9th Class Physics Notes 2026 – Chapter 1

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Class 9 Physics Notes – Chapter 1: Physical Quantities and Measurements

Punjab Board | 2026

A. Multiple Choice Questions (MCQs)

1.1. The instrument that is most suitable for measuring the thickness of a few sheets of cardboard is a:

(a) Metre rule
(b) Measuring tape
(c) Vernier Callipers
(d) Micrometer screw gauge

Correct Answer: (d) Micrometer screw gauge

Explanation: A micrometer screw gauge has a small least count, making it suitable for measuring very small thicknesses.

1.2. One femtometre is equal to:

(a) \(10^{-9}\) m
(b) \(10^{-15}\) m
(c) \(10^9\) m
(d) \(10^{15}\) m

Correct Answer: (b) \(10^{-15}\) m

Explanation: The SI prefix femto- represents \(10^{-15}\).

1.3. A light year is a unit of:

(a) Light
(b) Time
(c) Distance
(d) Speed

Correct Answer: (c) Distance

Explanation: A light year measures the distance light travels in one year.

1.4. Which one is a non-physical quantity?

(a) Distance
(b) Density
(c) Colour
(d) Temperature

Correct Answer: (c) Colour

Explanation: In this question, colour is treated as a non-physical quantity because it is a qualitative perception rather than a quantity expressed directly using a standard unit.

1.5. When using a measuring cylinder, one precaution to take is to:

(a) Check for the zero error
(b) Look at the meniscus from below the level of the water surface
(c) Take several readings by looking from more than one direction
(d) Position the eye in line with the bottom of the meniscus

Correct Answer: (d) Position the eye in line with the bottom of the meniscus

Explanation: Reading the meniscus at eye level helps prevent parallax error.

1.6. Volume of water consumed by you per day is estimated in:

(a) Millilitre
(b) Litre
(c) Kilogram
(d) Cubic metre

Correct Answer: (b) Litre

Explanation: Daily water consumption is commonly measured in litres.

1.7. A displacement can is used to measure:

(a) Mass of a liquid
(b) Mass of a solid
(c) Volume of a liquid
(d) Volume of a solid

Correct Answer: (d) Volume of a solid

Explanation: A displacement can is used to determine the volume of an irregular solid by measuring the water it displaces.

1.8. Two rods with lengths 12.321 cm and 10.3 cm are placed side by side. The difference in their lengths is:

(a) 2.02 cm
(b) 2.0 cm
(c) 2 cm
(d) 2.021 cm

Correct Answer: (b) 2.0 cm

Explanation: The calculated difference is \(12.321-10.3=2.021\) cm. Rounded to one decimal place, it is 2.0 cm.

1.9. Four students measure the diameter of a cylinder with Vernier Callipers. Which of the following readings is correct?

(a) 3.4 cm
(b) 3.475 cm
(c) 3.47 cm
(d) 3.5 cm

Correct Answer: (b) 3.475 cm

Explanation: The supplied answer identifies 3.475 cm as the correct reading. The actual instrument readings or figure are not included in the provided text.

1.10. Which of the following measurements is likely to represent the thickness of a sheet of this book?

(a) \(6\times10^{-25}\) m
(b) \(1\times10^{-4}\) m
(c) \(1.2\times10^{-15}\) m
(d) \(4\times10^{-2}\) m

Correct Answer: (b) \(1\times10^{-4}\) m

Explanation: \(1\times10^{-4}\) m is equal to 0.1 mm, a realistic thickness for a sheet of paper.

1.11. In a Vernier Callipers, ten smallest divisions of the Vernier scale are equal to nine smallest divisions of the main scale. If the smallest division of the main scale is half a millimetre, the Vernier constant is equal to:

(a) 0.5 mm
(b) 0.1 mm
(c) 0.05 mm
(d) 0.001 mm

Correct Answer: (c) 0.05 mm

Explanation:

One main scale division (MSD) = 0.5 mm

Ten Vernier scale divisions = Nine main scale divisions

\[
1\text{ VSD}=\frac{9}{10}\times0.5=0.45\text{ mm}
\]

\[
\text{Least Count}=1\text{ MSD}-1\text{ VSD}
\]

\[
=0.5-0.45=\boxed{0.05\text{ mm}}
\]

B. Short Answer Questions

1.1. Can a non-physical quantity be measured? If yes, then how?

Answer: According to the supplied solution, a non-physical quantity, such as feelings or beauty, cannot be measured because it cannot be quantified using numbers and standard units. Only measurable physical quantities can be expressed using standard units.

1.2. What is measurement? Name its two parts.

Answer: Measurement is the process of comparing an unknown physical quantity with a standard quantity called a unit. Its two essential parts are:

1. Numerical value or magnitude
2. Proper unit

1.3. Why do we need a standard unit for measurements?

Answer: Standard units are needed to communicate scientific observations and commercial measurements universally and unambiguously, without confusion.

