:2 or 1:5 are great because they match our counting system. Ratios like 1:3 are bad because they make it really hard to put the dots in the right spots between the lines.

📝 Teacher's Note: Explain that since our number system is decimal-based (factors of 10), ratios of 2 and 5 fit the grid lines perfectly. 3 and 7 create "infinite decimals" on the grid.

🎯 Exam Tip: Your graph should always cover more than half of the graph paper area. If it's too small, your scale is wrong.

Question 3. State three important precautions which must be followed while plotting points on a graph.
Answer: Precautions for plotting points on a graph :
1. The points marked on graph paper should be sharp, but not thick.
2. Ordinates of points should be written close to the plotted point.
3. It is not necessary that graph line should pass through all points. A best fit line should be drawn.
In simple words: Use a sharp pencil for tiny dots. Label the points. Draw a smooth line that goes through the middle of the dots rather than zig-zagging to hit every single one.

📝 Teacher's Note: The "Best Fit Line" is crucial. It shows the general trend and ignores small experimental errors (outliers).

🎯 Exam Tip: Always use a sharpened HB or 2H pencil for graph work. A thick line can hide the true result of your experiment.

Question 4. State two important precautions for drawing a graph line.
Answer: Precautions for drawing a graph line :
1. The graph line should be thin, single straight line and sharp.
2. It is not necessary that graph line should pass through all the points. A best fit graph line should be drawn.
In simple words: Your line should be as thin and straight as possible. It doesn't have to hit every dot if the dots are slightly messy; just draw the most accurate path through the center.

📝 Teacher's Note: If the data is meant to be a straight line but the points are scattered, the best-fit line is the "average" of those points.

🎯 Exam Tip: Use a long ruler (30 cm) so you can draw the whole graph line in one single stroke without lifting the pencil.

Question 5. (a) What is a best fit line for a graph?
(b) What does best fit line show regarding the variables plotted and the work of experimenter?
Answer:
(a) A best bit line for a graph means a line which either passes through maximum number of points or passes closest to the maximum number of points, which appear on either side of the line.
(b) A best fit line shows that two variable quantities are directly proportional to each other. With its help, experimenter can easily understand nature of proportional relations between two variable quantities.
In simple words: A best fit line is a "trend line." It ignores tiny mistakes and shows how the two things you measured are linked together (like if one goes up, the other goes up too).

📝 Teacher's Note: It averages out "random errors." If points are on both sides of the line, the experimenter's work is balanced.

🎯 Exam Tip: A best fit line passing through the origin (0,0) proves that the two variables are "directly proportional."

Question 6. (a) What do you understand by the term constant of proportionality?
(b) How can proportionality constant be determined from the best fit straight line graph?
Answer:
(a) Constant of proportionality : If a quantity say X is directly proportional to another quantity Y, then X is written as X = KY, where K is called constant of proportionality.
(b) Constant of proportionality : can be determined from the best fit straight line by calculating the slope of graph by using the formula. Slope of graph

\( = \frac{\text{Difference between the co-ordinates on y-axis for any two points on the graph line}}{\text{Difference between the co-ordinates on x-axis for any two points on the graph line}} \)
In simple words: The constant (K) is the "multiplier" that links two variables. You find it by measuring how steep the graph line is (the slope).

📝 Teacher's Note: Explain the formula as "Rise over Run." How much does Y go up (Rise) for every bit X moves to the right (Run)?

🎯 Exam Tip: When picking points for the slope, choose two points that are far apart on the line. This makes the calculation more accurate.

Question 7. State three uses of graph.
Answer: Uses of a graph :
(a) One can determine constant of proportionality by calculating slope of graph.
(b) It can be used to calculate mean average value of large number of observations.
(c) It can be used for verifying already known physical laws.
(d) It can also show the weakness of the experimenter at some particular instant during the course of experiment.
In simple words: Graphs help us find hidden patterns, average out mistakes, check if scientific rules are true, and see if we made a mess of our data at any point.

📝 Teacher's Note: Highlight part (d). If one dot is very far from the others, it tells the student exactly where they made a mistake during the lab.

🎯 Exam Tip: Identifying "outliers" or "wrong points" is a very practical and important use of graphs in science.

Question 8. How does a graph help in determining the proportional relationship between two quantities?
Answer: It has been found that if a graph is plotted between pressure of an enclosed gas at constant temperature, against its volume, the graph line is a smooth curve, which does not meet X-axis or Y- axis on extending. From the figure, it is clear that pressure of gas is not directly proportional to volume of gas. However, if a graph is plotted between pressure and inverse of volume, the graph line is a straight line. From the straight line graph we can say: Pressure is inversely proportional to volume.
Thus, \( P \propto \frac{1}{V} \). Similarly, if a graph is plotted between length and time period of a simple pendulum, the graph line is a curve. However, if a graph is plotted between length and \( (\text{Time})^2 \), the graph line is a straight line. Thus, we can say : length of a simple pendulum is directly proportional to the square of its time period. From the above discussion it is very clear that graph line helps to determine the nature of proportional relationship between two variable quantities.
In simple words: If a graph is a curve, the link isn't direct. But if we can turn that curve into a straight line by changing the numbers (like using 1/Volume), we have found the secret mathematical rule!

📝 Teacher's Note: This "linearization" of data is how scientists discover laws like Boyle's Law or the Pendulum Law. Finding a straight line is the ultimate goal.

🎯 Exam Tip: If the graph is a straight line through the origin, use the term "Directly Proportional." If it's a curve that drops down, use "Inversely Proportional."