1.4. Write the names of three base quantities and three derived quantities.

Answer:

Base quantities: Length, mass and time.

Derived quantities: Area, volume and speed.

1.5. Which SI unit will you use to express the height of your desk?

Answer: The SI unit of length is the metre (m). The height of a desk may also be expressed in centimetres (cm) for convenience.

1.6. Write the names and symbols of all SI base units.

Answer:

1. Length — metre (m)
2. Mass — kilogram (kg)
3. Time — second (s)
4. Electric current — ampere (A)
5. Thermodynamic temperature — kelvin (K)
6. Amount of substance — mole (mol)
7. Luminous intensity — candela (cd)

1.7. Why is a prefix used? Name three sub-multiple and three multiple prefixes with their symbols.

Answer: Prefixes are used to express very large or very small quantities conveniently, avoiding the need to write numerous zeros.

Sub-multiple prefixes: milli- (m), micro- (µ), nano- (n)

Multiple prefixes: kilo- (k), mega- (M), giga- (G)

1.8. What is meant by: (a) 5 pm (b) 15 ns (c) 6 µm (d) 5 fs?

Answer:

(a) \(5\text{ pm}=5\times10^{-12}\) metres (picometres)

(b) \(15\text{ ns}=15\times10^{-9}\) seconds (nanoseconds)

(c) \(6\text{ µm}=6\times10^{-6}\) metres (micrometres)

(d) \(5\text{ fs}=5\times10^{-15}\) seconds (femtoseconds)

1.9. (a) For what purpose is a Vernier Callipers used? (b) Name its two main parts. (c) How is the least count found? (d) What is meant by zero error?

Answer:

(a) Purpose: Vernier Callipers are used to measure small lengths, internal and external diameters, and depths with precision.

(b) Main parts: Main scale and Vernier scale.

(c) Least count:

\[
\text{Least Count}=1\text{ MSD}-1\text{ VSD}
\]

where MSD is one main scale division and VSD is one Vernier scale division.

(d) Zero error: Zero error occurs when the zero of the Vernier scale does not coincide with the zero of the main scale when the jaws are fully closed.

1.10. State the least count and Vernier scale reading shown in the figure and hence find the length.

Answer:

– Least count = 0.01 cm (or 0.1 mm)
– Main Scale Reading (MSR) = 3.1 cm
– Vernier coincidence = 4th division
– Vernier reading = \(4\times0.01=0.04\) cm

Total length:

\[
\text{Length}=\text{MSR}+\text{Vernier Reading}
\]

\[
=3.1+0.04=\boxed{3.14\text{ cm}}
\]

Note: The referenced figure was not included in the supplied text, so these readings are reproduced from the provided solution.

1.11. Which reading out of A, B and C shows the correct length and why?

Answer: Reading B shows the correct length because the line of sight is perpendicular to the scale mark. This prevents parallax error.

1.12. Why might a standard system of measurement be helpful to a tailor?

Answer: A standard system of measurement ensures precision, consistency and accurate communication of measurements, helping garments fit correctly.

1.13. The minimum main scale reading of a micrometer screw gauge is 1 mm and there are 100 divisions on the circular scale. What is the least count and thickness?

Answer:

Assuming the stated 1 mm is the pitch of the screw:

\[
\text{Least Count}=\frac{\text{Pitch}}{\text{Number of circular scale divisions}}
\]

\[
=\frac{1\text{ mm}}{100}=0.01\text{ mm}
\]

\[
=0.001\text{ cm}
\]

The thickness is calculated using:

\[
\text{Thickness}=\text{MSR}+(\text{CSR}\times\text{Least Count})
\]

\[
=\text{MSR}+(\text{CSR}\times0.01\text{ mm})
\]

Note: A numerical thickness cannot be calculated without the specific main scale reading (MSR) and circular scale reading (CSR).

1.14. You are provided a metre scale and a bundle of pencils. How can the diameter of a pencil be measured using the metre scale with the same precision as that of Vernier Callipers? Describe briefly.

Answer: Place (N) identical pencils tightly side by side in a bundle. Measure their total width using the metre scale and divide the result by (N). This method helps reduce the effect of measurement error when estimating the diameter of one pencil.

1.15. The end of a metre scale is worn out. Where will you place a pencil to find its length?

Answer: Do not use the worn-out zero mark. Start measuring from a clear, intact mark, such as 1.0 cm, and subtract the initial reading from the final reading.

1.16. Why is it better to place the object close to the metre scale?

Answer: Placing the object close to the scale helps minimise parallax error caused by the separation between the object and the scale markings.

1.17. Why is a standard unit needed to measure a quantity correctly?

Answer: A standard unit ensures uniformity, reproducibility and universal acceptance of scientific and commercial measurements.

1.18. Suggest some natural phenomena that could serve as a reasonably accurate time standard.

Answer: Examples include:

– Earth’s rotation on its axis, producing day and night.
– Earth’s revolution around the Sun, producing a year.
– Atomic vibrations, such as those associated with a caesium-133 atomic clock.

1.19. It is difficult to locate the meniscus in a wider vessel. Why?

Answer: In a wider vessel, the curvature of the liquid surface may be flatter, making the meniscus more difficult to identify and read accurately.

1.20. Which instrument can be used to measure: (i) the internal diameter of a test tube, and (ii) the depth of a beaker?

Answer:

(i) Internal diameter: Vernier Callipers, using the internal jaws.

(ii) Depth: Vernier Callipers, using the depth probe.

C. Constructed Response Questions

1.1. In what unit will you express each of the following?

Answer:

(a) Thickness of a five-rupee coin — millimetre (mm)

(b) Length of a book — centimetre (cm) or metre (m)

(c) Length of a football field — metre (m)

(d) Distance between two cities — kilometre (km)

(e) Mass of a five-rupee coin — gram (g)

(f) Mass of a school bag — kilogram (kg)

(g) Duration of a class period — minute (min) or second (s)

(h) Volume of petrol filled in a car’s tank — litre (L) or cubic metre (m³)

(i) Time taken to boil one litre of milk — minute (min)

D. Comprehensive Questions

1.1. What is meant by base and derived quantities? Give the names and symbols of SI base units.

Answer:

Base quantities: Fundamental quantities that are not expressed in terms of other physical quantities, such as length, mass and time.

Derived quantities: Quantities expressed in terms of base quantities, such as speed, area and volume.

SI base units:

– Metre (m)
– Kilogram (kg)
– Second (s)
– Ampere (A)
– Kelvin (K)
– Mole (mol)
– Candela (cd)

1.2. Give three examples of derived units in SI. How are they derived from base units? Describe briefly.

Answer:

1. Area: The SI unit is square metre (m²).

\[
\text{Area}=\text{Length}\times\text{Width}=m\times m=m^2
\]

2. Volume: The SI unit is cubic metre (m³).

\[
\text{Volume}=\text{Length}\times\text{Width}\times\text{Height}=m^3
\]

3. Speed: The SI unit is metre per second (m/s).

\[
\text{Speed}=\frac{\text{Distance}}{\text{Time}}=\frac{m}{s}
\]

1.3. State the similarities and differences between Vernier Callipers and a micrometer screw gauge.

Answer:

Similarities:

Both are precision measuring instruments used to measure small dimensions more accurately than an ordinary metre scale.

Differences:

– Vernier Callipers: Used to measure external and internal diameters, lengths and depths. The least count depends on the instrument and is commonly 0.1 mm or 0.01 cm.
– Micrometer screw gauge: Uses a screw mechanism and is commonly used to measure the thickness of sheets and the diameter of wires. Its least count is often 0.01 mm.

1.4. Identify and explain the reasons for human errors, random errors and systematic errors in experiments.

Answer:

Human (personal) errors: Caused by poor observation, parallax error or incorrect reading of an instrument.

Random errors: Caused by unpredictable variations in experimental conditions, such as fluctuating voltage or air drafts.

Systematic errors: Caused by faulty instrument calibration, zero error or a consistent experimental bias.

1.5. Differentiate between precision and accuracy of a measurement with examples.

Answer:

Precision: The closeness of repeated measurements to one another.

Accuracy: The closeness of a measurement to the true or accepted value.

Example: If the true length of a rod is 10.0 cm and three measurements are 10.5 cm, 10.51 cm and 10.5 cm, the readings are close to one another and therefore precise, but they are not accurate because they differ from the true value.

E. Numerical Problems

1.1. Calculate the number of seconds in (a) a day, (b) a week and (c) a month. State your answers using SI prefixes.

Answer:

(a) Day

\[
1\text{ day}=24\times60\times60=86,400\text{ s}
\]

\[
=86.4\times10^3\text{ s}=\boxed{86.4\text{ ks}}
\]

(b) Week

\[
7\text{ days}=7\times86,400=604,800\text{ s}
\]

\[
=604.8\times10^3\text{ s}=\boxed{604.8\text{ ks}}
\]

(c) Month (30 days)

\[
30\times86,400=2,592,000\text{ s}
\]

\[
=2.592\times10^6\text{ s}=\boxed{2.592\text{ Ms}}
\]

1.2. State the answers of Problem 1.1 in scientific notation.

Answer:

(a) Day:

\[
\boxed{8.64\times10^4\text{ s}}
\]

(b) Week:

\[
\boxed{6.048\times10^5\text{ s}}
\]

(c) Month:

\[
\boxed{2.592\times10^6\text{ s}}
\]

1.3. Solve the following addition or subtraction. State your answers in scientific notation.

(a)

\[
4\times10^{-4}\text{ kg}+3\times10^{-5}\text{ kg}
\]

\[
=4\times10^{-4}+0.3\times10^{-4}
\]

\[
=\boxed{4.3\times10^{-4}\text{ kg}}
\]

(b)

\[
5.4\times10^{-6}\text{ m}-3.2\times10^{-5}\text{ m}
\]

\[
=0.54\times10^{-5}-3.2\times10^{-5}
\]

\[
=\boxed{-2.66\times10^{-5}\text{ m}}
\]

1.4. Solve the following multiplication or division. State your answers in scientific notation.

(a)

\[
(5\times10^4\text{ m})(3\times10^{-2}\text{ m})
\]

\[
=(5\times3)\times10^{4-2}\text{ m}^2
\]

\[
=15\times10^2\text{ m}^2
\]

\[
=\boxed{1.5\times10^3\text{ m}^2}
\]

(b)

\[
\frac{6\times10^8\text{ kg}}{3\times10^4\text{ m}^3}
\]

\[
=\frac{6}{3}\times10^{8-4}\text{ kg m}^{-3}
\]

\[
=\boxed{2.0\times10^4\text{ kg m}^{-3}}
\]

1.5. Calculate the following and state your answer in scientific notation.

\[
\frac{(3\times10^2\text{ kg})(4.0\text{ km})}{5\times10^2\text{ s}^2}
\]

Convert kilometres into metres:

\[
4.0\text{ km}=4,000\text{ m}=4\times10^3\text{ m}
\]

Numerator:

\[
(3\times10^2)(4\times10^3)=12\times10^5\text{ kg m}
\]

Divide by the denominator:

\[
\frac{12\times10^5}{5\times10^2}
=\frac{12}{5}\times10^{5-2}
\]

\[
=\boxed{2.4\times10^3\text{ kg m s}^{-2}}
\]

The unit \(\text{kg m s}^{-2}\) is equivalent to the newton (N).

1.6. State the number of significant digits in each measurement.

Answer:

(a) \(0.0045\text{ m}\) — 2 significant digits. Leading zeros are not significant.

(b) \(2.047\text{ m}\) — 4 significant digits. Zeros between non-zero digits are significant.

(c) \(3.40\text{ m}\) — 3 significant digits. The trailing zero after the decimal point is significant.

(d) \(3.420\times10^4\text{ m}\) — 4 significant digits.

1.7. Write in scientific notation.

(a)

\[
0.0035\text{ m}=\boxed{3.5\times10^{-3}\text{ m}}
\]

(b)

\[
206.4\times10^2\text{ m}=\boxed{2.064\times10^4\text{ m}}
\]

1.8. Write using correct prefixes.

(a)

\[
5.0\times10^4\text{ cm}=5.0\times10^2\text{ m}
\]

\[
=500\text{ m}=\boxed{0.5\text{ km}}
\]

(b)

\[
580\times10^2\text{ g}=58,000\text{ g}
\]

\[
=\boxed{58\text{ kg}}
\]

(c)

\[
45\times10^{-4}\text{ s}=4.5\times10^{-3}\text{ s}
\]

\[
=\boxed{4.5\text{ ms}}
\]

1.9. A light year is a unit of distance used in astronomy. Taking the speed of light as \(3.0\times10^8\text{ m s}^{-1}\), calculate the distance travelled by light in one year.

Answer:

Distance = Speed × Time

Time for one year:

\[
365\times24\times3600=3.1536\times10^7\text{ s}
\]

Distance:

\[
=(3.0\times10^8)(3.1536\times10^7)\text{ m}
\]

\[
=9.4608\times10^{15}\text{ m}
\]

\[
\boxed{9.46\times10^{15}\text{ m}}
\]

1.10. Express the density of mercury given as \(13.6\text{ g cm}^{-3}\) in \(\text{kg m}^{-3}\).

Answer:

Given:

\[
\rho=13.6\text{ g cm}^{-3}
\]

Convert grams into kilograms and cubic centimetres into cubic metres:

\[
13.6\text{ g}=13.6\times10^{-3}\text{ kg}
\]

\[
1\text{ cm}^3=(10^{-2}\text{ m})^3=10^{-6}\text{ m}^3
\]

Therefore,

\[
\rho=\frac{13.6\times10^{-3}\text{ kg}}{10^{-6}\text{ m}^3}
\]

\[
=13.6\times10^3\text{ kg m}^{-3}
\]

\[
\boxed{1.36\times10^4\text{ kg m}^{-3}}
\]

Notes by Sir Dilawar

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